Knowing that α= 60° and that boom AC exerts on pin C a force directed along line AC, determine (a)the magnitude of t hat force, (b) the tension in cable BC.Figure:Cable BC is making an angle alpha with fixed support.

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.

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.

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

Answer: d(5) = 115m

Position after 5 seconds is 115m

Explanation:

Given;

Initial position d(0) = 0

Time = 5 sec

Velocity function vx(t) = 10t - 2

To determine its position after 5 sec we need to calculate the position function.

d(t) = integral of vx(t)

d(t) = ∫10t - 2

d(t) = (10/2)t^2 - 2t + c

d(t) = 5t^2 - 2t + d(0)

c = d(0) = 0

d(t) = 5t^2 - 2t

So, at time t = 5

d(5) = 5(5^2) -2(5) = 125 - 10

d(5) = 115m

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).

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.

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]

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

Answer:

1) 13.377 m/s

2) 12.044 m/s

3) 8.893 m

4) 32.85 m

5) 19.05 m/s

6) 3.25 m

Explanation:

1)

V_o,x = V_o * cos (Q)

V_o,x = 18 * cos (42)

V_o,x = 13.377 m/s

2)

V_o,y = V_o * sin (Q)

V_o,y = 18 * sin (42)

V_o,y = 12.044 m/s

3)

Maximum height is reached when V,y = 0

V,y = V_o,y + a*t

0 = 12.044 - 9.81t

Solve above equation for t:

t = 1.228 s

Compute S_y @t = 1.228 s

S_y = S_o,y + V_o,y*t + 0.5*a*t^2

S_y = 1.5 + 12.044*1.228 - 4.905*1.228^2

S_y = 8.893 m

4)

Time taken for the ball to complete path:

S_y = S_o,y + V_o,y*t + 0.5*a*t^2 = 1.5

V_o,y*t + 0.5*a*t^2 = 0

12.044*t - 4.905*t^2 = 0

t = 0, t = 2.455 s

Total distance traveled in horizontal direction S_x @ t = 2.455 s

S_x = S_o,x + V_o,x*t

S_x = 0 + 13.377*2.455 = 32.85 m

5)

S_y = S_o,y + V_o,y*t + 0.5*a*t^2 = 10

10 = 1.5 + V_o,y*t -4.905*t^2   .... Eq 1

Maximum height is reached when V,y = 0

V,y = V_o,y + a*t

0 =  V_o,y - 9.81t  .... Eq2

Solve Eq 1 and Eq 2 simultaneously

V_o,y = 9.81*t

10 = 1.5 + 9.81*t^2 -4.905*t^2

8.5 = 4.905*t^2

t = 1.316 s

V_o,y = 12.914 m/s

Compute Velocity

V = sqrt (V_o,x^2 + V_o,y^2)

V = sqrt (14^2 + 12.914^2)

V = 19.05 m/s

6)

Total distance traveled in horizontal direction between players is 32.85m

S_x = S_o,x + V_o,x*t

S_x = 0 + 14*t = 32.85 m

t = 2.3464 s

Compute Sy @ t = 2.3464 s

S_y = S_o,y + V_o,y*t + 0.5*a*t^2

S_y = 10  - 4.905*(1.1732)^2

S_y = 3.25 m

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

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]                      

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.

Answer:

(a) The speed of proton decreases as it moves from A to B.

(b) Plate B is at a higher potential.

(c) Plate B is positive, plate A is negative.

(d) Parallel lines parallel to the two plates.

    Parallel lines equally spaced.

Explanation:

The electric potential energy is given by the following formula:

[tex]U = \frac{1}{4\pi \epsilon_0}\frac{qq_0}{r}[/tex]

Alternatively, potential energy in a uniform electric field is

[tex]U = qEr[/tex]

where 'r' is the distance from negative to positive plates. This definition is analogues to that of gravitational potential energy, U = mgh.

If the positively charged proton is gaining potential energy as it gets closer to plate B, then plate B is charged positively.

(a) According to this information, the speed of proton decreases as it moves from A to B. This is similar to the speed of an object which is gaining potential energy by moving higher.

(b) By the same gravitational analogy, plate B is at a higher potential.

(c) As explained before, Plate A is negative and Plate B is positive.

(d) The equipotential lines are parallel to electric field lines which are perpendicular to the plates. So, the equipotential lines are parallel to the plates. Since the electric field between the plates is uniform, then the equipotential lines are equally seperated.

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]

Answer:

q1= +0.75 nC

Explanation:

As the electrostatic force is linear, we can apply the superposition principle to calculate the total force on q₃ due to q₂ and q1, according to Coulomb's Law, as follows:

F₃₂ = k*q₃*q₂/r₃₂² = 9*10⁹ N*m²/C²*+5nC*(-3 nC) / (0.04m)² = -84.4*10⁻⁶ N

F₃₁ = k*q₃*q₁ / r₃₁² = 9*10⁹ N*m²/C²*q₁*(+5 nC) / (0.02m)²

The total force on q₃ is just the sum of F₃₂ and F₃₁, which must add to 0, as follows:

F₃ = F₃₂ + F₃₁ = 0

⇒ -84.4* 10⁻⁶ N = -9*10⁹ N*m²/C²*q₁*(+5 nC) / (0.02m)²

Solving for q₁, we get:

q₁ = (84.4 / 11.25)*10⁻¹⁰ C = +0.75 nC

q₁ must be positive, in order to counteract the attractive force on q₃ due to q₂.

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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Answer: wavelength λ = 2.9Å

Explanation:

Using the particle in a box model. The energy level level increases with n^2

En = (n^2h^2)/ 8mL^2 .....1

For the ground state, n = 1 to level n= 5, the energy level changes from E1 to E5

∆E = (5^2 - 1^2)h^2/8mL^2

but 5^2 - 1^2 = 24.

so,

∆E = 24h^2/8mL^2 .....2

And the wavelength of the radiation can be derived from the equation below:

E = hc/λ

λ = hc/E .......3

Substituting equation 2 to 3

λ = hc/[(24h^2)/ 8mL^2]

λ = 8mcL^2/(24h)

λ = 8mcL^2/24h .....4

Where,

n = energy state

h = Planck's constant = 6.626 × 10^-34 Js

m= mass of electron = 9.1 × 10^-31 kg

L = length = 45.7pm = 45.7×10^-12 m

E = energy

c= speed of light = 3.0 ×10^8 m/s

λ= wavelength

Substituting the values into equation 4 above

λ = [(8×9.1×3×45.7^2)/(24×6.626)] × 10^(-31+8-24+34)

λ = 2868.285390884 × 10^-13 m

λ = 2.9 × 10^-10 m

λ = 2.9Å

The wavelength of the electromagnetic radiation required to excite an electron can be calculated using the formula wavelength = (2 * box length) / n, where n is the energy level. The wavelength of the electromagnetic radiation required is 9.14 pm.

To calculate the wavelength of the electromagnetic radiation required to excite an electron from the ground state to the level with n = 5 in a one-dimensional box, we can use the formula:

wavelength = (2 * box length) / n

Given that the box length is 45.7 pm and n = 5, we can substitute the values into the formula:

wavelength = (2 * 45.7 pm) / 5

Simplifying the expression gives us:

wavelength = 9.14 pm

Therefore, the wavelength of the electromagnetic radiation required is 9.14 pm.

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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

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]

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.

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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To solve this problem we will apply the linear motion kinematic equations. With the information provided we will calculate the time it takes for the object to fall. From that time, considering that the ascent rate is constant, we will take the reference distance and calculate the distance traveled while the object hit the ground, that is,

[tex]h = v_0 t -\frac{1}{2} gt^2[/tex]

[tex]-18 = 15*t + \frac{1}{2} 9.8*t^2[/tex]

[tex]t = 3.98s[/tex]

Then the total distance traveled would be

[tex]h = h_0 +v_0t[/tex]

[tex]h = 18+15*3.98[/tex]

[tex]h = 77.7m[/tex]

Therefore the railing will be at a height of 77.7m when it has touched the ground

The height  is the railing when the camera hits the ground should be considered as the 77.7 m.

Since we know that

[tex]h = vt - 1/2gt^2\\\\-18 = 15*1 + 1/2*9.8*t^2[/tex]

t = 3.98s

Now the total distance should be

[tex]= 18 + 15*3.98[/tex]

= 77.7 m

hence, The height  is the railing when the camera hits the ground should be considered as the 77.7 m.

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Answer:

string stretched is 1.02 cm

Explanation:

given data

length = 80-cm

diameter = 1.0-mm

tension = 2000 N

solution

we get here string stretched that will be as and here  

we know that young modulus for steel = 200 × [tex]10^{9}[/tex]

so here stress will be

stress = y × strain  .............1

that is express as

[tex]\frac{force}{area} = \frac{Y \Delta L}{L}[/tex]

ΔL = [tex]\frac{0.80*2000}{\pi * 0.0005^2*200*10^9}[/tex]

ΔL = 0.0102 m

ΔL = 1.02 cm

so string stretched is 1.02 cm

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

[tex]\sum F_y = 0\\N - mg = 0\\N = mg[/tex]

[tex]\sum F_x = 0\\N\mu - R = 0\\R = N\mu[/tex]

[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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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

Answer:

The boat moves away from the dock at 0.5 m/s.

Explanation:

Hi there!

Since no external forces are acting on the system boy-boat at the moment at which the boy lands on the boat, the momentum of the system is conserved (i.e. it remains constant).

The momentum of the system is calculated as the sum of the momentum of the boy plus the momentum of the boat. Before the boy lands on the boat, the momentum of the system is given by the momentum of the boy.

momentum of the system before the boy lands on the boat:

momentum of the boy + momentum of the boat

m1 · v1 + m2 · v2 = momentum of the system

Where:

m1 and v1: mass and velocity of the boy.

m2 and v2: mass and velocity of the boat.

Then:

50 kg · 2.0 m/s + 150 kg · 0 m/s = momentum of the system

momentum of the system = 100 kg m/s

After the boy lands on the boat, the momentum of the system will be equal to the momentum of the boat moving with the boy on it:

momentum of the system = (m1 + m2) · v (where v is the velocity of the boat).

100 kg m/s = (50 kg + 150 kg) · v

100 kg m/s / 200 kg = v

v = 0.5 m/s

The boat moves away from the dock at 0.5 m/s.

Final answer:

The child running off a dock and landing in a rowboat scenario involves applying the conservation of momentum principle to find the boat's final velocity. The speed of the rowboat is 0.67 m/s.

Explanation:

Given:

Child mass (m1) = 50 kg

Child velocity (v1) = 2.0 m/s

Boat mass (m2) = 150 kg

Let the final velocity of the boat be v2

Using the conservation of momentum:

m1v1 = (m1 + m2)v2

Substitute the values to find v2: 50 kg * 2.0 m/s = (50 kg + 150 kg) * v2

Solving for v2, we get v2 = 0.67 m/s

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Gauss' law states that the total charge contained within a closed surface immediately proportionately affects the electric flux through that surface. The flux across the spherical surface will quadruple if the charge is tripled.

The enclosed charge determines the electric flux through a closed surface, not the size or shape of the surface. The quantity of charge confined does not change when the sphere's volume is doubled, hence the flux does not change.

Similar to component (b), the flux is unaffected by changing the surface's shape as long as the enclosed charge stays constant.

The total charge contained by the surface is the only factor that affects the electric flux across a closed surface.

Thus, since there is no longer any charge contained by the closed surface if the charge is transported outside of it, there is no longer any electric flux through the surface.

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The total flux through a spherical surface surrounding a point charge depends on the charge, while the flux through the surface remains constant when the volume of the sphere or the shape of the surface is changed. If the charge is moved to another location inside the surface, the total flux remains constant, but if the charge is moved outside the surface, the total flux goes to zero.

(a) When the charge is tripled, the total flux through the surface also triples. This is because the total flux is directly proportional to the charge enclosed by the surface.

(b) When the volume of the sphere is doubled, the total flux through the surface remains constant. This is because the total flux is independent of the volume of the surface.

(c) When the surface is changed to a cube, the total flux through the surface remains constant. This is because the total flux is independent of the shape of the surface.

(d) When the charge is moved to another location inside the surface, the total flux through the surface remains constant. This is because the total flux is independent of the position of the charge within the surface.

(e) When the charge is moved outside the surface, the total flux through the surface goes to zero. This is because the total flux is zero when there are no charges enclosed by the surface.

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To develop this problem we will apply the linear motion kinematic equations. For this purpose we will define the change in speed as the product between acceleration and time.

[tex]v_f-v_i = at[/tex]

The relation between initial velocity final velocity and time is

[tex]v_f = v_i+at[/tex]

The acceleration is due to the acceleration due to gravity, then we have

[tex]v_f = v_i-gt[/tex]

At the maximum height the final velocity is zero. Then we have that

[tex]0 = v_i-gt[/tex]

[tex]t = \frac{v_i}{g}[/tex]

The time the player must wait before touching he ball is

[tex]t = \frac{4.41}{9.8}[/tex]

[tex]t = 0.45s[/tex]

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

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

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%.

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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Answer:

I = 70.2mA

V1 = 3.09V

V2 = 1.19V

V3 = 7.72V

Explanation:

Total resistance for a series connection = R1 + R2 + R3 = 44ohm + 17ohm + 110ohm = 171ohm

From ohm's law

Voltage (V) = current (I) × resistance (R)

I = V/R = 12/171 = 0.0702A = 0.0702×1000mA = 70.2mA

V1 = I×R1 = 0.0702×44 = 3.09V

V2 = I×R2 = 0.0702×17 = 1.19V

V3 = I×R3 = 0.0702×110 = 7.72V

Answer:

a) I = 0.0702A = 70.2mA

b) V1 = 3.09V

c) V2 = 1.19V

d) V3 = 7.72V

Explanation:

Given that the circuit consist of three series resistors.

For resistors arranged in series, the total resistance R can be given as:

R = R1 + R2 + R3

R1 = 44 ohms

R2 = 17 ohms

R3 = 110 ohms.

R = 44 + 17 + 110 = 171 ohms

V = 12 V

a) The current of a circuit is given by;

Potential difference V = current × total resistance

V = IR

Making I the subject of formula,

I = V/R

I = 12/171 = 0.0702A

I = 7.02mA

b) the potential difference across any resistors is given by:

V = IR

Since the arrangement is parallel, the same current flows through each of the resistors.

V1 = IR1

V1 = 0.0702 × 44

V1 = 3.09V

c) applying the same rule as b above:

V2 = IR2

V2 = 0.0702 × 17

V2 = 1.19V

d) applying the same rule as b above.

V3 = IR3

V3 = 0.0702 × 110

V3 = 7.72V