To solve this problem we will apply the concepts related to the conservation of momentum. That is, the final momentum must be the same final momentum. And in each state, the momentum will be the sum of the product between the mass and the velocity of each object, then
[tex]\text{Initial Momentum} = \text{Final Momentum}[/tex]
[tex]m_1u_1 +m_2u_2 = m_1v_1+m_2v_2[/tex]
Here,
[tex]m_{1,2}[/tex]= Mass of each object
[tex]u_{1,2}[/tex]= Initial velocity of each object
[tex]v_{1,2}[/tex]= Final velocity of each object
When they position the final velocities of the bodies it is the same and the car is stationary then,
[tex]m_2u_2 = (m_1+m_2)v_f[/tex]
Rearranging to find the final velocity
[tex]v_f = \frac{m_2u_2}{ (m_1+m_2)}[/tex]
[tex]v_f = \frac{ 12500*7.8}{ 12500+7430}[/tex]
[tex]v_f = 4.8921ft/s[/tex]
The expression for the impulse received by the first car is
[tex]I = m_1 (v-u)[/tex]
[tex]I = \frac{W}{g} (v-u)[/tex]
Replacing,
[tex]I = \frac{12500}{32.2}(4.89-7.8)[/tex]
[tex]I = -1129.65lb\cdot s[/tex]
The negative sign show the opposite direction.
A small metal sphere has a mass of 0.19 g and a charge of -21.0 nC. It is 10.0 cm directly above an identical sphere that has the same charge. This lower sphere is fixed and cannot move. If the upper sphere is released, it will begin to fall.
What is the magnitude of its initial acceleration?
The magnitude of the initial acceleration is 11.52 m/s^2.
Acceleration is a fundamental concept that describes how an object's velocity changes over time. Velocity includes both speed and direction, so acceleration can involve changes in speed, direction, or both. When an object speeds up, slows down, or changes direction, it experiences acceleration.
To find the magnitude of the initial acceleration, we need to use Newton's Law of Gravitation and Coulomb's Law. The weight and electric force acting on the sphere will determine its acceleration. The weight of the sphere is given by the formula W = mg, where m is the mass and g is the acceleration due to gravity. The electric force is given by the formula[tex]F = k(q1q2/r^2)[/tex], where k is the Coulomb's constant, q1 and q2 are the charges, and r is the distance between the spheres.
First, let's calculate the weight of the sphere:
[tex]W = (0.19 g)(9.8 m/s^2) = 1.862 N[/tex]
Next, let's calculate the electric force:
[tex]F = (8.99 x 10^9 Nm^2/C^2)(21.0 x 10^-9 C)^2 / (0.10 m)^2 = 0.4083 N[/tex]
Now, we can calculate the acceleration using Newton's second law:
F = ma, so a = F/m
a = (1.862 N + 0.4083 N) / 0.19 g
a = 11.52 m/s^2
Therefore, the magnitude of the initial acceleration is [tex]11.52 m/s^2.[/tex]
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A bottle with a volume of 193 U. S. fluid gallons is filled at the rate of 1.9 g/min. (Water has a density of 1000 kg/m3, and 1 U.S. fluid gallon = 231 in.3.) How long does the filling take?
Answer:
0.732 years
Explanation:
We are given that
Volume of fluid filled in bottle=193 U.S
1 U.S fluid gallon=231 cubic inch
193 U.S gallon=[tex]231\times 193=44583in^3[/tex]
[tex]1 in^3=1.63871\times 10^{-5}m^3[/tex]
[tex]44583in^3=44583\times 1.63871\times 10^{-5}=0.731m^3[/tex]
Density of water=[tex]1000kg/m^3[/tex]
Mass=[tex]Volume\times density[/tex]
Using the formula
Mass=[tex]0.731\times 1000=731kg[/tex]
1kg =1000g
[tex]731kg=731\times 1000=731\times 10^3g[/tex]
By using [tex]1000=10^3[/tex]
1.9 g of fluid takes time to fill in the bottle=1 min
1 g of fluid takes time to fill in the bottle=[tex]\frac{1}{1.9}min[/tex]
[tex]731\times 10^3[/tex]g of fluid takes time to fill in the bottle=[tex]\frac{1}{1.9}\times 731\times 10^3[/tex]min
Time=[tex]384.7\times 10^3[/tex]min
[tex]1min=\frac{1}{60\times 24\times 365}[/tex]years
Time=[tex]384.7\times 10^3\times \frac{1}{60\times 24\times 365}years[/tex]
Time =0.732 years
Hence, it takes to filling the 193 U.S fluid gallons in bottle=0.732 years
A basketball has a mass of 567 g. Heading straight downward, in the direction, it hits the floor with a speed of 2 m/s and rebounds straight up with nearly the same speed. What was the momentum change ?
Final answer:
The change in momentum of the basketball after impacting the floor and rebounding at the same speed is 2.268 kg·m/s upward.
Explanation:
You asked what the momentum change was when a basketball with a mass of 567 g hits the floor at a speed of 2 m/s and rebounds at nearly the same speed. To find the change in momentum, we need to consider the initial and final momentums of the basketball. The momentum of an object is given by the product of its mass and velocity. Given the initial velocity (vi) is 2 m/s downward and the final velocity (vf) is 2 m/s upward, and the direction is important in momentum calculations.
Momentum before impact = mass × velocity = 0.567 kg × (-2 m/s) = -1.134 kg·m/s (negative sign indicates downward direction)
Momentum after impact = mass × velocity = 0.567 kg × (2 m/s) = 1.134 kg·m/s (positive sign indicates upward direction)
To find the momentum change, Δp, we subtract the initial momentum from the final momentum:
Δp = pf - pi = 1.134 kg·m/s - (-1.134 kg·m/s) = 1.134 kg·m/s + 1.134 kg·m/s = 2.268 kg·m/s.
Therefore, the momentum change is 2.268 kg·m/s in the upward direction.
Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject material as high as 500 km (or even higher) above the surface. Io has a mass of 8.93×1022 kg and a radius of 1821 km . For this calculation, ignore any variation in gravity over the 500-km range of the debris.A. How high would this material go on earth if it were ejected with the same speed as on Io?
The height that an object reaches in a gravitational field is dependent on its initial speed and the strength of the field. With the same initial speed, an object in a weaker gravitational field will reach a greater height. Therefore, material ejected with the same speed on Io and Earth will reach a greater height on Io due to its weaker gravitational field.
Explanation:To understand how high material would go on Earth if ejected at the same speed as on Io, we first need to consider that the height that an object reaches in a gravitational field (ignoring air resistance) depends on its initial speed and the strength of the gravitational field. Specifically, the maximum height is given by the equation H = (v^2)/(2g), where v is the initial speed, and g is the acceleration due to gravity.
When an object is thrown upwards, it slows down under the force of the Earth's gravity until its speed decreases to zero. At that point, it begins to fall back down. Therefore, if two objects are thrown upwards with the same speed but under the influence of different gravitational fields (one stronger and one weaker), the object in the weaker gravitational field will reach a greater height.
In the case of Io, Jupiter's moon, the strength of the gravitational field is much less than that on Earth because its mass is much smaller. That's why volcanic material can be ejected to such high altitudes. Now, if we were to eject material with the same speed on Earth, which has a stronger gravitational field compared to Io, it would not reach the same height.
To calculate the specific height that material would reach on Earth assuming the same ejection speed as on Io, we would need to know that ejection speed. Unfortunately, the question does not provide this information.
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material ejected from Io with the same speed on Earth would reach a height of approximately [tex]\( 3.41 \times 10^8 \, \text{m} \) or \( 341 \, \text{km} \)[/tex] above the surface of Earth.
To determine how high material ejected from Io would go on Earth if it were ejected with the same speed, we can use the concept of escape velocity and the energy conservation principle.
The escape velocity [tex](\( v_e \))[/tex] of a celestial body is the minimum speed an object must have to break free from its gravitational pull and escape into space. It is given by the formula:
[tex]\[ v_e = \sqrt{\frac{2GM}{r}} \][/tex]
Where:
- [tex]\( v_e \)[/tex] is the escape velocity,
- ( G ) is the universal gravitational constant[tex](\( 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)),[/tex]
- ( M \ is the mass of the celestial body,
- ( r ) is the distance from the center of the celestial body.
Let's calculate the escape velocity of Io first:
[tex]\[ v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \times 8.93 \times 10^{22} \, \text{kg}}{1821 \times 10^3 \, \text{m}}} \]\[ v_e = \sqrt{\frac{2 \times 6.67 \times 8.93}{1821}} \times 10^{11} \, \text{m/s} \]\[ v_e = \sqrt{2 \times 3.34} \times 10^{11} \, \text{m/s} \]\[ v_e \approx 8.19 \times 10^4 \, \text{m/s} \][/tex]
Now, let's assume that material ejected from Io is given this speed of [tex]\( 8.19 \times 10^4 \, \text{m/s} \)[/tex] on Earth. We can calculate how high it would go using energy conservation principles.
The kinetic energy ( KE ) of the material is given by:
[tex]\[ KE = \frac{1}{2} m v^2 \][/tex]
Where:
- \( m \) is the mass of the material, and
- \( v \) is the velocity.
The potential energy (\( PE \)) gained by the material at height \( h \) above the surface of Earth is given by:
[tex]\[ PE = mgh \][/tex]
Where:
- ( g ) is the acceleration due to gravity on Earth[tex](\( 9.81 \, \text{m/s}^2 \))[/tex],
- ( h ) is the height above the surface of Earth.
At the maximum height, the kinetic energy is converted entirely into potential energy. So we have:
[tex]\[ \frac{1}{2} m v^2 = mgh \]\[ \frac{1}{2} v^2 = gh \]\[ h = \frac{v^2}{2g} \][/tex]
Substitute the values:
[tex]\[ h = \frac{(8.19 \times 10^4 \, \text{m/s})^2}{2 \times 9.81 \, \text{m/s}^2} \]\[ h = \frac{6696.61 \times 10^8 \, \text{m}^2/\text{s}^2}{19.62 \, \text{m/s}^2} \]\[ h \approx 3.41 \times 10^8 \, \text{m} \][/tex]
So, material ejected from Io with the same speed on Earth would reach a height of approximately [tex]\( 3.41 \times 10^8 \, \text{m} \) or \( 341 \, \text{km} \)[/tex] above the surface of Earth.
The electric force is exerted by a 40-nC charged particle located at the origin of a Cartesian coordinate system on a 15-nC charged particle located at (2.0 m, 2.0 m ).
Determine the direction of the force.
a. ?y direction
b. +x direction
c. ?x direction
d. 45? clockwise from the +x axis.
e. 45? counterclockwise from the +x axis.
f. +y direction
Answer:
Explanation:
Check the attachment for solution
The direction of the electric force is exerted by a 40-nC charged particle located at the origin of a cartesian coordinate system on a 15-nC charged particle located at (2.0 m, 2.0 m ) is 45° counterclockwise from the +x axis.
What is electric force?Electric force is the attracting or repulsive interaction between any two charged things. Similar to any force, Newton's laws of motion describe how it affects the target body and how it does so. One of the many forces that affect objects is the electric force.
The given two charges are: 40 nC and 15 nC. As both of them are positive, the electric force is repulsive in nature.
15-nC charged particle located at (2.0 m, 2.0 m ).
So, the direction of repulsive force = tan⁻¹(2.0/2.0) = 45⁰.
Hence, The direction of the electric force is 45° counterclockwise from the +x axis. Option (e) is correct.
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How much time elapses from when the first bit starts to be created until the conversion back to analog begins
The question is not complete. Kindly find the complete question below:
Host A converts analog to digital at a = 58 Kbps Link transmission rate R = 1.9 Mbps Host A groups data into packets of length L = 112 bytes Distance to travel d = 931.9 km Propagation speed s = 2.5 x 108 m/s Host A sends each packet to Host B as soon as it gathers a whole packet. Host B converts back from digital to analog as soon as it receives a whole packet. How much time elapses from when the first bit starts to be created until the conversion back to analog begins? Give answer in milliseconds (ms) to two decimal places, normal rounding, without units (e.g. 1.5623 ms would be entered as "1.56" without the quotes)
Answer / Explanation
The answer is 19.65
Two plastic spheres each with radius 1.30 cm and mass 9.75 g, have uniform charge densities of 0.100 C/m^3 and -0.100 C/m^3, respectively. The spheres are released from rest when their centers are 80.0 cm apart. What is the speed of each sphere at the moment they collide? Ignore polarization of the spheres.
To develop this problem we will proceed to use the principle of energy conservation. For this purpose we will have that the change in the electric potential energy and kinetic energy at the beginning must be equal at the end. Our values are given as shown below:
[tex]m = 9.75g = 0.00975kg[/tex]
[tex]r = 1.3cm = 0.013m[/tex]
[tex]q_1 = 0.1 C/m^3 * \frac{4}{3} \pi r^3[/tex]
[tex]q_1 = 9.2*10^{-7}C[/tex]
[tex]q_2 = -9.2*10^{-7}C[/tex]
Applying energy conservation equations
[tex]U_1+K_1 = U_2+K_2[/tex]
[tex]\frac{k q_1q_2}{d} +0 = \frac{kq_1q_2}{2r}+ \frac{1}{2} (2m)v^2[/tex]
Replacing,
[tex]9*10^{9} (9.2*10^{-7})^2(\frac{1}{0.026}-\frac{1}{0.8}) = v^2[/tex]
Solving for v,
[tex]v = 9.2*10^{-7} (\frac{9*10^9}{0.00975}(\frac{1}{0.026}-\frac{1}{0.8}))^{1/2}[/tex]
[tex]v = 5.4 m/s[/tex]
Therefore the speed of each sphere at the moment they collide is 5.4m/s
A heat engine uses two containers held at different temperatures. One container is at 294 K 294 K , while the other is kept at 552 K 552 K . What is the maximum possible efficiency for this engine?
Answer:
Explanation:
Given
Lower Temperature [tex]T_L=294 \K[/tex]
Higher Temperature [tex]T_H=522 \K[/tex]
Maximum Possible efficiency is achieved when the engine works as carnot Engine
i.e. [tex]\eta _{max}=1-\frac{T_L}{T_H}[/tex]
[tex]\eta_{max}=1-\frac{294}{522}[/tex]
[tex]\eta _{max}=\frac{228}{522}=0.436[/tex]
[tex]\eta _{max}=43.64\ %[/tex]
The maximum possible efficiency of a heat engine can be determined using the temperatures of the hot and cold reservoirs in the formula Effc = 1 - Tc / Th. Applying this formula to the given temperatures of 294 K and 552 K results in a maximum efficiency of 46.8%.
Explanation:The maximum possible efficiency of a heat engine (also known as the Carnot efficiency) can be calculated using the temperatures of the heat source (hot reservoir) and the heat sink (cold reservoir). This efficiency can be determined by using the formula Effc = 1 - Tc / Th, where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir. Both these temperatures should be in Kelvin.
In the given problem, Tc is 294 K and Th is 552 K. Substituting these values into the formula gives the maximum possible efficiency:
Effc = 1 - 294 / 552 = 0.468. Thus, the maximum possible efficiency of this engine is approximately 46.8%.
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The pressure reading from a barometer is 742 mm Hg. Express this reading in kilopascals, kPa. (Use 760 mm Hg = 1.013 x 105 Pa)
Answer:
98.9 kPa
Explanation:
given,
Pressure reading of barometer = 742 mm Hg
we know,
760 mm Hg = 1.013 x 10⁵ Pa
[tex]1\ mm\ Hg = \dfrac{1.013\times 10^5}{760}[/tex]
[tex]742\ mm\ Hg = \dfrac{1.013\times 10^5}{760}\times 742[/tex]
= 0.989 x 10⁵ Pa
= 98.9 x 10³ Pa
= 98.9 kPa
the reading of the barometer is equal to 98.9 kPa
The pressure reading from the barometer expressed in kilopascal is 98.9kPa.
Given that;
Pressure reading from the barometer; [tex]P = 742mmHg[/tex]
Pressure reading from the barometer in kilopascals; [tex]x = \ ?[/tex]
First we convert the units from Millimeter of Mercury (mmHg) to Pascal (Pa)
We are to use;
[tex]760 mm Hg = 1.013 * 10^5 Pa\\\\\frac{760mmHg}{760} = \frac{1.013 * 10^5 Pa}{760} \\\\1mmHg = 1.33289 * 10^2 Pa[/tex]
So,
Pressure reading is pascal
[tex]P = 742 * [1.33289*10^2Pa]\\\\P = 98900.438Pa[/tex]
Next we convert to kilopascal
We know that; [tex]1\ Pascal = 0.001\ kilopascal[/tex]
so
[tex]P = 98900.438 * 0.001 kPa\\\\P = 98.9kPa[/tex]
Therefore, the pressure reading from the barometer expressed in kilopascal is 98.9kPa.
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An object is thrown upwards with a speed of 14 m/s. How long does it take to reach a height of 5.0 m above the projection point while descending?
Answer:
2.43 s
Explanation:
Using newton's equation of motion.
T = (v-u)/g
Where T = time taken for the object to return to the point of projection , u = initial velocity, v = final velocity, g = acceleration due to gravity.
Given: v =-14 m/s, u = 14 m/s, g = -9.8 m/s²
T = (-14-14)/-9.81
T = 2.85 s
Note: We look for the object's speed at 5.0 m.
using
v² = u²+2gs.................................... Equation 1
Where v = final velocity, u = initial velocity, g = acceleration due to gravity, s = distance.
Given: u = 14 m/s, g = -9.81 m/s², s = 5.0 m
Substitute into equation 1
v² = 14²+(-9.81×5×2)
v² = 196-98.1
v = √97.9
v = 9.89
We look for the time taken for the velocity to decrease from 14 m/s to 9.89 m/s.
using
v = u+gt
t =(v-u)/g........................... Equation 2
Where t = time taken for the object to decrease it velocity from 14 m/s to 9.89 m/s
Given: v = 9.89 m/s, u =14 m/s g = -9.81 m/s²
t = (14-9.89)/-9.81
t = -4.11/-9.81
t = 0.42 s
Thus,
Time taken to reach 5.0 m above projection point = T-t
=2.85-0.42
2.43 s
It takes approximately 1.03 seconds for the object to reach a height of 5.0 m above the projection point while descending.
The time it takes for an object to reach a certain height when thrown upwards can be determined using equations of projectile motion.
In this case, with an initial speed of 14 m/s and a height of 5.0 m, we can use the equation :
[tex]$$y = y_0 + v_{oy}t - \frac{1}{2}gt^2$$[/tex]
Substituting the known values, we get:
[tex]$$5.0 = 0 + 14t - \frac{1}{2}(9.8)t^2$$[/tex]
By rearranging the equation and solving it,
we find that it takes approximately 1.03 seconds for the object to reach a height of 5.0 m above the projection point while descending.
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A spelunker is surveying a cave. She follows a passage 120 mm straight west, then 250 mm in a direction 45ââ east of south, and then 280 mm at 30ââ east of north. After a fourth unmeasured displacement, she finds herself back where she started.
(A) Use a scale drawing to determine the magnitude of the fourth displacement. Express your answer using two significant figures.
(B) Determine the direction of the fourth displacement.Express your answer using two significant figures.
Answer:
R = 207.45 mm , θ_return = 18.47 south west
Explanation:
This vector addition exercise is schematized in the attachment where the displacements are
1 d1 = 120 mm west
2 d2 = 250mm at 45 south east
3 d3 = 280 mm at 30 east of nort.
R is the final displacement that takes the goat to its initial point (origin)
The analytical way to perform this exercise is to find the components of each displacement and add them
Decompose the displacement using trigonometry
Displacement d1
d1ₓ = 120 cos 180 = -120 mm
Displacement d2, with the angle measured from the axis this θ = 270 + 45
sin 45 = [tex]d2_{y}[/tex] / d2
cos 45 = d2ₓ / d2
[tex]d2_{y}[/tex] = d2 sin45
[tex]d2_{y}[/tex] = 250 sin (270 + 45)
[tex]d2_{y}[/tex] = -176.77 mm
d2ₓ = d2 cos (270 + 45)
d2ₓ = 176.77 mm
displacement d3, for half the angle from the east axis θ = 90-30 = 60
sin 60 = [tex]d3_{y}[/tex] / d3
cos 60 = d3ₓ / d3
[tex]d3_{y}[/tex] = d3 sin 60
d3ₓ = d3 cos 60
[tex]d3_{y}[/tex] = 280 sin 60 = 242.49 mm
d3ₓ = 280 cos 60 = 140 mm
Having all the displacement components we can find the total displacement
Rₓ = d1ₓ + d2ₓ + d3ₓ
Ry = [tex]d1_{y}[/tex] + [tex]d2_{y}[/tex] + [tex]d3_{y}[/tex]
Rₓ = -120 + 176.77 +140
Rₓ = 196.77 mm
Ry = 0 -176.77 +242.49
Ry = 65.72 mm
Therefore the displacement you must make to return to the starting point is
R = RA Rx2 + Ry2)
R = RA (196.77 2 + 65.72 2)
R = 207.45 mm
We used trigonometry
tan tea = RY / Rx
tea = tan-1 Ry / Rx
ea = tan-1 (65.72 / 196.77)
tea = 18.47
This is the point where the girl is, to return to its origin this path must be serial, but in the opposite direction,
θ_return = 18.47 south west
The problem involves adding vectors to find the total displacement, then finding the negative of this displacement to calculate the magnitude and direction of the fourth displacement.
Explanation:This problem involves the mathematical concept of vectors, particularly in determining the resultant vector, which in this context represents the spelunker's path. First, we convert the movements into rectangular coordinates: going 120 mm west is -120i, going 250 mm 45° east of south is -250 cos(45)i - 250 sin(45)j, and going 280 mm 30° east of north is 280 cos(30)i + 280 sin(30)j. Adding these vectors together, we get the spelunker's total displacement vector. The negatives of this sum will represent the fourth displacement needed to get the spelunker back to where she started.
(A) The magnitude of the fourth displacement is the sum of these vectors taken as negatives, because she has to go back, which is equivalent to the length of the path taken. This can be calculated using Pythagoras' theorem for two dimensions.
(B) The direction of the fourth displacement is calculated by finding the angle made by the resulting vector with respect to one of the axes (for instance, the x-axis). For this, we take the inverse tangent of the y-coefficient over the x-coefficient of the vector.
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Light illuminating a pair of slits contains two wavelengths, 600 nm and an unknown wavelength. The 10th bright fringe of the unknown wavelength overlaps the 9th bright fringe of the 600 nm light.
1. What is the unknown wavelength?
Answer:
unknown wavelength is 540 nm
Explanation:
given data
wavelengths = 600 nm
10th bright fringe overlap = 9th bright fringe of 600 nm
to find out
unknown wavelength
solution
we get here first nth bright fringe from the central maximum at distance y
y = n ×[tex]\frac{\lambda D }{d}[/tex] ...............1
and we have given 10th bright fringe overlap = 9th bright fringe of 600 nm
so here
10 × [tex]\frac{\lambda D }{d}[/tex] = 9 × 600 × [tex]\frac{D}{d}[/tex]
solve it we get [tex]\lambda[/tex]
[tex]\lambda[/tex] = 9 × [tex]\frac{600}{10}[/tex]
[tex]\lambda[/tex] = 540 nm
so unknown wavelength is 540 nm
The unknown wavelength is equal to 540 nanometers.
Let the unknown wavelength be W.Given the following data:
Wavelength = 600 nmTo determine the unknown wavelength:
At the maximum distance, the nth bright fringe is given by the formula:
[tex]Y = \frac{n\lambda D}{d}[/tex]
From the question, we were told that the 10th bright fringe of the unknown wavelength overlaps the 9th bright fringe of the 600 nm light.
Mathematically, this can be expressed as follows:
[tex]Y_{10} = Y_9\\\\\frac{10WD}{d} =\frac{9\times 600D}{d} \\\\10W=5400\\\\W=\frac{5400}{10}[/tex]
W = 540 nanometers.
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You were driving a car with velocity <19, 0, 23> m/s. You quickly turned and braked, and your velocity became <14, 0, 26> m/s. The mass of the car was 1300 kg. (a) What was the (vector) change in momentum during this maneuver? Pay attention to signs. < -6500 , 0 , 3900 > kg·m/s (b) What was the (vector) impulse applied to the car?
Answer:
a.<-6500,0,3900>kgm/s
b.<-6500,0,3900>kgm/s
Explanation:
We are given that
Initial velocity of car,u=[tex]<19,0,23>[/tex]m/s
Final velocity of car=[tex]v=<14,0,26>m/s[/tex]
Mass of the car=m=1300 kg
a.We have to find the change in momentum during this manuver.
Change in momentum=[tex]\Delta P=m(v-u)[/tex]
Using the formula
[tex]\Delta P=1300(<14,0,26>-<19,0,23>)=1300(<-5,0,3>)=<-6500,0,3900>kgm/s[/tex]
Hence, the change in momentum during this maneuver=<-6500,0,3900>kgm/s
b.Impulse =Change in momentum of car
Impulse applied to the car=<-6500,0,3900>kgm/s
You are hiking in the mountains and find a shiny gold nugget. It might be the element gold, or it might be "fool's gold ," which is a nickname for iron pyrite,
FeS2
Which of the following physical properties do you think would help determine if the shiny nugget is really gold?
a. appearance,
b. melting point,
c.. density, or
d. physical state
Final answer:
Option C.) To determine if the nugget is gold, you can consider its density, melting point, and appearance.
Explanation:
To determine if the shiny nugget is really gold, one physical property that would be helpful is its density. Gold has a high density of approximately 19 g/cm³, which means it is much denser than most other minerals. Another physical property that could be useful is its melting point. Gold has a relatively low melting point of 1,064 degrees Celsius, while iron pyrite has a much higher melting point. Lastly, the appearance of the nugget can also provide some clues. Gold has a distinctive yellow color, while fool's gold (iron pyrite) has a brassy or pale yellow color.
A boat is able to move throught still water at 20m/s. It makes a round trip to a town 3.0km upstrea. If the river flows at 5m/s, the time required for this round trip is
Answer:
t=320s
Explanation:
Given Data
Boat speed=20 m/s
River flows=5 m/s
Total trip of distance d=3.0km = 3000m
To find
Total time taken
Solution
As
[tex]Velocity=distance/time\\time=distance/velocity\\[/tex]
Here we have two conditions
First when boat moves upward and the river pushing back.then velocity is given as
velocity=20m/s-5m/s
velocity=15 m/s
Time for that velocity
[tex]t_{1} =distance/velocity\\t_{1}=\frac{3000m}{15m/s}\\ t_{1}=200s[/tex]
Now for second condition when river flows and boat speed on same direction
velocity=20m/s+5m/s
velocity=25 m/s
Time taken for that velocity
[tex]t_{2}=distance/velocity\\t_{2}=\frac{3000m}{25m/s}\\ t_{2}=120m/s[/tex]
Now the total time
[tex]t=t_{1}+t_{2}\\t=(200+120)s\\t=320s[/tex]
Two tiny, spherical water drops, with identical charges of −8.00 ✕ 10^(−17) C, have a center-to-center separation of 2.00 cm.
What is the magnitude of the electrostatic force acting between them?
Answer:
F=1.4384×10⁻¹⁹N
Explanation:
Given Data
Charge q= -8.00×10⁻¹⁷C
Distance r=2.00 cm=0.02 m
To find
Electrostatic force
Solution
The electrostatic force between between them can be calculated from Coulombs law as
[tex]F=\frac{kq^{2} }{r^{2} }[/tex]
Substitute the given values we get
[tex]F=\frac{(8.99*10^{9} )*(-8.00*10^{-17} )^{2} }{(0.02)^{2} }\\ F=1.4384*10^{-19} N[/tex]
A particle is moving along a straight line such that its' acceleration is defined as
a(v) = (-2v) m/s^2 where v is in meters per second.
If v = 20 m/s when s= 0 and t=0 find:
1. The particles position as a function of time
2. The particles velocity as a function of time
3. The particles acceleration as a function of time
Final answer:
To find the particle's dynamics, we need to integrate the given acceleration function in terms of velocity and apply the initial conditions. The velocity function v(t) is √(400 - 2t), the position function s(t) would result from further integration, and the acceleration a(t) is -2√(400 - 2t).
Explanation:
A particle is moving in a straight line with its acceleration a(v) given as a(v) = (-2v) m/s², where v is the velocity in meters per second. To find the particle's position, velocity, and acceleration as functions of time, we'll need to integrate the acceleration function.
Finding the velocity as a function of time:
We have the acceleration in terms of velocity; thus, we can write:
'(v) = a(v) = -2v
(v) = -∫ 2v dv = -v² + C
When t=0, v = 20 m/s:
C = 20² = 400
v(t) = √(400 - 2t)
Finding the particle's position as a function of time:
s(t) = ∫ v(t) dt
s(t) = ∫ √(400 - 2t) dt
The integration will give us the position, s(t), in terms of t, which needs to be integrated carefully using appropriate techniques such as substitution.
Finding the acceleration as a function of time:
a(t) can be found by substituting the expression for v(t) into a(v), which gives us a(t) = -2√(400 - 2t).
If you punched thousands of holes in the aluminum foil of the scope (so there were more "holes" than "foil"), how many imageswould you see in the viewer?
The correct answer is (a.) thousands of small images. Each hole in the aluminum foil acts as a tiny pinhole camera, allowing light to pass through and create an image on the other side. Since there are numerous holes, each one forms a separate image, resulting in a multitude of small, overlapping pictures. This phenomenon is known as a pinhole array or pinhole sieve.
If you punched thousands of holes in the aluminum foil of the scope (so there were more “holes” than “foil”), the resulting images in the viewer would be thousands of small images. Each hole acts as a pinhole camera, allowing light to pass through and create an image on the other side. Since there are numerous holes, each one would form a separate image, resulting in a multitude of small, overlapping pictures. This phenomenon is known as a pinhole array or pinhole sieve.
complete question;
If you punched thousands of holes in the aluminum foil of the scope (so there were more "holes" than "foil"), how many images would you see in the viewer?
Choices are:
a. thousands of small images
b. a few bright images
c. one large blurry image
d. no images at all, the light waves would cancel
A mass m at the end of a spring oscillates with a frequency of 0.83 Hz.When an additional 730 gmass is added to m, the frequency is 0.65 Hz.What is the value of m?
Answer:
m will be equal to 1158.73 gram
Explanation:
We have given mass m when frequency is 0.83 Hz
So mass [tex]m_1=m[/tex] and frequency [tex]f_!=f[/tex] let spring constant of the spring is KK
Frequency of oscillation of spring is given by [tex]f=\frac{1}{2 \pi }\sqrt{\frac{k}{m}}[/tex]
From above relation we can say that [tex]{\frac{f_1}{f_2}}=\sqrt{\frac{m_2}{m_1}}[/tex]
It is given that when an additional 730 gram is added to m then frequency become 0.65 Hz , [tex]f_2=0.65Hz[/tex]
So [tex]m_2=m+730[/tex]
So [tex]\frac{0.93}{0.65}=\sqrt{\frac{m+730}{m}}[/tex]
[tex]\frac{m+730}{m}=1.63[/tex]
[tex]0.63m=730[/tex]
m= 1158.73 gram
To find the value of m, we can set up a proportion using the formula for frequency and solve for m. The value of m is 0.93 kg.
Explanation:To solve this problem, we can use the formula for the frequency of an object in simple harmonic motion:
f = 1 / T
Where f is the frequency and T is the period. Let's denote the mass of the object as m, and the original frequency as f1. When the additional mass is added, the frequency becomes f2. We can set up a proportion to solve for the value of m:
f1 / f2 = m / (m + 0.73)
Solving for m, we have:
m = (f1 / f2) * 0.73
Substituting the values f1 = 0.83 Hz and f2 = 0.65 Hz, we can find the value of m:
m = (0.83 / 0.65) * 0.73 = 0.93 kg
Therefore, the value of m is 0.93 kg.
A charged sphere with 1 × 10 8 units of negative charge is brought near a neutral metal rod. The half of the rod closer to the sphere has a surplus charge of 4 × 10 4 units. What is the charge on the half of the rod farther from the sphere? 0 − 1 × 10 8 1 × 10 8 − 4 × 10 4 4 × 10 4
The charge on the half of the rod farther from the sphere brought near to it, which had initially a neutral charge, is -4×104 units. This is due to a process known as charging by induction.
Explanation:The charge on the half of the rod farther from the sphere is -4×104 units. This occurs due to a process called charging by induction. Basically, when a charged object is brought near to an initially neutral conductor, it polarizes the conductor. Negative charges are attracted towards the charged sphere, leaving the far side of the rod positively charged.
However, the problem statement tells us that there is a surplus of charge on the closer half of the rod, hence the further half must have a deficiency of charge by the same amount, resulting in -4×104 units of charge. Remember, in a neutral object, the total charge is zero. Thus, if we develop a surplus (+4×104) on one side, we must have an equal amount of deficit (-4×104) on the other side to maintain the total charge at zero.
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A cylindrical shell of length 220 m and radius 4 cm carries a uniform surface charge density of σ = 14 nC/m^2.
1. What is the total charge on the shell?
Answer:
Explanation:
Given
Length of shell [tex]L=220\ m[/tex]
radius of cylindrical shell [tex]r=4\ cm[/tex]
surface charge density [tex]\sigma =14\ nC/m^2[/tex]
Total charge on the shell [tex]Q=surface\ area\times surface\ charge\ density[/tex]
surface area [tex]A=2\pi r\cdot L=2\pi \cdot 0.04\cdot 220=55.299\ m^2[/tex]
[tex]Q=\sigma \cdor A[/tex]
[tex]Q=14\times 10^{-9}\times 55.299[/tex]
[tex]Q=7.741\times 10^{-7}\ C[/tex]
If you double the radius, for a fixed angular velocity, the tangential velocity will:a. Decrease by halfb. Stay the samec. Increase by a factor of twod. Increase by a factor of four
Answer:
C. Increase by a factor of two.
Explanation:
Lets take angular speed = [tex]\omega \ rad/s[/tex]
Initially radius = r m
The tangential velocity = v m/s
We know that
[tex]v=\omega r[/tex]
If the radius becomes double ,lets say r'
r'= 2 r
Then
The tangential velocity = v' m/s
[tex]v'=\omega r'[/tex]
[tex]v'=\omega (2r)[/tex]
[tex]v'=2\omega r[/tex]
v' = 2 v
Then we can say that velocity will become double.
Therefore the answer will be C.
C. Increase by a factor of two.
Final answer:
Doubling the radius while maintaining a fixed angular velocity results in the tangential velocity increasing by a factor of two.
Explanation:
When the radius of an object is doubled while keeping the angular velocity constant, the tangential velocity will increase by a factor of two. This is because the tangential velocity (v) is calculated by multiplying the angular velocity (
ω) by the radius (r) of the circle, resulting in v = r x
ω. Therefore, if we double the radius (2r), the new tangential velocity becomes 2r x
ω, which is twice the original tangential velocity.
For a constant angular velocity, the relationship between tangential velocity and radius is direct; as one increases, so does the other. So, if you double the radius, the tangential velocity also doubles, not decreases. This explains why the correct answer to the student's question is c. Increase by a factor of two.
A 19.2 kg person climbs up a uniform ladder
with negligible mass. The upper end of the
ladder rests on a frictionless wall. The bottom
of the ladder rests on a floor with a rough
surface where the coefficient of static friction
is 0.1 . The angle between the horizontal and
the ladder is θ . The person wants to climb
up the ladder a distance of 0.49 m along the
ladder from the ladder’s foot.
What is the minimum angle θmin (between
the horizontal and the ladder) so that the
person can reach a distance of 0.49 m without
having the ladder slip? The acceleration of
gravity is 9.8 m/s^2
Answer in units of ◦
Answer:
63°
Explanation:
Draw a free body diagram of the ladder. There are 4 forces:
Normal force N pushing up at the base of the ladder.
Friction force Nμ pushing right at the base of the ladder.
Weight force mg pushing down a distance x up the ladder.
Reaction force R pushing left at the top of the ladder.
Sum of forces in the x direction:
∑Fₓ = ma
Nμ − R = 0
R = Nμ
Sum of forces in the y direction:
∑Fᵧ = ma
N − mg = 0
N = mg
Sum of moments about the base of the ladder:
∑τ = Iα
R (L sin θ) − mg (x cos θ) = 0
R (L sin θ) = mg (x cos θ)
Substituting:
Nμ (L sin θ) = mg (x cos θ)
mgμ (L sin θ) = mg (x cos θ)
μ (L sin θ) = x cos θ
tan θ = x / (μL)
θ = atan(x / (μL))
Given x = 0.49 m, μ = 0.1, and L = 2.5 m:
θ = atan(0.49 m / (0.1 × 2.5 m))
θ ≈ 63°
To solve the problem we will first calculate the reaction and the normal force.
The angle of the ladder should be 63°.
Given to us
Distance the person wants to travel, x = 0.49 m,the coefficient of static friction, μ = 0.1, Length of the ladder, L = 2.5 m:Free Body DiagramThere are 4 forcesReaction force R pushing left at the top of the ladder,Normal force N pushes up the ladder at the base of the ladder,Friction force Nμ pushing right at the base of the ladder,Weight force of the person pushing down = mg,Sum of Vertical vector forces,[tex]\sum F_y = 0\\N - mg = 0\\N = mg[/tex]
Sum of Horizontal vectors forces,[tex]\sum F_x = 0\\N\mu - R = 0\\R = N\mu[/tex]
Sum of moments at the base of the ladder[tex]R (L\ sin\theta) - mg (x\ cos\theta) = 0\\R (L\ sin\theta) = mg (x\ cos\theta)[/tex]
Substituting the values of R and N,
[tex]N\mu (L\ sin \theta) = mg (x\ cos \theta)\\mg\mu (L\ sin \theta) = mg (x\ cos\theta)\\\mu (L\ sin \theta) = x\ cos \theta\\tan \theta = \dfrac{x}{\mu L}\\\theta = tan^{-1}( \dfrac{x}{\mu L})[/tex]
Substituting the values,
[tex]\theta = tan^{-1}(\dfrac{0.49\ m}{0.1\times 2.5 m})\\\\\theta = tan^{-1} (0.96)\\\\\theta = 62.969^o \approx 63^o[/tex]
Hence, the angle of the ladder should be 63°.
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A 5.60 g bullet moving at 501.8 m/s penetrates a tree trunk to a depth of 4.59 cm. a) Use work and energy considerations to find the magnitude of the force that stops the bullet. Answer in units of N
Answer:
Explanation:
Given
mass of bullet [tex]m=5.6\ gm[/tex]
velocity of bullet [tex]v=501.8\ m/s[/tex]
Depth of penetration [tex]d=4.59\ cm[/tex]
According to the work energy theorem work done by all the force will be equal to change in kinetic energy of Particle
Suppose F is the force which is opposing the bullet motion
change in kinetic Energy [tex]\Delta K=\frac{1}{2}mv^2-0[/tex]
[tex]\Delta K=\frac{1}{2}mv^2=\frac{1}{2}\times 5.6\times 10^{-3}\times (501.8)^2[/tex]
[tex]\Delta K=705.049\ J[/tex]
[tex]\Delta K=F\cdot d[/tex]
[tex]F=\frac{\Delta K}{d}[/tex]
[tex]F=\frac{705.049}{4.59\times 10^{-2}}[/tex]
[tex]F=15,360.54\ N[/tex]
[tex]F=1.536\times 10^4\ N[/tex]
Pluto's atmosphere. As recently observed by the New Horizons mission, the surface pressure of Pluto is about 11 microbar. The surface temperature is about 37 K.
(a) What is the number density (in units of number per cubic centimeter) of molecules at Pluto's surface (Hint: use ideal gas law)? The radius of Pluto is about 1187 km and the surface gravity is about 0.62 m s. What is the total mass of the atmosphere in terms of Kg?
(b) Calculate the saturation vapor pressure (in units of Pa) of ethane at Pluto's surface. The saturated vapor pressure of ethane can be assumed as: log1o(P)-10.01-1085.0/(T-0.561). T is temperature in K and the vapor pressure (P) in units of millimeters of Hg (~133.32 Pa).
(c) If the volume mixing ratio of ethane on Pluto is about 1%, what is mass mixing ratio of ethane (assume the mean molecular weight is 28 g mol')? What is the partial pressure of ethane at the surface? (Hint: should you use volume mixing ratio or mass mixing ratio to calculate the partial pressure? Think about the physical meaning of gas pressure.) Finally, is ethane condensable at Pluto's surface)
Answer:
a) The number density is 3.623 × 10⁻³ [tex]\frac{mol}{m^{3} }[/tex]
The mass of the atmosphere is 1.3 × 10²²Kg
b) The pressure is 10⁻²⁰ Millimeter of mercury
c) The mass mixing ratio is 0.0107
The partial pressure of ethane is 0.01114 Pa
Yes it is condensable because it boiling point is -88.5 C which is equivalent to 184.5 K i.e is adding 273 to -88.5C and the temperature of the atmosphere is 37 K.
Explanation:
The explanation is on the first and second uploaded image
An object has a mass of 20 kg. Determine its weight, in N, at a location where the acceleration of gravity is 9.78 m/s2.
Answer:
Weight of the object will be 195.6 N
Explanation:
We have given mass of the object m = 20 kg
And acceleration due to gravity at any location is given [tex]g=9.78m/sec^2[/tex]
We have to find the weight of the object in N at the location where value of acceleration due to gravity is [tex]g=9.78m/sec^2[/tex]
Weight of the object is given by W = mg , here W is weight , m is mass and g is acceleration due to gravity
So weight W = 20×9.78 = 195.6 N
So weight of the object will be 195.6 N
The weight of an object is calculated by multiplying its mass with the acceleration due to gravity. Given a mass of 20 kg and gravitational acceleration of 9.78 m/s2, the weight of the object is 195.6 N.
Explanation:The subject of this question is Physics, specifically dealing with the concept of weight, which is a force. The weight of an object can be calculated by multiplying its mass by the acceleration due to gravity. Given in this problem, the object's mass is 20 kg and the place's gravitational acceleration is 9.78 m/s2.
To find the object's weight, we use the formula: Weight = mass x gravity. Applying the values, we get: Weight = 20 kg x 9.78 m/s2 = 195.6 N.
Therefore, the weight of the object at the given location is 195.6 Newtons.
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A golf ball is hit with an angle of elevation 30∘ and speed 20????????/????. Find the horizontal and vertical components of the velocity vector.
The horizontal and vertical components of the velocity vector for the golf ball hit with an angle of elevation of 30° and speed of 20 m/s are determined to be 17.32 m/s and 10 m/s, respectively.
Explanation:Velocity Vector Components
The components of the velocity vector, which are horizontal and vertical, are calculated by multiplying the speed of the object by cos(θ) for the horizontal component and sin(θ) for the vertical component, where θ is the angle of elevation.
Given that the angle of elevation is 30 degrees and the speed is 20 m/s, we can use these equations to find:
Horizontal Component: Vx = speed * cos(θ) = 20 * cos(30) = 17.32 m/sVertical Component: Vy = speed * sin(θ) = 20 * sin(30) = 10 m/sLearn more about Velocity Vector Components here:https://brainly.com/question/37190332
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Dan is gliding on his skateboard at 4.00m/s . He suddenly jumps backward off the skateboard, kicking the skateboard forward at 6.00m/s . Dan's mass is 60.0kg and the skateboard's mass is 7.00kg .
How fast is Dan going as his feet hit the ground?
To solve this problem we will apply the concepts related to the conservation of the Momentum. For this purpose we will define the momentum as the product between mass and velocity, and by conservation the initial momentum will be equal to the final momentum. Mathematically this is,
[tex]m_1u_1+m_2u_2 = m_1v_1+m_2v_2[/tex]
Here,
[tex]m_{1,2}[/tex] = Mass of Dan and Skateboard respectively
[tex]u_{1,2}[/tex] = Initial velocity of Dan and Skateboard respectively
[tex]v_{1,2}[/tex] = Final velocity of Dan and Skateboard respectively
Our values are:
Dan's mass
[tex]m_1 = 60kg[/tex]
Mass of the skateboard
[tex]m_2 = 7.0kg[/tex]
Both have the same initial velocity, then
[tex]u_1= u_2 = 4m/s[/tex]
Final velocity of Skateboard is
[tex]v_2 = 6m/s[/tex]
Rearranging to find the final velocity of Dan we have then,
[tex]m_1u_1+m_2u_2 = m_1v_1+m_2v_2[/tex]
[tex]m_1v_1+m_2v_2 = (m_1+m_2)u_1[/tex]
[tex]v_1 = \frac{ (m_1+m_2)u_1 -m_2v_2}{m_1}[/tex]
Replacing,
[tex]v_1 = \frac{(60+7)(4)-(7)(6)}{60}[/tex]
[tex]v_1 = 3.76m/s[/tex]
Therefore Dan will touch the ground at a speed of 3.76m/s
A thread is being pulled off a spool at the rate of 57.3 cm per sec. Find the radius of the spool if it makes 141 revolutions per min. Round to two decimal places as needed.
Answer: r= 0.04meter
Explanation:
The velocity V and angular speed w is related to the radio by
V=w*r
r=V/w
But the standard unit for velocity is m/s so we convert 57.3cm/sec to m/s
57.3/100= 0.57m/s
Also Angular speed w is rad/secs
So we convert rev/min to rad/sec
0.1rad/sec equals 1 rev/mins
141rev/mins * 0.1rad/secs
=14.1rad/sec
r = V/w
= 0.573/14.1
= 0.04m to 2d.p
The radius of the spool is 12.3 cm.
Explanation:To find the radius of the spool, we need to use the formula for the linear speed of a point on the edge of a rotating object: v = ωr. Here, v is the linear speed, ω is the angular velocity, and r is the radius. By converting the given information, we can calculate the linear speed:
Linear speed (v) = (57.3 cm/sec) / (100 cm/m) = 0.573 m/sec
The angular velocity can be found by converting the rate of revolution to radians per second:
Angular velocity (ω) = (141 rev/min) * (2 π rad/rev) / (60 sec/min) = 4.676 rad/sec
Now we can rearrange the formula to solve for the radius:
Radius (r) = v / ω = 0.573 m/sec / 4.676 rad/sec = 0.123 m = 12.3 cm
Two balloons (m = 0.012 kg) are separated by a distance of d = 15 m. They are released from rest and observed to have an instantaneous acceleration of a = 1.9 m/s2 toward each other at the moment they are released. Assume the two balloons are point charges and have equal but opposite charges.
Answer:
1.492*10^14 electrons
Explanation:
Since we know the mass of each balloon and the acceleration, let’s use the following equation to determine the total force of attraction for each balloon.
F = m * a = 0.012 * 1.9 = 0.0228 N
Gravitational forces are negligible
Charge force = 9 * 10^9 * q * q ÷ 225
= 9 * 10^9 * q^2 ÷ 225 = 0.0228
q^2 = 5.13 ÷ 9 * 10^9
q = 2.387 *10^-5
This is approximately 2.387 *10^-5 coulomb of charge. The charge of one electron is 1.6 * 10^-19 C
To determine the number of electrons, divide the charge by this number.
N =2.387 *10^-5 ÷ 1.6 * 10^-19 = 1.492*10^14 electrons