Answer: 3 ft
Explanation:
Given:
- the ball is initially at rest before it start moving at constant acceleration.
a = constant
u = initial speed = 0
Using the formula for displacement.
d = ut + 0.5at^2 .....1
Since u = 0, equation 1 becomes;
d = 0.5at^2 .....3
making a the subject of formula
a = 2d/t^2 ....2
Where; a = acceleration, d = distance travelled, t = time taken, u = initial speed
d = 1 ft, t =1 s
Substituting into equation 2 above.
a = (2×1)/1^2 = 2ft/s^2
Since the acceleration is 2ft/s^2, at t= 2 sec the distance it would have covered will be:
Using equation 3;
d = 0.5(2 × 2^2) = 0.5(8) = 4ft
therefore, the distance travelled between t = 1 to t= 2
Is :
d2 = d - d1
Where d = total distance (t=0 to t= 2) =4 ft
d1 = distance covered between (t=0 to t=1) = 1 ft
d2 = 4 - 1 = 3ft
The distance traveled between time 1s and time 2s is 3 ft.
The given parameters:Initial velocity of the ball, u = 0Acceleration = constantThe distance traveled by the ball at time t is calculated as follows;
[tex]s = ut + \frac{1}{2} at^2\\\\s = 0 + \frac{1}{2} at^2\\\\s = \frac{1}{2} at^2\\\\a = \frac{2s}{t^2}[/tex]
when the distance, s = 1 ft and time, t = 1 s
[tex]a = \frac{2(1)}{(1)^2} \\\\a = 2 \ ft/s^2[/tex]
When the time of motion, t = 2s, the distance traveled is calculated as;
[tex]s = \frac{1}{2} (2) (2)^2\\\\s = 4 \ ft[/tex]
The distance traveled between t =1 and t = 2 is calculated as follows;
[tex]s = s_2 - s_1\\\\s = 4 \ ft - 1\ ft\\\\s = 3 \ ft[/tex]
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One end of a stretched ideal spring is attached to an airtrack and the other is attached to a glider with a mass of 0.370 kg . The glider is released and allowed to oscillate in SHM. If the distance of the glider from the fixed end of the spring varies between 2.05 m and 1.06 m, and the period of the oscillation is 2.15 s, find the force constant of the spring, the maximum mass of the glider and the magnitude of the maximum acceleration of the glider.
The force constant of the spring is 1.932 N/m. The maximum mass of the glider can be calculated using Newton's second law. The magnitude of the maximum acceleration of the glider can be calculated using the equation for acceleration in SHM.
Explanation:In simple harmonic motion (SHM), the force exerted by an ideal spring is directly proportional to the displacement from its equilibrium position. Let's calculate the force constant of the spring by using Hooke's Law. We know that the force exerted by the spring is given by F = kx, where F is the force, k is the force constant, and x is the displacement. At the maximum displacement, the force exerted is F = k * 2.05 m. At the minimum displacement, the force exerted is F = k * 1.06 m. Equating these two forces, we get k * 2.05 = k * 1.06. Solving this equation, we find the force constant of the spring to be k = 2.05 / 1.06 = 1.932 N/m.
The maximum mass of the glider can be found by calculating the maximum force that can be exerted by the spring. The maximum force is given by F = k * x, where F is the force, k is the force constant, and x is the maximum displacement. Plugging in the values, we get F = 1.932 N/m * 2.05 m = 3.961 N. Using Newton's second law, F = m * a, where F is the force, m is the mass, and a is the acceleration, we can calculate the maximum mass. Rearranging the equation gives m = F / a. Since the acceleration in SHM is given by a = 4π^2 * x / T^2, where a is the acceleration, x is the amplitude, and T is the period, we can plug in the given values to find the maximum mass.
The magnitude of the maximum acceleration of the glider can be found by using the equation a = 4π^2 * x / T^2, where a is the acceleration, x is the amplitude, and T is the period. Plugging in the given values, we can calculate the magnitude of the maximum acceleration.
The electrostatic force of attraction between two charged objects separated by a distance of 1.0 cm is given by F. If the distance between the objects were increased to 5.0 cm, what would be the electrostatic force of attraction between them?
Answer:
New force between them will become [tex]\frac{1}{36}[/tex] times
Explanation:
Let the charge on both the object are [tex]q_1\ and\ q_2[/tex] and distance between them is is given 1 cm
So r = 1 cm = 0.01 m
Electric force between them is given F
According to Coulomb's between two charges is given by
[tex]F=\frac{1}{4\pi \epsilon _0}\frac{q_1q_2}{r^2}=\frac{Kq_1q_2}{r^2}[/tex]
According to question [tex]F=\frac{Kq_1q_2}{0.01^2}=\frac{Kq_1q_2}{10^{-4}}[/tex]-----------eqn 1
Now distance is increased by 5 cm so new distance = 5+1 = 6 cm = 0.06 m
So new force [tex]F_{new}=\frac{Kq_1q_2}{0.06^2}=\frac{Kq_1q_2}{36\times 10^{-4}}[/tex]------------------eqn 2
Comparing eqn 1 and eqn 2
[tex]F_{new}=\frac{F}{36}[/tex]
(b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)
Complete Question: A charge q1 = 2.2 uC is at a distance d= 1.63m from a second charge q2= -5.67 uC. (b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)
Answer:
d= 0.46 m
Explanation:
The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.
For a point charge, the electric potential, at a distance r from it, according to Coulomb´s Law and the definition of potential, can be expressed as follows:
V = [tex]\frac{k*q}{r}[/tex]
We have two charges, q₁ and q₂, and we need to find a point between them, where the electric potential due to them, be zero.
If we call x to the distance from q₁, the distance from q₂, will be the distance between both charges, minus x.
So, we can find the value of x, adding the potentials due to q₁ and q₂, in such a way that both add to zero:
V = [tex]\frac{k*q1}{x}[/tex] +[tex]\frac{k*q2}{(1.63m-x)}[/tex] = 0
⇒[tex]k*q1* (1.63m - x) = -k*q2*x[/tex]:
Replacing by the values of q1, q2, and k, and solving for x, we get:
⇒ x = (2.22 μC* 1.63 m) / 7.89 μC = 0.46 m from q1.
How much work must I do to assemble a charge distribution consisting of three point charges of -1.00 nC, 2.00nC, and 3.00 nC, located at the vertices of an equilateral triangle? The edges of the triangle are each 20.0 cm in length. Initially the three point charges are infinitely far apart.
Answer:
[tex]W_{total}=4.5*10^{-8}J[/tex]
Explanation:
Remember that electric potential can be written as:
[tex]V=\frac{kQ}{r}[/tex],
Where V is the electric potential, k is Coulomb's constant, Q is a point charge, and r is the distance from the point charge. Also, we can write the electric potential as:
[tex]V=\frac{W}{q}[/tex],
where W is the work made to move a charge from infinitely far apart to a certain distance, and q the point charge were are moving.
From all this we can get an expression for the work:
[tex]W=\frac{kQq}{r}[/tex]
We are going to let
[tex]q_{1}=-1.00nC\\q_{2}=2.00nC\\q_{3}=3.00nC[/tex]
To take the first charge [tex]q_{1}[/tex] from infinitely far apart to one of the vertices of the triangle, since there is no electric field and charges, we make no work.
Next, we will move [tex]q_{2}[/tex] . Now, [tex]q_{1}[/tex] is a vertice of the triangle, and we want [tex]q_{2}[/tex] to be 20.cm apart from [tex]q_{1 }[/tex] so the work we need to do is
[tex]W_{12}=\frac{kq_{1}q_{2}}{(0.20)}\\\\W_{12}=\frac{(9*10^{9})(-1*10^{-9})(2*10^{-9})}{0.20}\\\\W_{12}=-9*10^{-8}J[/tex]
Now, we move the last point charge. Here, we need to take in account the electric potential due to [tex]q_{1}[/tex] and [tex]q_{2}[/tex]. So
[tex]W=W_{13}+W_{23}\\\\\\W=\frac{kq_{1}q_{3}}{0.20}+\frac{kq_{2}q_{3}}{0.20}\\\\W=q_{3}k(\frac{q_{1}}{0.20}+\frac{q_{2}}{0.20})\\\\W=(3*10^{-9})(9*10^{9})(\frac{-1*10^{-9}}{0.20}+\frac{2*10^{-9}}{0.20})\\\\[/tex]
[tex]W=1.35*10^{-7}J[/tex]
Now, the only thing left to do is to find the total work, this can be easily done by adding [tex]W_{12}[/tex] and [tex]W[/tex]:
[tex]W_{total}=W_{12}+W\\W_{total}=1.35*10^{-7}-9*10^{-8}\\W_{total}=4.5*10^{-8}J[/tex]
Two pans of a balance are 25.1 cm apart.
The fulcrum of the balance has been shifted
1.4 cm away from the center by a dishonest
shopkeeper.
By what percentage is the true weight of the
goods being marked up by the shopkeeper?
Assume the balance has negligible mass.
Answer in units of %.
Answer:
25.1%
Explanation:
Let's say m is the true weight and M is the weight of the counterweight.
The original distance between the pans and the center is 25.1 cm / 2 = 12.55 cm. The fulcrum is then moved 1.4 cm towards the counterweight.
Sum of the moments about the fulcrum:
m (12.55 + 1.4) − M (12.55 − 1.4) = 0
13.95 m − 11.15 M = 0
13.95 m = 11.15 M
M / m = 13.95 / 11.15
M / m = 1.251
The true weight is being marked up by 25.1%.
The true weight of the goods being marked up by the shopkeeper is 25.1%.
What is lever?A lever is a simple machine made up of a beam or rigid rod that pivots at fulcrum. A lever is a rigid body that can rotate on a single point. The lever is classified into three types based on the locations of the fulcrum, load, and effort.
Let the true mass is m.
and the reading f the mass is M.
The initial distance between the two pan is 25.1 cm.
So, The initial distance between the pan and the fulcrum is = 25.1 cm / 2 = 12.55 cm.
The fulcrum is moved 1.4 cm towards the counterweight by the dishonest shopkeeper.
At equilibrium: the moment of force of two hands from fulcrum is equal. Hence,
mg (12.55 + 1.4) = Mg (12.55 − 1.4)
13.95 m = 11.15 M
M / m = 13.95 / 11.15
M / m = 1.251
Hence, the true weight is being marked up by = {(M-m)/m}×100%
= {(1.251-1)×100%}
= 25.1%.
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The amplitude of an electric field is 0.4 v/m 23 km from a radio transmitter. What is the total power emitted by the transmitter, if one assumes the radiation is emitted isotropically. The impedence of free space is 377 ohms. Answer in units of W.
The total power emitted by a radio transmitter can be calculated by first determining the intensity from the electric field and then multiplying by the surface area of the sphere covered by the isotropic radiation.
Explanation:The total power emitted by a radio transmitter can be calculated by using the relation between the electric field and the intensity of the radiation. The intensity (I) can be obtained by squaring the electric field strength (E) and dividing it by two times the impedance of free space (Z0). The formula is I = E²/2Z0. After finding the intensity, we can multiply it by the surface area of the sphere which is 4πr² (as it's isotropically radiated, it forms a sphere). The formula becomes Power (P) = I * 4πr². Substituting the given values, we calculate the total power.
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A monkey weighs 6.00 x 102 N and swings from vine to vine. As the monkey grabs a new vine, both vines make an angle of 35.0° with the vertical. In such a condition of static equilibrium, what is the net force on the new vine?
To solve this problem we will apply the considerations of equilibrium on a body, we will decompose the tensions into their respective components and apply the sum of forces and solving equations for the given system.
According to the graph, the sum of the horizontal components would be
[tex]\sum T_x = 0[/tex]
[tex]TSin35\° -TSin35\° = 0[/tex]
[tex]0 = 0[/tex]
The sum of the vertical components is
[tex]\sum T_y = 0[/tex]
[tex]2TCos35\° - mg = 0[/tex]
[tex]2TCos35\° = 600N[/tex]
[tex]T = \frac{600N}{2Cos35\°}[/tex]
[tex]T = 366.2N[/tex]
Therefore the net force on the new vine is 366.2N
In a hydrogen atom (i.e., one stationary proton and one orbiting electron), the angular velocity of the electron is 10 x 10^6 radians per second. What is the radius of the electron orbit? Assume uniform circular motion.
Assuming the centripetal force is provided by the Coulomb attraction between the electron and the proton we have that,
[tex]F_c = k \frac{e^2}{R^2}[/tex]
Here,
k = Coulomb's Constant
R = Distance
e = Electron charge
And by the centripetal force,
[tex]F_c = m_e R\omega^2[/tex]
[tex]m_e[/tex] = Mass of electron
R = Radius
[tex]\omega[/tex] = Angular velocity
Equation both expression,
[tex]k \frac{e^2}{R^2} = m_e R\omega^2[/tex]
Replacing,
[tex]R^3 = \frac{(9*10^9)(1.6*10^{-19})^2}{(9.11*10^{-31})(10^{6})^2}[/tex]
[tex]R^3 = 2.52908*10^{-10} m[/tex]
[tex]R = 6.32394*10^{-4}m[/tex]
Therefore the radius of the electron orbit is [tex]6.32394*10^{-4}m[/tex]
A Carnot heat engine receives 3,200 kJ/s of heat from a high temperature source at 825 ℃ and rejects heat to a cold temperature sink at 15 ℃. Please solve following questions:
a. What is the thermal efficiency of this engine?
b. What is the power delivered by the engine in watts?
c. At what rate is heat rejected to the cold temperature sink?
d. What is the entropy change of the sink?
Explanation:
Given:
source temperature T1= 825° C = 1099 K
Sink temperature T2= 15°C = 288 K
Heat Supplied to the engine Qs= 3200 KW
a) Efficiency of a Carnot engine is
[tex]\eta =1-\frac{T_1}{T_2}[/tex]
[tex]=1-\frac{1099}{288}[/tex]
=0.7379
=73.79%
b)
[tex]\eta= \frac{Power delivered}{power recieved}[/tex]
let W be the powe delivered
[tex]0.7379= \frac{W}{3200}[/tex]
W= 2361.408 KW
c) Heat rejected to the cold temperature Qr= Qs-W
= 3200-2361.408=838.60 KW
d) Entropy change in the sink
[tex]\Delta S= \frac{Qr}{T_{sink}}[/tex]
=838.60/288= 2.911 W/K
Answer:
a. [tex]\eta_{th}=0.7377=73.77\%[/tex]
b. [tex]P=2360655.7377\ W[/tex]
c. [tex]Q_o=839344.2622\ W[/tex]
d. [tex]\Delta s=2914.3897\ J.K^{-1}[/tex]
Explanation:
Given:
temperature of the source, [tex]T_H=825+273=1098\ K[/tex]heat received from the source, [tex]Q_i=3200000\ J.s^{-1}[/tex]temperature of the sink,[tex]T_L=15+273=288\ K[/tex]a.
Since the engine is a Carnot engine:
[tex]\eta_{th}=1-\frac{T_L}{T_H}[/tex]
[tex]\eta_{th}=1-\frac{288}{1098}[/tex]
[tex]\eta_{th}=0.7377=73.77\%[/tex]
b.
Now the power delivered:
since here we are given with the rate of heat transfer
[tex]P=\eta_{th}\times Q_i[/tex]
[tex]P=0.7377\times 3200000[/tex]
[tex]P=2360655.7377\ W[/tex]
c.
rate of heat rejection to the sink:
[tex]Q_o=Q_i-P[/tex]
[tex]Q_o=3200000-2360655.73770491808[/tex]
[tex]Q_o=839344.2622\ W[/tex]
d.
[tex]\Delta s=\frac{Q_o}{T_L}[/tex]
[tex]\Delta s=\frac{839344.2622}{288}[/tex]
[tex]\Delta s=2914.3897\ J.K^{-1}[/tex]
A hemispherical surface (half of a spherical surface) of radius R is located in a uniform electric field of magnitude E that is parallel to the axis of the hemisphere. What is the magnitude of the electric flux through the hemisphere surface?
Answer:
π*R²*E
Explanation:
According to the definition of electric flux, it can be calculated integrating the product E*dA, across the surface.
As the electric field E is uniform and parallel to the hemisphere axis, and no charge is enclosed within it, the net flux will be zero, so, in magnitude, the flux across the opening defining the hemisphere, must be equal to the one across the surface.
The flux across the open surface can be expressed as follows:
[tex]\int\ {E} \, dA = E*A = E*\pi *R^{2}[/tex]
As E is constant, and parallel to the surface vector dA at any point, can be taken out of the integral, which is just the area of the surface, π*R².
⇒Flux = E*π*R²
Final answer:
The electric flux through a hemisphere surface is calculated using Gauss's Law, where the electric field lines are parallel to the area element vector on the surface. The flux through a closed spherical surface is independent of its radius. Choosing a Gaussian surface with appropriate symmetry aids in evaluating the flux integral.
Explanation:
The magnitude of the electric flux through the hemisphere surface can be calculated using Gauss's Law. For a hemispherical surface of radius R in a uniform electric field,
The electric field lines are parallel to the area element vector on the surface. Electric flux is the product of the electric field magnitude and the area.The flux through a closed spherical surface is independent of the radius of the surface. Every field line piercing one radius also pierces another, resulting in constant flux.Choosing a suitable Gaussian surface simplifies the flux integral evaluation. A spherical surface is ideal due to uniform field angle and magnitude.Which statement might be correct based on dimensional analysis?
a. The height of the Transamerica Pyramid is 332 m2.
b. The time duration of a fortnight is 66 m/s.
c. The speed of the train is 9.8 m/s2
Answer:
none
Explanation:
a. the unit of height does not carry a square( m^2 is unit of area)
b. time cannot be in m/s. it is unit of speed
c. m/s^2 are the units of acceleration
None of the options are correct on the basis of dimensional analysis.
(a)The height of the Transamerica Pyramid is 332 m².
the dimension of height is not correct, it must be m instead of m².
(b)The time duration of a fortnight is 66 m/s.
the dimension of time is s ( seconds) not m/s (meter/second).
(c) The speed of the train is 9.8 m/s².
The dimentsion of speed is m/s (meter/second) not m/s².
Hence, none of the options are correct.
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A 90 kg man lying on a surface of negligible friction shoves a 69 g stone away from himself, giving it a speed of 4.0 m/s.
What speed does the man acquire as a result?
Answer:
v = 3.06 x 10⁻³ m/s
Explanation:
given,
mass of man,M = 90 kg
mass of stone , m= 69 g = 0.069 Kg
speed of the stone, u = 4 m/s
speed of the man, v = ?
using conservation of momentum
initial velocity of the man and stone is equal to zero
(M + m)V = M v + m u
(M + m) x 0 = 90 x v + 0.069 x 4
90 v = 0.276
v = 0.00306 m/s
v = 3.06 x 10⁻³ m/s
speed of the man on the frictionless surface is equal to 3.06 x 10⁻³ m/s
The man's velocity after shoving the stone is -0.003067 m/s, indicating movement in the direction opposite to the stone, as per the conservation of momentum.
Explanation:This problem is a classic example of the conservation of momentum, which is a fundamental concept in physics. According to the law of conservation of momentum, if no external forces are acting on a system, the total momentum of the system remains constant. In this scenario, the man and the stone together constitute an isolated system because the surface is frictionless and there are no external forces acting on the system.
The total momentum before the man shoves the stone is zero since both are initially at rest. When the man shoves the stone, the stone acquires momentum in one direction, and the man must acquire momentum in the opposite direction to conserve the total momentum of the system. The magnitude of the momentum gained by the stone and the man will be equal because their initial total momentum was zero.
The man's velocity (v) can be calculated using the formula:
Momentum of stone = mass of stone (m_stone) * velocity of stone (v_stone)
Momentum of man = mass of man (m_man) * velocity of man (v)
Since total momentum is conserved, Momentum of stone = -Momentum of man.
We can then solve for the man's velocity:
(m_stone * v_stone) = -(m_man * v)
To find v, we rearrange the equation:
v = - (m_stone * v_stone) / m_man
Plugging in the given values:
v = - (0.069 kg * 4.0 m/s) / 90 kg
v = - (0.276 kg·m/s) / 90 kg
v = -0.003067 m/s
The negative sign indicates that the man's velocity is in the opposite direction of the stone's velocity.
A spaceship is on a straight-line path between Earth and its moon. At what distance from Earth is the net gravitational force on the spaceship zero?
To solve this problem we will start from the constants given on the two celestial bodies, that is, the earth and the moon. From there we will seek the balance of forces, where the gravitational force is zero. That distance will remain incognito and will be the one we will seek to find, this is
Given that,
[tex]M_E =5.97*10^24 kg[/tex]
[tex]m_m = 7.36*10^22 kg[/tex]
[tex]r_E = 6370 km[/tex]
[tex]r_m = 1737 km[/tex]
The distance between Earth and Moon
[tex]x = 384403*10^3 m[/tex]
As the net gravitational force is zero we have that
[tex]\frac{GMm'}{d^2} = \frac{Gmm'}{(x -d)^2}[/tex]
[tex]\frac{M}{d^2} = \frac{m}{(x -d)^2}[/tex]
Replacing with our values
[tex]\frac{(5.97*10^{24} )}{d^2} = \frac{(7.36*10^{22})}{(x -d)^2}[/tex]
[tex]\frac{(x- d)^2}{d^2}= 0.0123[/tex]
[tex]x^2 + d^2 - 2xd = 0.0123 d^2[/tex]
Replacing x as the distance betwen moon and earth,
[tex](384403 x 10^3)^2+d^2-2(384403 x 10^3)d=0.0123 d^2[/tex]
Solving the polynomial we have that
[tex]d =4.3387*10^8 \text{ or } 3.4506*10^{8}[/tex]
So the minimum value is [tex]3.4506*10^{8}m[/tex]
Two parallel square metal plates that are 1.5 cm and 22 cm on each side carry equal but opposite charges uniformly spread out over their facing surfaces. How many excess electrons are on the negative surface if the electric field between the plates has a magnitude of 18,000 N/C? (k = 1/4 pi epsilon_0 = 9.0 times 10^9 N m^2/C^2, e = 1.6 times 10^-19 C)
Answer:
The number of excess electrons are on the negative surface is [tex]4.80\times10^{10}\ electrons[/tex]
Explanation:
Given that,
Distance =1.5 cm
Side = 22 cm
Electric field = 18000 N/C
We need to calculate the capacitance in the metal plates
Using formula of capacitance
[tex]C=\dfrac{\epsilon_{0}A}{d}[/tex]
Put the value into the formula
[tex]C=\dfrac{8.85\times10^{-12}\times(22\times10^{-2})^2}{1.5\times10^{-2}}[/tex]
[tex]C=0.285\times10^{-10}\ F[/tex]
We need to calculate the potential
Using formula of potential
[tex]V=Ed[/tex]
Put the value into the formula
[tex]V=18000\times1.5\times10^{-2}\ V[/tex]
[tex]V=270\ V[/tex]
We need to calculate the charge
Using formula of charge
[tex]Q=CV[/tex]
Put the value into the formula
[tex]Q=0.285\times10^{-10}\times270[/tex]
[tex]Q=76.95\times10^{-10}\ C[/tex]
Here, the charge on both the positive and negative plates
[tex]Q=+76.95\times10^{-10}\ C[/tex]
[tex]Q=-76.95\times10^{-10}\ C[/tex]
We need to calculate the number of excess electrons are on the negative surface
Using formula of number of electrons
[tex]n=\dfrac{q}{e}[/tex]
Put the value into the formula
[tex]n=\dfrac{76.95\times10^{-10}}{1.6\times10^{-19}}[/tex]
[tex]n=4.80\times10^{10}\ electrons[/tex]
Hence, The number of excess electrons are on the negative surface is [tex]4.80\times10^{10}\ electrons[/tex]
two forces equal in magnitude and opposite in direction act at the same point on an object. is it possible for there to be a net toeque on the object?
Final answer:
Even if two forces are equal in magnitude and opposite in direction, they can still create a net torque if they are applied at different points that are at different distances from the object's pivot point.
Explanation:
In physics, particularly when discussing Newton's third law, it is essential to recognize that while two forces may be equal in magnitude and opposite in direction, they do not necessarily result in equilibrium if they are acting on different systems. When considering the possibility of a net torque, it is crucial to understand that torque is not just about the magnitude of the force, but also its point of application and the distance from the pivot point (or the axis of rotation).
To produce a torque, a force must be applied in such a manner that it causes the object to rotate around a pivot point. If two equal and opposite forces are applied at different points on an object, and those points are at different distances from the pivot point, then a net torque can occur.
For example, imagine a seesaw with equal forces pushing down on either end. If these forces are applied at different distances from the fulcrum (pivot point), the seesaw will rotate due to the net torque, despite the forces being equal and opposite. This demonstrates that even with equal and opposite forces, the effect on the object's rotational motion depends on the forces' points of application and their distances from the pivot point.
A monkey weighs 6.00 x 102 N and swings from vine to vine. As the monkey grabs a new vine, both vines make an angle of 35.0° with the vertical. In such a condition of static equilibrium, what is the net force on the new vine? ANS: 366 N
Answer:
[tex]T=366.23\ N[/tex]
Explanation:
Given:
mass of monkey, [tex]w=600\ N[/tex]angle of vine from the vertical, [tex]\theta=35^{\circ}[/tex]Now follow the schematic to understand the symmetry and solution via Lami's theorem.
The weight of the monkey will be balanced equally by the tension in both the vines:
Using Lami's Theorem:
[tex]\frac{w}{sin\ 70^{\circ}} =\frac{T}{sin\ 145^{\circ}}[/tex]
[tex]\frac{600}{sin\ 70^{\circ}} =\frac{T}{sin\ 145^{\circ} }[/tex]
[tex]T=366.23\ N[/tex]
An 8900-pF capacitor holds plus and minus charges of 1.85×10−7 C . Part A What is the voltage across the capacitor?
The voltage across the capacitor is 20.79 V.
Explanation:Capacitance is a measure of how much charge can be stored in a capacitor per unit potential difference. The voltage across the capacitor can be calculated using the formula V = Q/C, where V is the voltage, Q is the charge, and C is the capacitance. In this case, the charge is 1.85×10-7 C and the capacitance is 8900 pF. Converting the capacitance to farads, we get C = 8900 × 10-12 F. Plugging in these values, we have V = 1.85×10-7 C / (8900 × 10-12 F), which simplifies to V = 20.79 V.
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You have a foam rod rubbed with felt and a small aluminum foil ball attached to a thread. Describe what happens when you slowly approach the ball with the rod and then touch the ball. Explain why this happens.
Answer:
as soon as they touch the sheet, it returns to its initial position due to the decrease in the load on the rubber ball.
Explanation:
When you rub a foam rod that is an insulator with felt it acquires some loads, when separated these two the barrel is loaded
When approaching the loaded rubber rod, suppose positively, to the aluminum ball that is a metal, the positive charge attracts the free electors of the aluminum until a balance of the electric force and the tension of the wire in which it is hung is established The aluminum foil therefore the aluminum foil approaches the rubber ball, due to the electrostatic attraction, the closer the aluminum wave deflects.
When the ball touches the sheet, the free charges of the metal pass to the insulator neutralizing the existing charges, whereby the electric field and the electric force decrease. Consequently the aluminum sheet descending and moving away from the rubber ball.
In summary, as soon as they touch the sheet, it returns to its initial position due to the decrease in the load on the rubber ball.
Due do the friction charged develop in the ball, the foil and ball attract.
When the ball and rod contact the charge neutralized and the attraction force decreased, which separate the ball and rod.
What is charge by friction?
When the insulating material are rubbed against each other, the electric charge generate between them.
The effect of friction charge between rod and ball leads to,
In these two materials, the martial which gains the electron besoms negatively charged and the martial which losses the electron becomes positively charged.Now the foam rob rubbed with felt it losses the electron and becomes positively charged. Now when the aluminum foil ball is approached with the charged foam rob the attracted to the foil, as the positive charge attracts the aluminum foil.This happens due to negative charged electron of the aluminum foil which attracts towards the ball.As the rod touches the ball the charges passes thought the insulator and the charged over the ball neutralized due to this contact. Now the attraction between them is disappear and they ball start to move.
Hence due to the friction charged develop in the ball, the foil and ball attract.
When the ball and rod contact the charge neutralized and the attraction force decreased, which separate the ball and rod.
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What are the main reasons there are so few impact craters on Earth compared to the Moon
The main reasons for having few impact craters on Earth compared to moon is due to the presence of atmosphere, water bodies and tectonic activities.
Explanation:
Impact craters are formed on the surface of any object when another very high velocity object hits the surface of the particular object. As force exhibited by any object is directly proportional to the product of mass and acceleration of the object.
In case of moon or other natural satellites, there are no means of reducing the acceleration or mass of the incident high velocity object. But in earth, the presence of different layers of atmosphere causes sufficient friction to reduce the size as well as acceleration of the impeding high velocity object.
Then the presence of larger volume of water bodies on the Earth's surface reduces the chance of hitting on land areas. The third important factor reducing the impact craters are the variation in tectonic plates which helps in keeping only recent impacts on the Earth's surface and erase the old impacts.
So the presence of different layers of atmosphere, water bodies and tectonic plats are the main reasons for having fewer impact craters on Earth compared to the Moon.
If columbia is orbiting at 0.3*10^6 m above the earths surface, what is the accceleratoin of Columbia
due to the Earth’s gravity?
(Radius of Earth = 6.4 x 106 m, mass of Earth = 6.0 x 1024 kg)
Final answer:
The acceleration due to gravity at the altitude of the Columbia shuttle is approximately 8.95 m/s^2, which is about 91% of Earth's surface gravity.
Explanation:
To determine the acceleration due to gravity at the altitude at which Columbia is orbiting, we apply the universal law of gravitation, which is represented by the formula:
F = G Mm/r²
From this formula, we can derive the expression for acceleration due to gravity (g):
g = GM/r²
Given:
Radius of Earth (rE) = 6.4 × 106 m
Altitude of Columbia (h) = 0.3 × 106 m
Mass of Earth (M) = 6.0 × 1024 kg
Gravitational constant (G) = 6.67408 × 10-11 m3kg-¹s-²
Total distance from the center of Earth (r) = rE + h
The resulting equation for the acceleration due to gravity at the altitude of Columbia is:
g = (6.67408 × 10-11 m3kg-1s-2) × (6.0 × 1024 kg) / (6.7 × 106 m)²
Upon calculating, this yields an acceleration due to gravity of approximately 8.95 m/s². This acceleration is about 91% of the gravitational acceleration at Earth's surface, which is 9.8 m/s².
On Planet X, a ball on a massless, rigid rod oscillates as a simple pendulum with a period of 2.0s. If the pendulum is taken to the moon of Planet X, where the free-fall acceleration g is half as big, the period will be
Answer:
Time period at the moon will be 2.828 sec
Explanation:
Let initially the length of the pendulum is l and acceleration due to gravity is g
And time period is given T = 2 sec
So time period of the pendulum is equal to [tex]T=2\pi \sqrt{\frac{l}{g}}[/tex]-------------eqn 1
Now on the planet acceleration due to gravity [tex]g_{new}=\frac{g}{2}[/tex]
So time period of the pendulum at this planet [tex]T_{new}=2\pi \sqrt{\frac{l}{g_{new}}}[/tex]--------eqn 2
Now dividing eqn 1 by eqn 2
[tex]\frac{T}{T_{new}}=\sqrt{\frac{\frac{g}{2}}{g}}[/tex]
[tex]T_{new}=\sqrt{2}T=1.414\times 2=2.828sec[/tex]
So time period at the moon will be 2.828 sec
If the acceleration due to gravity decreases by a half, the period of a simple pendulum increases by √2, or approximately 1.41. Therefore, the period of the pendulum on the moon of Planet X, where gravity is half, would be 2.8 seconds.
Explanation:The period of a simple pendulum in physics is influenced by its length and the acceleration due to gravity (g). It is calculated by the formula: T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the gravity. To find the effect of a change in g on the period of the pendulum, we can look at the relationship between them in the formula. Because the formula for the period of a pendulum involves the square root of the gravity, if gravity is reduced by a half, the period will increase by a factor of √2, which is approximately 1.41.
Thus, if the pendulum's period on Planet X was 2.0 s, then on its moon, where gravity is halved, the period will be approximately 2.0 s * 1.41 = 2.8 s.
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In the context of energy transfers with hot and cold reservoirs, the sign convention is that _______________.
A. QC > 0; QH > 0
B. QC > 0; QH < 0
C. QC = 0; QH > 0
D. QC < 0; QH > 0
E. QC < 0; QH < 0
Answer:
B. QC > 0; QH < 0
Explanation:
Given that there are two reservoir of energy.
Sign convention for heat and work :
1.If the heat is adding to the system then it is taken as positive and if heat is going out from the system then it is taken as negative.
2. If the work is done on the system then it is taken as negative and if the work is done by the system then it is taken as positive.
From hot reservoir heat is going out that is why it is taken as negative
[tex]Q_H<0[/tex]
From cold reservoir heat is coming inside the reservoir that is why it is taken as positive
[tex]Q_C>0[/tex]
That is why the answer will be
[tex]Q_H<0[/tex] ,[tex]Q_C>0[/tex]
In the context of energy transfers with hot and cold reservoirs, the sign convention is
Option A (QC > 0; QH > 0)Given that there are two reservoir of energy. Sign convention for heat and work :
1) If the heat is adding to the system then it is taken as positive and if heat is going out from the system then it is taken as negative.
2) If the work is done on the system then it is taken as negative and if the work is done by the system then it is taken as positive.
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Assuming the same percent uncertainty and the same measurement device, what is the uncertainty in a blood pressure measurement of 71 mm Hg ?
Final answer:
The percent uncertainty of a blood pressure measurement of 120 mm Hg with an uncertainty of ±2 mm Hg is 1.67%. Using this percentage, the uncertainty in a blood pressure measurement of 71 mm Hg would be ±1.2 mm Hg, after rounding to one decimal place.
Explanation:
To calculate the uncertainty in a blood pressure measurement of 71 mm Hg, given the percent uncertainty is the same as that for a measurement of 120 mm Hg with an uncertainty of ± 2 mm Hg, we first find the percent uncertainty of the original measurement.
The percent uncertainty is calculated by dividing the uncertainty by the actual measurement and then multiplying by 100 to get a percentage:
Percent Uncertainty = (Uncertainty / Measurement) × 100
For the measurement of 120 mm Hg, the percent uncertainty is:
(2 mm Hg / 120 mm Hg) × 100 = 1.67%
Assuming the same percent uncertainty for a blood pressure measurement of 71 mm Hg, you would use the same percentage to find the uncertainty:
Uncertainty = Percent Uncertainty × Measurement / 100
Uncertainty in 71 mm Hg = 1.67% × 71 mm Hg / 100 = 1.187 mm Hg
Since we typically round to the nearest whole unit or significant digit the uncertainty would be ± 1.2 mm Hg (rounded to one decimal place).
A mass-spring system has k = 56.8 N/m and m = 0.46 kg.
A) Find the frequency of its simple harmonic motion. Express your answer with the appropriate units.
B) Find the angular frequency of its simple harmonic motion. Express your answer with the appropriate units.
C) Find the period of its simple harmonic motion. Express your answer with the appropriate units.
Answer:
A. [tex]\omega=11.1121\ rad.s^{-1}[/tex]
B. [tex]f=1.7685\ Hz[/tex]
C. [tex]T=0.5654\ s[/tex]
Explanation:
Given:
spring constant, [tex]k=56.8\ N.m^{-1}[/tex]mass attached, [tex]m=0.46\ kg[/tex]A)
for a spring-mass system the frequency is given as:
[tex]\omega=\sqrt{\frac{k}{m} }[/tex]
[tex]\omega=\sqrt{\frac{56.8}{0.46}}[/tex]
[tex]\omega=11.1121\ rad.s^{-1}[/tex]
B)
frequency is given as:
[tex]f=\frac{\omega}{2\pi}[/tex]
[tex]f=\frac{11.1121}{2\pi}[/tex]
[tex]f=1.7685\ Hz[/tex]
C)
Time period of a simple harmonic motion is given as:
[tex]T=\frac{1}{f}[/tex]
[tex]T=0.5654\ s[/tex]
A simple pendulum in a science museum entry hall is 3.58 m long, has a 1.23 kg bob at its lower end, and swings with an amplitude of 10.9 ∘. How much time does the pendulum take to swing from its extreme right side to its extreme left side?
Answer:
1.897825 seconds
Explanation:
L = Length of pendulum = 3.58 m
g = Acceleration due to gravity = 9.81 m/s²
The amplitude of a pendulum is given by
[tex]T=2\pi \sqrt{\dfrac{L}{g}}\\\Rightarrow T=2\pi\sqrt{\dfrac{3.58}{9.81}}\\\Rightarrow T=3.79565\ s[/tex]
It can be seen that the time period does not depend on amplitude
Time period includes the time to go from extreme left to right and back to extreme left.
So, time taken to go from extreme right to left is [tex]\dfrac{T}{2}=\dfrac{3.79565}{2}=1.897825\ s[/tex]
The period of a simple pendulum is determined by its length and the acceleration due to gravity. It is independent of other factors such as mass and amplitude.
Explanation:The period of a simple pendulum is determined by its length and the acceleration due to gravity. The period is independent of other factors such as mass and amplitude. For a small amplitude of less than 15 degrees, the period of a simple pendulum can be calculated using the formula:
T = 2π√(L/g)
where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
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Estimate and determine the order of magnitude of the circumference of the Earth in miles and the speed of a sailboat in miles per hour. The Earth's circumference is 10^4 mile and the the speed of a sailboat is 10^1 mile/hour.
Answer:
The time it would take to sail around the world in a sailboat is 1000 hour.
Explanation:
It is given that,
The circumference of the Earth, [tex]d=10^4\ mile[/tex]
The speed of a sailboat, [tex]v=10\ mile/hour[/tex]
We need to find the time it would take to sail around the world in a sailboat. It can be calculated using the definition of velocity. It is given by :
[tex]v=\dfrac{d}{t}[/tex]
[tex]t=\dfrac{d}{v}[/tex]
[tex]t=\dfrac{10^4\ mile}{10\ mile/hour}[/tex]
t = 1000 hour
So, the time it would take to sail around the world in a sailboat is 1000 hour. Hence, this is the required solution.
A Capacitor is a circuit component that stores energy and can be charged when current flows through it. A current of 3A flows through a capacitor that has an initial charge of 2μC (micro Coulombs). After two microseconds, how much is the magnitude of the net electric charge (in μC) of the capacitor?
Answer:
[tex]8\mu C[/tex]
Explanation:
t = Time taken = [tex]2\mu s[/tex]
i = Current = 3 A
q(0) = Initial charge = [tex]2\mu C[/tex]
Charge is given by
[tex]q=\int_0^t idt+q(0)\\\Rightarrow q=\int_0^{2\mu s} 3dt+2\mu C\\\Rightarrow q=3(2\mu s-0)+2\mu C\\\Rightarrow q=6\mu C+2\mu C\\\Rightarrow q=8\mu C[/tex]
The magnitude of the net electric charge of the capacitor is [tex]8\mu C[/tex]
You have a box of filled with gas and by some process you double the number of atoms in the box, but now each one only has (on average) half as much energy as before the process started.
What has happened to the thermal energy and temperature of the gas, respectively?
-Thermal energy increased, temperature decreased.
-Thermal energy decreased, temperature unchanged.
-Thermal energy unchanged, temperature unchanged.
-Thermal energy unchanged, temperature decreased.
Answer:
Option-(D):Thermal energy unchanged, temperature decreased.
Explanation:
Thermal energy and Temperature,T:
Thermal energy of the system is remained unchanged while the temperature,T of the objects present inside the system is more associated with the objects fundamental properties which each matter shows or exhibits in nature.
While the thermal energy is the flow or exchange of thermal or heat energy between the different objects in interaction to each other.
A 0.105-g sample of X2 contains 8.92x1020 molecules. The compound MX3 consists of 54.47% X by mass. What are the identities of M and X, and what is the correct name for MX3?
Answer:
The element X is chlorine.
The element M is Ytterium.
Ytterium (III) chloride is the correct name for [tex]YCl_3[/tex]
Explanation:
Molecules of [tex]X_2, N = 8.92\times 10^{20} [/tex]
Moles of [tex]X_2, n= ?[/tex]
[tex]N=n\times N_A[/tex]
[tex]n=\frac{N}{N_A}=\frac{8.92\times 10^{20}}{6.022\times 10^{23} mol^{-1}}[/tex]
n = 0.001481 mol
Mass of [tex]X_2, m=0.105 g[/tex]
Molar mass of [tex]X_2=M[/tex]
[tex]m=M\times n[/tex]
[tex]M=\frac{m}{n}=\frac{0.105 g}{0.001481 mol}[/tex]
M = 70.87 g/mol
Atomic mass of X = [tex]\frac{70.87 g/mol}{2}=35.435 g/mol[/tex]
X is a chlorine atom.
The compound [tex]MX_3[/tex] consists of 54.47% X by mass.
M' = Molar mass of compound [tex]MX_3[/tex]
Percentage of X in compound = 54.47%
[tex]54.47\%=\frac{3\times 35.435 g/mol}{M'}\times 100[/tex]
M' = 195.16 g/mol
Atomic mass of M = a
[tex]195.16 g/mol=a+3\times 35.435 g/mol[/tex]
[tex]a = 195.16 g/mol-3\times 35.435 g/mol[/tex]
a = 88.86 g./mol
The element M is Ytterium.
Ytterium (III) chloride is the correct name for [tex]YCl_3[/tex]
The element X is Chlorine and the element M is Ytterium.
Ytterium Chloride is the correct name for [tex]MX_3[/tex] that is [tex]YCl_3[/tex]
Number of molecules of [tex]X_2={8.92*10^{20}[/tex]
Number of moles of X:
[tex]n=\frac{8.92*10^{20}}{6.023*10^{23}}[/tex]
n = 0.001481 mol = 0.105g
Molar mass of [tex]X_2[/tex] molecule
[tex]M= 0.105/0.001481\\\\M=70.87g/mol[/tex]
Therefore molar mass of X atom = molar mass of [tex]X_2[/tex]/2
m = 35.435g/mol
X is a chlorine atom.
Given that the compound consists of 54.47% X by mass.
Let M' = Molar mass of compound [tex]MX_3[/tex]
Percentage of X in compound = 54.47%
54.47 = {amount of X}/{mola mass of compound}*100
[tex]54.47 = \frac{3*35.435}{M_{'} }*100\\\\M^{'}= 195.16g/mol[/tex]
let Atomic mass of M = A
A = 195.16 - 3 × 35.435
A = 88.86 g/mol
The element M is Ytterium.
Therefore the compound [tex]MX_3[/tex] is Ytterium Chloride [tex]YCl_3[/tex]
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An electron and a proton are each placed at rest in a uniform electric field of magnitude 560 N/C. Calculate the speed of each particle 46.0 ns after being released. electron 4.5e^-6 Incorrect.
Answer:
The speed of electron is [tex]v=4.52\times 10^6\ m/s[/tex] and the speed of proton is 2468.02 m/s.
Explanation:
Given that,
Electric field, E = 560 N/C
To find,
The speed of each particle (electrons and proton) 46.0 ns after being released.
Solution,
For electron,
The electric force is given by :
[tex]F=qE[/tex]
[tex]F=1.6\times 10^{-19}\times 560=8.96\times 10^{-17}\ N[/tex]
Let v is the speed of electron. It can be calculated using first equation of motion as :
[tex]v=u+at[/tex]
u = 0 (at rest)
[tex]v=\dfrac{F}{m}t[/tex]
[tex]v=\dfrac{8.96\times 10^{-17}}{9.1\times 10^{-31}}\times 46\times 10^{-9}[/tex]
[tex]v=4.52\times 10^6\ m/s[/tex]
For proton,
The electric force is given by :
[tex]F=qE[/tex]
[tex]F=1.6\times 10^{-19}\times 560=8.96\times 10^{-17}\ N[/tex]
Let v is the speed of electron. It can be calculated using first equation of motion as :
[tex]v=u+at[/tex]
u = 0 (at rest)
[tex]v=\dfrac{F}{m}t[/tex]
[tex]v=\dfrac{8.96\times 10^{-17}}{1.67\times 10^{-27}}\times 46\times 10^{-9}[/tex]
[tex]v=2468.02\ m/s[/tex]
So, the speed of electron is [tex]v=4.52\times 10^6\ m/s[/tex] and the speed of proton is 2468.02 m/s. Therefore, this is the required solution.