A team of astronauts is on a mission to land on and explore a large asteroid. In addition to collecting samples and performing experiments, one of their tasks is to demonstrate the concept of the escape speed by throwing rocks straight up at various initial speeds. With what minimum initial speed ????esc will the rocks need to be thrown in order for them never to "fall" back to the asteroid? Assume that the asteroid is approximately spherical, with an average density ????=2.02×106 g/m3 and volume ????=3.09×1012 m3 . Recall that the universal gravitational constant is ????=6.67×10−11 N·m2/kg2 .

Both light and matter can possess particle properties and wave properties at the same time.

This quantum behavior was proposed by the French physicist Louis De Broglie in 1924 in his doctoral thesis when proposing that all matter has an associated wave, and light did not escape from that behavior. Then this was empirically tested in various experiments where light behaved as a photon (particle) and in others as particles.

Einstein also demonstrated this fact, however, in spite of being observed this behavior several times, it could not have been done simultaneously, until in 2015 a group of scientists from the Federal Polytechnic School of Lausanne (Switzerland), succeeded to perform an experiment and capture for the first time the image of the wave-particle duality of light.

Answer:

-78.4 cm

Explanation:

We can solve the problem by using the lens equation:

[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}[/tex]

where

p is the distance of the object from the lens

q is the distance of the image from the lens

f is the focal length

Here we have

p = 80.0 cm

q = -39.6 cm (negative, because the image is on the same side as the object , so it is a virtual image)

Substituting, we find f:

[tex]\frac{1}{f}=\frac{1}{80.0 cm}+\frac{1}{-39.6 cm}=-0.0128 cm^{-1}[/tex]

[tex]f=\frac{1}{0.0128 cm^{-1}}=-78.4 cm[/tex]

(a) [tex]1.24\cdot 10^{-11} F[/tex]

The general formula for the capacitance of a capacitor is:

[tex]C=\frac{Q}{V}[/tex]

where

Q is the charge stored on the capacitor

V is the potential difference across the capacitor

In this problem, we have

[tex]Q=3.10 nC = 3.10\cdot 10^{-9} C[/tex] is the charge stored

V = 250 V is the potential difference

Substituting, we find

[tex]C=\frac{3.10\cdot 10^{-9}C}{250 V}=1.24\cdot 10^{-11} F[/tex]

(b) 0.0294 m

The capacitance of a spherical capacitor is given by

[tex]C=\frac{4\pi \epsilon_0}{\frac{1}{a}-\frac{1}{b}}[/tex]

where

a is the radius of the inner shell

b is the radius of the outer shell

Here we have

b = 4.00 cm = 0.04 m

and the capacitance is

[tex]C=1.24\cdot 10^{-11} F[/tex]

So we can re-arrange the equation to find a, the radius of the inner sphere:

[tex]a=(\frac{4\pi \epsilon_0}{C}+\frac{1}{b})^{-1} =(\frac{4\pi (8.85\cdot 10^{-12}F/m)}{1.24\cdot 10^{-11}F}+\frac{1}{0.04 m})^{-1}=0.0294 m[/tex]

(c) [tex]3.225 \cdot 10^4 V/m[/tex]

The electric field just outside the surface of the inner sphere with charge Q is equal to the electric field produced by a single point charge Q at a distance of r = a:

[tex]E=\frac{Q}{4\pi \epsilon_0 a^2}[/tex]

where

[tex]Q=3.10 \cdot 10^{-9} C[/tex] is the charge on the sphere

a = 0.0294 m is the radius of the inner sphere

Substituting the data into the formula, we find:

[tex]E=\frac{3.10\cdot 10^{-9}C}{4\pi (8.85\cdot 10^{-12}F/m) (0.0294 m)^2}=3.225 \cdot 10^4 V/m[/tex]

1. 13,500 cal

First of all, we need to find the amount of heat needed to raise the temperature of the ice from -20°C to 0°C. This is given by

[tex]Q_1 = m C_i \Delta T[/tex]

where

m = 150 g is the mass of the ice

C_i = 0.5 cal/g·C° is the specific heat capacity of the ice

[tex]\Delta T=0 C-(-20 C)=20^{\circ}C[/tex] is the change in temperature of the ice

Substituting,

[tex]Q_1 = (150 g)(0.5 cal/gC)(20 C)=1500 cal[/tex]

Now we have to find the amount of heat needed to melt the ice, which is

[tex]Q_2 = m \lambda_f[/tex]

where

m = 150 g is the mass of the ice

[tex]\lambda_f = 80 cal/g[/tex] is the latent heat of fusion

Substituting,

[tex]Q_2 = (150 g)(80 cal/g)=12,000 cal[/tex]

So the total heat required is

[tex]Q_3 = 1500 cal + 12,000 cal = 13,500 cal[/tex]

2. 3750 cal

The additional amount of heat required to heat the water to 25°C is

[tex]Q_4 = m C_w \Delta T[/tex]

where

m = 150 g is the mass of water

C_w = 1 cal/g·C is the speficic heat capacity of water

[tex]\Delta T=25 C-0 C=25^{\circ}C[/tex] is the change in temperature

Substituting,

[tex]Q_4 = (150 g)(1 cal/gC)(25 C)=3,750 cal[/tex]

3. 9200 cal

First of all, we need to find the amount of heat needed to raise the temperature of the ice from -20°C to 0°C. As at point 1., this is given by

[tex]Q_1 = m C_i \Delta T[/tex]

where

m = 80 g is the mass of the ice

C_i = 0.5 cal/g·C° is the specific heat capacity of the ice

[tex]\Delta T=0 C-(-20 C)=20^{\circ}C[/tex] is the change in temperature of the ice

Substituting,

[tex]Q_1 = (80 g)(0.5 cal/gC)(20 C)=800 cal[/tex]

Now we have to find the amount of heat needed to melt the ice:

[tex]Q_2 = m \lambda_f[/tex]

where

m = 80 g is the mass of the ice

[tex]\lambda_f = 80 cal/g[/tex] is the latent heat of fusion

Substituting,

[tex]Q_2 = (80 g)(80 cal/g)=6,400 cal[/tex]

Finally, the amount of heat required to heat the water to 25°C is

[tex]Q_3 = m C_w \Delta T[/tex]

where

m = 80 g is the mass of water

C_w = 1 cal/g·C is the speficic heat capacity of water

[tex]\Delta T=25 C-0 C=25^{\circ}C[/tex] is the change in temperature

Substituting,

[tex]Q_3 = (80 g)(1 cal/gC)(25 C)=2,000 cal[/tex]

So the total heat required is

[tex]Q=Q_1+Q_2+Q_3=800 cal+6,400 cal+2,000 cal=9,200 cal[/tex]

4. No

Explanation:

The total heat required for this process consists of 3 different amounts of heat:

1- The heat required to bring the ice at melting temperature

2- The heat required to melt the ice, while its temperature stays constant

3- The heat required to raise the temperature of the water

However, computing how much heat is required to melt the ice and adding the amount of heat required to raise the temperature of 80 g of water by 45°C is not equivalent: in fact, the calculation of point 1) requires to use the specific heat capacity of ice, not that of water, therefore the two are not equivalent.

The rate of change of atmospheric pressure with respect to altitude is proportional to the current pressure. Using this information, we can calculate the pressure at different altitudes.

To solve this problem, we can use the fact that the rate of change of atmospheric pressure with respect to altitude is proportional to the current pressure. We can set up a proportion using the given information to find the constant of proportionality. Then, we can use this constant to find the pressure at different altitudes.

(a) Let's use the given information to find the constant of proportionality. We have P = kP, where k is the constant of proportionality. Using the values at sea level and 1000m, we can set up the proportion 102.1/87.8 = k. Solving for k, we find k ≈ 1.16.

Now, we can use this constant to find the pressure at an altitude of 4500m. We set up the proportion 102.1/x = 1.16, where x is the pressure at 4500m. Solving for x, we find x ≈ 122.0 kPa.

(b) We can use the same constant of proportionality to find the pressure at the top of a mountain that is 6165m high. We set up the proportion 102.1/x = 1.16, where x is the pressure at the top of the mountain. Solving for x, we find x ≈ 89.2 kPa.

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Answer:

Explanation:

The energy stored in the spring = the kinetic energy at the bottom of the incline

1/2 kx² = 1/2 mv²

kx² = mv²

(500 N/m) (0.75 m)² = (10 kg) v²

v ≈ 5.3 m/s

The energy stored in the spring = the initial potential energy - work done by friction

1/2 kx² = mgh - W

1/2 (500 N/m) (0.75 m)² = (10 kg) (9.8 m/s²) (2.0 m) - W

W ≈ 55 J

Since the horizontal surface is frictionless, the object will have the same speed at the bottom of the incline as before.

v ≈ 5.3 m/s

Answer:

In this problem we need to use the conservation of energy theorem, which is about to constant exchange between kinetic and potential energy. In this case, we have two important points to analyse. The first one is at the initial point, where the mass is just released at 2 m from the ground. The final point would be when the mass hits the massless spring.

So, in the initial point the total energy of the mass would be only potential energy due to its position and its lack of speed. Then, the mass will be released gaining speed, until reaches the spring, where all the kinetic energy is transformed in potential energy but due to elastic forces (spring). So, in the final point, both energy are the same, the kinetic and the elastic potential energy, which we express like this:

[tex]K_{f}=U_{f}[/tex]

But, the definition of each one is:

[tex]K=\frac{1}{2}mv^{2}  \\U=\frac{1}{2}kx^{2}[/tex]

Where k is the constant of the spring. So, we replace this in the initial equation and solve for v:

[tex]\frac{1}{2}mv_{f} ^{2}=\frac{1}{2}kx^{2}\\v_{f}=\sqrt{\frac{kx^{2}}{m}}\\v_{f}=\sqrt{\frac{500(0.75)^{2} }{10}}=5.30m/s[/tex]

Therefore, the speed in the spring point is 5.30 m/s.

Now, to calculate the work due to friction, we have to analyse first, because the problem doesn't offer a friction coefficient, but, we can analyse according to energies again, because work and energy are magnitudes close related, and they have the same dimension.

So, if we thought about it, we'll find that the work due to friction is what slows down the mass, subtracting its potential energy from the initial point, giving as a result the final potential energy:

[tex]U_{f}=U_{i}-W_{k}[/tex]

[tex]\frac{1}{2}kx^{2}  =mgy-W_{k} \\W_{k}=10(9.8)(2)-\frac{1}{2}(500)(0.75)^{2}\\W_{k}=196-140.625=55.38J[/tex]

Therefore, the word due to friction is 55.38 J.

Lastly, the option c is asking about the speed of the object when it reaches the vase of the incline, which is the horizontal ground. The problem specify that this portion of the trajectory is frictionless, which mean that nothing slows down the mass.

Therefore, the speed in the horizontal part is the same 5.30 m/s.

a. 37500;  299,792,457.9 m/s

The formula for the length contraction for a particle moving close to the speed of light is:

[tex]L' = \frac{L}{\gamma}[/tex]

where

L' is the length observed by the particle moving

L is the length observed by an observer at rest

In this problem, we have

[tex]L = 3 km = 3000 m[/tex] is the length of the SLAC measured by an observer at rest

[tex]L' = 8cm = 0.08 m[/tex] is the length measured by the electrons moving

Substituting into the formula, we find the gamma factor

[tex]\gamma = \frac{L}{L'}=\frac{3000 m}{0.08 m}=37,500[/tex]

The formula for the gamma factor is

[tex]\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

where v is the electron's speed and c is the speed of light. Re-arranging the equation, we find v:

[tex]1-\frac{v^2}{c^2}=\frac{1}{\gamma^2}\\v=c \sqrt{ 1-\frac{1}{\gamma^2}}=(299 792 458 m/s)\sqrt{1-\frac{1}{(37500)^2}}=299,792,457.9 m/s[/tex]

b. [tex]1.0\cdot 10^{-5}s[/tex]

For an observer at rest in the laboratory, the electron is moving at a speed of

v = 299,792,457.9 m/s

and it covers a total distance of

L = 3000 m

which is the length of the SLAC measured by the observer. Therefore, the time it takes for the electron to travel down the tube is

[tex]t=\frac{L}{v}=\frac{3000 m}{299,792,457.9 m/s}=1.0\cdot 10^{-5}s[/tex]

c. [tex]2.67\cdot 10^{-10}s[/tex]

From the electron's point of view, the length of the SLAC is actually contracted, so the electron "sees" a total distance to cover of

[tex]L' = 0.08 m[/tex]

And this means that the total time of travel of the electron, in its frame of reference will be shorter; in particular it is given by the formula:

[tex]t' = \frac{t}{\gamma}[/tex]

where

[tex]t=1.0\cdot 10^{-5}s[/tex] is the time measured by the observer at rest

[tex]\gamma=37500[/tex]

Substituting,

[tex]t' = \frac{1.0\cdot 10^{-5}s}{37500}=2.67\cdot 10^{-10}s[/tex]

Answer:

magnitude: 21.6; direction: 33.7 degrees

Explanation:

When we multiply a vector by a scalar, we have to multiply each component of the vector by the scalar number. In this case, we have

vector: (-3,-2)

Scalar: -6

so the vector multiplied by the scalar will have components

[tex](-3\cdot (-6), -2 \cdot (-6))=(18,12)[/tex]

The magnitude is given by Pythagorean's theorem:

[tex]m=\sqrt{18^2+12^2}=21.6[/tex]

and the direction is given by the arctan of the ratio between the y-component and the x-component:

[tex]\theta = tan^{-1} (\frac{12}{18})=33.7^{\circ}[/tex]

The average speed and kinetic energy of gas molecules decrease with decreasing temperature, and there are gas molecules that move slower than the average.

The correct statements are:

The average speed of gas molecules decreases with decreasing temperature. This is because at lower temperatures, the molecules have less kinetic energy and move slower.

The average kinetic energy of gas molecules decreases with decreasing temperature. As temperature decreases, the average kinetic energy of the gas molecules also decreases.

There are gas molecules that move slower than the average. In a sample of gas, there will always be molecules that have speeds below the average.

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Several statements about gas molecules and temperature are true, including the decrease in average speed and kinetic energy with decreasing temperature. Molecular speeds and kinetic energies vary, ensuring not all molecules have the same energy and some move slower than average. The kinetic energy of a molecule can determine its speed.

Let's analyze each statement one by one to determine which are true:

Answer:

kinetic energy from wind turns blades of wind turbines → wind turbine blades turn generator → generator changes kinetic energy to electricity

Answer:

kinetic energy from wind turns blades of wind turbines → wind turbine blades turn generator → generator changes kinetic energy to electricity

Explanation:

Wind energy is the energy harnessed from the velocity of wind. It is obtained with help of wind mills that have large fan type structure. The fan type structure has blades fixed on frame. When the high velocity wind blows, it cause the fan blades to rotate. The rotating fan causes the generator to turn. This rotating generator produces electricity.  Now according to the explanation second option is the correct answer that is kinetic energy from wind turns blades of wind turbines → wind turbine blades turn generator → generator changes kinetic energy to electricity

Answer:

no

Explanation:

because will the cool air will be leaving the fridge the motor in the bottom will be running and eventually the mixture of the cool and hot air will balance out

'a' is the correct statement.

I don’t know what the answer is I Wish I could help

(a) [tex]3.1\cdot 10^7 J[/tex]

The total mechanical energy of the space probe must be constant, so we can write:

[tex]E_i = E_f\\K_i + U_i = K_f + U_f[/tex] (1)

where

[tex]K_i[/tex] is the kinetic energy at the surface, when the probe is launched

[tex]U_i[/tex] is the gravitational potential energy at the surface

[tex]K_f[/tex] is the final kinetic energy of the probe

[tex]U_i[/tex] is the final gravitational potential energy

Here we have

[tex]K_i = 5.0 \cdot 10^7 J[/tex]

at the surface, [tex]R=3.3\cdot 10^6 m[/tex] (radius of the planet), [tex]M=5.3\cdot 10^{23}kg[/tex] (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is

[tex]U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J[/tex]

At the final point, the distance of the probe from the centre of Zero is

[tex]r=4.0\cdot 10^6 m[/tex]

so the final potential energy is

[tex]U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J[/tex]

So now we can use eq.(1) to find the final kinetic energy:

[tex]K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J[/tex]

(b) [tex]6.3\cdot 10^7 J[/tex]

The probe reaches a maximum distance of

[tex]r=8.0\cdot 10^6 m[/tex]

which means that at that point, the kinetic energy is zero: (the probe speed has become zero):

[tex]K_f = 0[/tex]

At that point, the gravitational potential energy is

[tex]U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J[/tex]

So now we can use eq.(1) to find the initial kinetic energy:

[tex]K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J[/tex]

Answer:

Move further apart

Explanation:

The negative charge on the rod repels the electrons on the cap thus they concentrate at the leaves causing an increase in the amount of charge at the leaves. Since like charges repel, the two leaves of the electroscope move further apart.

Final answer:

When a negatively-charged rod is brought near a negatively-charged electroscope, the leaves of the electroscope will move farther apart due to increased electrostatic repulsion.

Explanation:

When a negatively-charged rod is brought close to an already negatively-charged electroscope, the electrostatic repulsion between the like charges will cause the two leaves of the electroscope to move farther apart. This is due to an increase in the density of negative charges within the leaves of the electroscope, which causes the leaves to repel each other even more.

Charging by induction occurs when a charged object is brought near, but not in contact with, a conductor. However, since the electroscope in this case is already negatively charged, bringing another negatively charged object close will simply enhance the repulsion between charges.

Answer:

93 m

Explanation:

We need to find the resultant displacement.

Let's take north as positive y-direction and east as positive direction. We have

- Displacement 1 is 51 m to the north, so the two components are

[tex]x_1 =0\\y_1 = 51 m[/tex]

- Displacement 2 is 45 m [tex]60^{\circ}[/tex] north of east, so the two components are

[tex]x_1 = (45)cos 60^{\circ}=22.5 m\\y_1 = (45) sin 60^{\circ}=39.0 m[/tex]

So to find the resultant displacement we have to sum the components along each direction:

[tex]x=x_1 + x_2 = 0+22.5 m = 22.5 m\\y = y_1 + y_2 = 51 m +39.0 m = 90.0 m[/tex]

And the magnitude of the resultant displacement is

[tex]m=\sqrt{x^2+y^2}=\sqrt{(22.5 m)^2+(90.0 m)^2}=92.8 m \sim 93 m[/tex]

Answer:

[tex]42.1^{\circ}C[/tex]

Explanation:

The heat extracted from the aluminium is given by:

[tex]Q=mC_s (T_f-T_i)[/tex]

where we have

Q = -1050 J is the heat extracted

m = 65.0 g = 0.065 kg is the mass

Cs = 900 J/kgC is the specific heat of aluminum

[tex]T_i = 60.0^{\circ}[/tex] is the initial temperature

By solving for [tex]T_f[/tex], we find the final temperature:

[tex]T_f = \frac{Q}{m C_s}+T_i=\frac{-1050 J}{(0.065 kg)(900 J/kg C)}+60.0^{\circ}C=42.1^{\circ}C[/tex]

To find the final temperature, you use the formula for heat transfer, Q = mcΔT, and specific heat value. Given that 1050J of heat is extracted, which is negative heat transfer (-1050 J), plug in the values to get ΔT and then add this to the initial temperature to get the final temperature.

The subject of the question is the concept of heat transfer and specific heat in Physics. The specific heat of a substance is the amount of heat required to raise one gram of the substance by one degree Celsius, which is a property intrinsic to the substance. Different substances require different amounts of heat to change their temperature.

To find the final temperature, you use the formula for heat transfer, Q = mcΔT, where Q is heat transferred, m is mass, c is the specific heat, and ΔT is the change in temperature (final temperature - initial temperature). Here, we're asked to find the final temperature of a 65 g (or 0.065 kg) mass of aluminum initially at 60.0°C when 1050 J of heat energy is extracted.

The sign of Q needs to represent that heat energy is extracted from the aluminum, so Q is -1050 J. Now, you can plug in the values and solve the equation for ΔT first, then add the initial temperature to find the final temperature.

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1. [tex]1.69\cdot 10^{11} m[/tex]

The Schwarzschild radius of an object of mass M is given by:

[tex]r_s = \frac{2GM}{c^2}[/tex] (1)

where

G is the gravitational constant

M is the mass of the object

c is the speed of light

The black hole in the problem has a mass of

[tex]M=5.7\cdot 10^7 M_s[/tex]

where

[tex]M_s = 2.0\cdot 10^{30} kg[/tex] is the solar mass. Substituting,

[tex]M=(5.7\cdot 10^7)(2\cdot 10^{30}kg)=1.14\cdot 10^{38} kg[/tex]

and substituting into eq.(1), we find the Schwarzschild radius of this black hole:

[tex]r_s = \frac{2(6.67\cdot 10^{-11})(1.14\cdot 10^{38} kg)}{(3\cdot 10^8 m/s)^2}=1.69\cdot 10^{11} m[/tex]

2) 242.8 solar radii

We are asked to find the radius of the black hole in units of the solar radius.

The solar radius is

[tex]r_S = 6.96\cdot 10^5 km = 6.96\cdot 10^8 m[/tex]

Therefore, the Schwarzschild radius  of the black hole in solar radius units is

[tex]r=\frac{1.69\cdot 10^{11} m}{6.96\cdot 10^8 m}=242.8[/tex]

Final answer:

In the classical model of the hydrogen atom, the velocity of the electron as it orbits around the nucleus can be deduced using the equation for the centripetal force. The magnitude of the net force towards the center, Fc, is equal to mv^2/r, and solving for v gives the velocity of the electron.

Explanation:

In the classical model of the hydrogen atom, the electron orbits the proton in a circular path. The magnitude of the net force towards the center, also known as the centripetal force, is equal to the mass of the electron times its velocity squared divided by the radius of the orbit (Fc = mv^2 / r). In this case, we have the mass of the electron, the radius of the orbit, and we need to find the velocity of the electron as it orbits around the nucleus.

Using the given equation of the centripetal force, we can rearrange it to solve for the velocity (v). So, v = sqrt(Fc * r / m). Plugging in the values for the mass of the electron, the radius of the orbit, and the known value of the centripetal force, we can calculate the velocity.

For example, if the radius of the orbit is 5.28 x 10^-11 m and the mass of the electron is 9.11 x 10^-31 kg, the velocity of the electron would be v = sqrt(Fc * r / m) = sqrt(m * v^2 / r * r / m) = v = sqrt(v^2) = v = 2.18 x 10^6 m/s.

Answer:

6.214g/cm³

Explanation:

The question is on density of a material

Density=mass/volume

Given, mass=87grams   and volume= 14 cm³  density=?

Density=m/v 87/14 =6.214g/cm³

Answer:

22.10 kg

Explanation:

The worker wants to recast the model from iron to gold, so the volume of the model made of iron and made of gold will be the same:

[tex]V_i = V_g[/tex]

the volume of each model can be rewritten as the ratio between the mass of the model, m, and the density of the material, d:

[tex]V=\frac{m}{d}[/tex]

so we can rewrite the first equation as

[tex]\frac{m_i}{d_i}=\frac{m_g}{d_g}[/tex]

where we have

[tex]m_i = 9.00 kg[/tex] is the mass of the model in iron

[tex]d_i = 7.86\cdot 10^3 kg/m^3[/tex] is the density of iron

[tex]m_g[/tex] is the mass of gold needed

[tex]d_g = 19.30\cdot 10^3 kg/m^3[/tex] is the density of gold

Solving for [tex]m_g[/tex], we find

[tex]m_g = d_g \frac{m_i}{d_i}=(19.30\cdot 10^3 kg/m^3) \frac{9.00 kg}{7.86\cdot 10^3 kg/m^3}=22.10 kg[/tex]

The force of gravity on earth is 9.807, therefore you multiply earth’s gravitational force by the 5.7 kilograms which will give you 5.7 x 9.807=55.86 Newtons

(a) 50 rad/s

The angular speed of the CD is related to the linear speed by:

[tex]\omega=\frac{v}{r}[/tex]

where

[tex]\omega[/tex] is the angular speed

v is the linear speed

r is the distance from the centre of the CD

When scanning the innermost part of the track, we have

v = 1.25 m/s

r = 25.0 mm = 0.025 m

Therefore, the angular speed is

[tex]\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s[/tex]

(b) 21.6 rad/s

As in part a, the angular speed of the CD is given by

[tex]\omega=\frac{v}{r}[/tex]

When scanning the outermost part of the track, we have

v = 1.25 m/s

r = 58.0 mm = 0.058 m

Therefore, the angular speed is

[tex]\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s[/tex]

(c) 5550 m

The maximum playing time of the CD is

[tex]t =74.0 min \cdot 60 s/min = 4,440 s[/tex]

And we know that the linear speed of the track is

v = 1.25 m/s

If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:

[tex]d=vt=(1.25 m/s)(4,440 s)=5,550 m[/tex]

(d) [tex]-6.4\cdot 10^{-3} rad/s^2[/tex]

The angular acceleration of the CD is given by

[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]

where

[tex]\omega_f = 21.6 rad/s[/tex] is the final angular speed (when the CD is scanned at the outermost part)

[tex]\omega_i = 50.0 rad/s[/tex] is the initial angular speed (when the CD is scanned at the innermost part)

[tex]t=4440 s[/tex] is the time elapsed

Substituting into the equation, we find

[tex]\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2[/tex]

Answer:  (a) 50 rad/s, (b) 21.6 rad/s, (c) 5550 m, (d) [tex]-6.4\cdot 10^{-3} rad/s^2[/tex]

A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outward toward the rim of the disc; the inner and outer radii of this spiral are 25.0 mm and 58.0 mm, respectively. As the disc spins inside a CD player, the track is scanned at a constant linear speed of 1.25 m/s

Part

A What is the angular speed of the CD when scanning the innermost part of the track?

B What is the angular speed of the CD when scanning the outermost part of the track?

C The maximum playing time of a CD is 74.0 min. What would be the length of the track on such a maximum-duration CD if it were stretched out in a straight line?

D What is the average angular acceleration of a maximum-duration CD during its 74.0-min playing time?

(a) 50 rad/s

[tex]\omega=\frac{v}{r}[/tex]

where

[tex]\omega[/tex] is the angular speed

v is the linear speed,  v = 1.25 m/s

r is the distance from the centre of the CD, r = 25.0 mm = 0.025 m

Therefore, the angular speed

[tex]\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s[/tex]

(b) 21.6 rad/s

The angular speed of the CD is

[tex]\omega=\frac{v}{r}[/tex]

When scanning the outermost part of the track

v = 1.25 m/s

r = 58.0 mm = 0.058 m

Therefore, the angular speed is

[tex]\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s[/tex]

(c) 5550 m

[tex]t =74.0 min \cdot 60 s/min = 4,440 s[/tex]

the linear speed of the track is  v = 1.25 m/s

the total length of the track would be:

d=vt=(1.25 m/s)(4,440 s)=5,550 m

(d) [tex]-6.4\cdot 10^{-3} rad/s^2[/tex]

The angular acceleration of the CD is given by[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]

where

[tex]\omega_f = 21.6 rad/s[/tex] is the final angular speed (when the CD is scanned at the outermost part)

[tex]\omega_i = 50.0 rad/s[/tex] is the initial angular speed (when the CD is scanned at the innermost part)

[tex]t=4440 s[/tex] is the time elapsed

Substituting into the equation, we find

[tex]\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2[/tex]

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(a) [tex]E(r) = \frac{q}{4\pi \epsilon_0 r^2}[/tex]

We can solve the different part of the problem by using Gauss theorem.

Considering a Gaussian spherical surface with radius r<a (inside the shell), we can write:

[tex]E(r) \cdot 4\pi r^2 = \frac{q}{\epsilon_0}[/tex]

where q is the charge contained in the spherical surface, so

[tex]q=5.00 C[/tex]

Solving for E(r), we find the expression of the field for r<a:

[tex]E(r) = \frac{q}{4\pi \epsilon_0 r^2}[/tex]

(b) 0

The electric field strength in the region a < r < b is zero. This is due to the fact that the charge +q placed at the center of the shell induces an opposite charge -q on the inner surface of the shell (r=a), while the outer surface of the shell (r=b) will acquire a net charge of +q.

So, if we use Gauss theorem for the region  a < r < b, we get

[tex]E(r) \cdot 4\pi r^2 = \frac{q'}{\epsilon_0}[/tex]

however, the charge q' contained in the Gaussian sphere of radius r is now the sum of the charge at the centre (+q) and the charge induced on the inner surface of the shell (-q), so

q' = + q - q = 0

And so we find

E(r) = 0

(c) [tex]E(r) = \frac{q}{4\pi \epsilon_0 r^2}[/tex]

We can use again Gauss theorem:

[tex]E(r) \cdot 4\pi r^2 = \frac{q'}{\epsilon_0}[/tex] (1)

where this time r > b (outside the shell), so the gaussian surface this time contained:

- the charge +q at the centre

- the inner surface, with a charge of -q

- the outer surface, with a charge of +q

So the net charge is

q' = +q -q +q = +q

And so solving (1) we find

[tex]E(r) = \frac{q}{4\pi \epsilon_0 r^2}[/tex]

which is identical to the expression of the field inside the shell.

(d) [tex]-12.3 C/m^2[/tex]

We said that at r = a, a charge of -q is induced. The induced charge density will be

[tex]\sigma_a = \frac{-q}{4\pi a^2}[/tex]

where [tex]4 \pi a^2[/tex] is the area of the inner surface of the shell. Substituting

q = 5.00 C

a = 0.18 m

We find the induced charge density:

[tex]\sigma_a = \frac{-5.00 C}{4\pi (0.18 m)^2}=-12.3 C/m^2[/tex]

(e) [tex]-1.9 C/m^2[/tex]

We said that at r = b, a charge of +q is induced. The induced charge density will be

[tex]\sigma_b = \frac{+q}{4\pi b^2}[/tex]

where [tex]4 \pi b^2[/tex] is the area of the outer surface of the shell. Substituting

q = 5.00 C

b = 0.46 m

We find the induced charge density:

[tex]\sigma_b = \frac{+5.00 C}{4\pi (0.46 m)^2}=-1.9 C/m^2[/tex]

(a) [tex]2.7\cdot 10^{25} kg[/tex]

The acceleration due to gravity on the surface of the planet is given by

[tex]g=\frac{GM}{R^2}[/tex] (1)

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

Here we know:

[tex]g=22.4 m/s^2[/tex]

[tex]d=1.8\cdot 10^7 m[/tex] is the diameter, so the radius is

[tex]R=\frac{d}{2}=\frac{1.8\cdot 10^7 m}{2}=9\cdot 10^6 m[/tex]

So we can re-arrange eq.(1) to find M, the mass of the planet:

[tex]M=\frac{gR^2}{G}=\frac{(22.4 m/s^2)(9\cdot 10^6 m)^2}{6.67\cdot 10^{-11}}=2.7\cdot 10^{25} kg[/tex]

(b) [tex]4.8\cdot 10^{31}kg[/tex]

The planet is orbiting the star, so the centripetal force is equal to the gravitational attraction between the planet and the star:

[tex]m\frac{v^2}{r}=\frac{GMm}{r^2}[/tex] (1)

where

m is the mass of the planet

M is the mass of the star

v is the orbital speed of the planet

r is the radius of the orbit

The orbital speed is equal to the ratio between the circumference of the orbit and the period, T:

[tex]v=\frac{2\pi r}{T}[/tex]

where

[tex]T=402 days = 3.47\cdot 10^7 s[/tex]

Substituting into (1) and re-arranging the equation

[tex]m\frac{4\pi r^2}{rT^2}=\frac{GMm}{r^2}\\\frac{4\pi r}{T^2}=\frac{GM}{r^2}\\M=\frac{4\pi r^3}{GMT^2}[/tex]

And substituting the numbers, we find the mass of the star:

[tex]M=\frac{4\pi^2 (4.6\cdot 10^{11} m)^3}{(6.67\cdot 10^{-11})(3.47\cdot 10^7 s)^2}=4.8\cdot 10^{31}kg[/tex]

Answer:

[tex]2.54\cdot 10^{-34}m[/tex]

Explanation:

First of all, we need to find the final velocity of the ball just before it reaches the ground. Since the ball is in free-fall motion, its final velocity is given by

[tex]v^2 = u^2 +2gh[/tex]

where

v is the final velocity

u = 0 is the initial velocity

g = 9.8 m/s^2 is the acceleration due to gravity

h = 39.8 m is the height of the building

Solving for v,

[tex]v=\sqrt{2gh}=\sqrt{2(9.8 m/s^2)(39.8 m)}=27.9 m/s[/tex]

Now we can calculate the ball's momentum:

[tex]p=mv=(0.0935 kg)(27.9 m/s)=2.61 kg m/s[/tex]

And now we can calculate the De Broglie's wavelength of the ball:

[tex]\lambda = \frac{h}{p}=\frac{6.63\cdot 10^{-34} Js}{2.61 kg m/s}=2.54\cdot 10^{-34}m[/tex]

a) See part a of the attached picture.

There are two forces involved here:

- The weight of the block attached to the spring, of magnitude

W = mg

where m is the mass of the block and g = 9.8 m/s^2 is the acceleration due to gravity, pointing downward

- The restoring force of the spring,

F = kx

with k being the spring constant and x the displacement of the spring with respect to the equilibrium position, pointing upward

b) [tex]L' = L+\frac{mg}{k}[/tex]

The spring is in equilibrium when the restoring force of the spring is equal to the weigth of the block:

[tex]kx = mg[/tex]

where

x is the displacement of the spring with respect to the natural length of the spring, L. Solving for x,

[tex]x=\frac{mg}{k}[/tex]

And since the natural length is L, the equilibrium length of the spring is

[tex]L' = L+x=L+\frac{mg}{k}[/tex]

c) [tex]y(t) = y cos(\sqrt{\frac{k}{m}} t + \pi)[/tex]

Let's assume that

y = 0

corresponds to the equilibrium position of the spring (which is stretched by an amount [tex]L+\frac{mg}{k}[/tex]). If doing so, the vertical position of the mass at time t is given by

[tex]y(t) = y cos(\omega t + \pi)[/tex]

where

[tex]\omega = \sqrt{\frac{k}{m}}[/tex] is the angular frequency

t is the time

y is the initial displacement with respect to the equilibrium position

[tex]\pi[/tex] is the phase shift, so that the position at time t=0 is negative:

y(0) = -y

So rewriting the angular frequency:

[tex]y(t) = y cos(\sqrt{\frac{k}{m}} t + \pi)[/tex]

d) [tex]T=2 \pi \sqrt{\frac{m}{k}}[/tex]

The period of oscillation is given by

[tex]T=\frac{2\pi}{\omega}[/tex]

where

[tex]\omega = \sqrt{\frac{k}{m}}[/tex] is the angular frequency

Substituting [tex]\omega[/tex], we find an expression for the period

[tex]T=2 \pi \sqrt{\frac{m}{k}}[/tex]

e) k' = 2k (see part e of attached picture)

Here the two half springs of spring constant k' are connected in series, so the sum of the stretchings of the two springs is equal to the total stretching of the spring, x:

[tex]x=x_1 + x_2[/tex]

Also, since the two springs are identical, their stretching will be the same:

[tex]x_1 = x_2 = x'[/tex]

so we have

[tex]x=x'+x'=2x'[/tex]

[tex]x'=\frac{x}{2}[/tex]

Substituting x find in part (b),

[tex]x'=\frac{mg}{2k}[/tex] (1)

Hooke's law for each spring can be written as

[tex]F=k' x'[/tex] (2)

where

F = mg (3)

is still the weight of the block

Using (1), (2) and (3) together, we find an expression for the spring constant k' of each spring

[tex]mg=k'\frac{mg}{2k}\\k' =2k[/tex]

So, the spring constant of each half-spring is twice the spring constant of the original spring.

fa) See part f) of attached picture

This time we have the block hanging from both the two half-springs, each with spring constant k'. So at equilibrium, the weight of the block is equal to the sum of the restoring forces of the two springs:

[tex]k' x_1 + k' x_2 = mg[/tex]

fb) [tex]x'=\frac{L}{2}+\frac{mg}{2k'}[/tex]

For the two springs in parallel, the sum of the restoring forces of the two springs must be equal to the weight of the block:

[tex]F_1 + F_2 = mg\\k_1 x_1 + k_2 x_2 = mg[/tex]

The two springs are identical, so they have same spring constant:

[tex]k_1 = k_2 = k'[/tex]

So (1) can be rewritten as

[tex]k' (x_1 + x_2) = mg[/tex]

And since the two springs are identical, their stretching x' is the same:

[tex]x_1 = x_2 = x'[/tex]

so we can rewrite this as

[tex]k' (2x') = mg[/tex]

[tex]x'=\frac{mg}{2k'}[/tex]

and so the equilibrium length of each spring will be

[tex]x'=\frac{L}{2}+\frac{mg}{2k'}[/tex]

fc) [tex]y(t) = y cos(\sqrt{\frac{2k'}{m}} t + \pi)[/tex]

The system of two springs in parallel can be treated as a system of a single spring with equivalent spring constant given by

[tex]k_{eq}=k_1 + k_2 = 2k'[/tex]

where k' is the spring constant of each spring.

So, let's assume again that

y = 0

corresponds to the equilibrium position as calculated in the previous part. If doing so, the vertical position of the mass at time t is given by:

[tex]y(t) = y cos(\omega t + \pi)[/tex]

where this time we have

[tex]\omega = \sqrt{\frac{k_{eq}}{m}}[/tex] is the angular frequency of the system

t is the time

y is the initial displacement with respect to the equilibrium position

[tex]\pi[/tex] is the phase shift, which we put so that the position at time t=0 is negative:

y(0) = -y

If we rewrite the angular frequency,

[tex]\omega = \sqrt{\frac{2k'}{m}}[/tex]

the position of the mass is

[tex]y(t) = y cos(\sqrt{\frac{2k'}{m}} t + \pi)[/tex]

fd) [tex]T=2 \pi \sqrt{\frac{m}{2k'}}[/tex]

Similarly to part d), the period of oscillation is

[tex]T=\frac{2\pi}{\omega}[/tex]

where

[tex]\omega = \sqrt{\frac{k_{eq}}{m}}=\sqrt{\frac{2k'}{m}}[/tex] is the angular frequency

Substituting [tex]\omega[/tex], we find

[tex]T=2 \pi \sqrt{\frac{m}{2k'}}[/tex]

(a) 5.35 s, 84.65 s

The distance covered by the rocket during the first part of the motion (accelerated motion) is

[tex]d_1 = \frac{1}{2}at_1 ^2[/tex] (1)

where

a = +35 ft/s^2 is the acceleration

While the distance covered during the second part of the motion (uniform motion) is

[tex]d_2 = v t_2[/tex]

where v is the constant speed reached after the first part of the motion, which is given by

[tex]v=at_1[/tex]

So we can rewrite the equation as

[tex]d_2 = a t_1 t_2[/tex] (2)

We also know:

[tex]d= d_1 + d_2 = 16350[/tex] (3) is the total distance

[tex]t= t_1 +t_2 = 90[/tex] (4) is the total time

We can rewrite (3) as

[tex]\frac{1}{2}at_1^2 + a t_1 t_2 = 16350[/tex] (5)

And (4) as

[tex]t_2 = 90-t_1[/tex]

Substituting into (5), we get

[tex]\frac{1}{2}at_1^2 + a t_1 (90-t_1) = 16350\\0.5 at_1^2 + 90at_1 - at_1^2 = 16350\\-0.5at_1^2 +90at_1 = 16350[/tex]

Substituting a=+35 ft/s^2, we have

[tex]-17.5 t_1^2 +3150 t_1 - 16350 = 0[/tex]

which has two solutions:

[tex]t_1 = 174.65 s[/tex] --> discarded, since the total time cannot be greater than 90 s

[tex]t_1 = 5.35 s[/tex] --> this is the correct solution

And therefore [tex]t_2[/tex] is

[tex]t_2 = 90-t_1 = 90 -5.35 s=84.65 s[/tex]

(b) 187.25 ft/s

The velocity v is the velocity at the end of the accelerated motion of the rocket, so after

[tex]t_1 = 5.35 s[/tex]

The acceleration is

a = +35 ft/s^2

Therefore, the velocity after the first part is:

[tex]v=at_1 = (+35 ft/s^2)(5.35 s)=187.25 ft/s[/tex]

(c) 16,585 ft

At the 15750 ft mark, the velocity of the sled is

[tex]v=187.25 ft/s[/tex]

Then, it starts decelerating with acceleration

a = -21 ft/s^2

So, its distance covered during this part of the motion will be given by:

[tex]v_f ^ 2 -v^2 = 2ad_3[/tex]

where

vf = 0 is the final speed

v = 187.25 ft/s is the initial speed

a = -21 ft/s^2 is the acceleration

d3 is the distance covered during this part

Solving for d3,

[tex]d_3 = \frac{v_f^2 -v^2}{2a}=\frac{0-(187.25 ft/s)^2}{2(-21 ft/s^2)}=835 ft[/tex]

So, the final position of the sled will be

[tex]x_3 = 15750 ft + 835 ft =16,585 ft[/tex]

(d) 93.92 s

The duration of the first part of the motion is

[tex]t_1 = 5.35 s[/tex]

The distance covered during this part is

[tex]d_1 = \frac{1}{2}at_1^2=\frac{1}{2}(+35 ft/s^2)(5.35 s)^2=501 ft[/tex]

In the second part (uniform motion), the sled continues with constant speed until the 15750 ft mark, so the distance covered is

[tex]d_2 = 15750- 835 =14915[/tex]

And so the duration of the second part is

[tex]t_2 =\frac{d_2}{v}=\frac{14915}{187.25}=79.65 s[/tex]

While the duration of the third part (decelerated motion) is

[tex]a=\frac{v_f-v}{t_3}\\t_3 = \frac{v_f-f}{a}=\frac{0-(187.25 ft/s)}{-21 ft/s^2}=8.92 s[/tex]

So the total duration is

t = 5.35 s+79.65 s+8.92 s=93.92 s

This physics problem involving kinematics can be solved returning to the kinematic equations and applying them to the scenarios described. Acceleration, constant velocity and deceleration periods of motion each have specific times and these can be used to derive the specifics of the journey such as times, velocity and final position.

The subject of this question is Physics, specifically, it is related to the concept of Kinematics which falls under Classical Mechanics.

In the first scenario, the sled accelerates from rest under the power of the rocket engine. We can use the kinematic equation: d = (1/2)at2 to solve for time t1, where a is the acceleration and d is the distance. The distance travelled under the acceleration is total distance minus the distance it goes with constant speed.

After finding t1, the time the sled moves with constant speed t2 is total time minus t1.

The constant speed v can be found by the distance traveled under constant speed divided by t2. When the sled hits the 15750ft mark, it starts to decelerate at -21 ft/s2. We can use the equation v2 = u2 + 2ad to get the distance it travels before it comes to rest.

Add this distance to 15750ft to get the final position of the sled when it comes to rest. Since we have the acceleration and final velocity is zero, one may get the time of this stage by using equation v = u + at . And the total duration is the sum of all time segments.

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Answer:

Explanation:

The rate of conductive heat transfer in watts is:

q = (k/s) A ΔT

where k is the heat conductivity, s is the thickness, A is the area, and ΔT is the temperature difference.

a)

Given k = 52.4 W/m/K, s = 0.014 m, and ΔT = 350-140 = 210 K, we can find q/A:

q/A = (52.4 / 0.014) (210)

q/A = 786,000 W/m²

b)

Given that A = 0.42 m², we can find q:

q = (0.42 m²) (786,000 W/m²)

q = 330,120 W

A watt is a Joule per second.  Convert to Joules per hour:

q = 330,120 J/s * 3600 s/hr

q = 1.19×10⁹ J/hr

c)

If we change k to 1.8 W/m/K:

q = (k/s) A ΔT

q = (1.8 / 0.014) (0.42) (210)

q = 11,340 J/s

q = 4.08×10⁷ J/hr

d)

If k is 52.4 W/m/K and s is 0.024 m:

q = (k/s) A ΔT

q = (52.4 / 0.024) (0.42) (210)

q = 192,570 J/s

q = 6.93×10⁸ J/hr

The heat flux is 785,600 W/m^2, and the heat loss per hour is 1,110,912,000 J/h for the first metal. Using a material with a thermal conductivity of 1.8 W/m-K, the heat loss is 42,504,000 J/h. With increased thickness to 24 mm, the heat loss is 491,820,000 J/h.

Firstly, you would calculate the heat flux through the metal sheet using Fourier's Law of Heat Conduction, which is expressed as:

q = -k * (T_2 - T_1) / d

In this case, the temperature difference (T_2 - T_1) is (350°C - 140°C), the thickness (d) of the metal sheet is 14 mm (which is 0.014 meters), and the thermal conductivity (k) is given as 52.4 W/m-K. Substituting these values in:

q = -52.4 * (350 - 140) / 0.014 = 785,600 W/m^2

For part (b), to calculate the heat loss per hour, we multiply the heat flux by the area of the sheet and convert seconds to hours:

Q/t = q * A = 785,600 * 0.42 m^2 * 3600 seconds = 1,110,912,000 J/h

For part (c), using a material with a thermal conductivity of 1.8 W/m-K, the process is similar, and the heat loss per hour would be:

Q/t = -1.8 * (350 - 140) / 0.014 * 0.42 * 3600 = 42,504,000 J/h

For part (d), with the increased thickness of 24 mm (0.024 meters), the heat loss per hour would be:

Q/t = -52.4 * (350 - 140) / 0.024 * 0.42 * 3600 = 491,820,000 J/h

Answer:

[tex]d=\frac{V}{E}[/tex]

Explanation:

The electric field strength between the plates of a parallel plate capacitor is given by:

[tex]E=\frac{V}{d}[/tex]

where

V is the potential difference

d is the distance between the plates

If we re-arrange the formula, we find a new equation that lets us calculate d:

[tex]d=\frac{V}{E}[/tex]

We can at this point substitute the values:

V = 112 V

E = 1.1 kV/cm = 1100 V/cm = 110,000 V/m

To find the distance:

[tex]d=\frac{112 V}{110,000 V/m}=1.02\cdot 10^{-3} m[/tex]

The distance between capacitor plates, using the formula d = V/E, is approximately 1 cm

To calculate the distance d between the plates of a parallel plate capacitor, we use the relationship between the electric field strength E, the voltage V, and the distance d: E = V/d. Given:

V = 112 V (voltage across the plates)

E = 1.1 kV/cm = 1.1 * 10⁵ V/m (electric field strength)

The formula can be rearranged to solve for d:

d = V/E

Where:

Substituting the given values into the equation will allow us to calculate the distance between the plates.

Given:

Voltage (V) = 112 V

Electric field strength (E) = 1.1 kV/cm = 1.1 × 10⁴ V/m

Distance (d) = ?

Formula: E = V/d

Substitute: 1.1 × 10⁴ = 112/d

Solve for d: d = 112 / (1.1 × 10⁴)

[tex]\approx[/tex] 0.01 m [tex]\approx[/tex] 1 cm

The distance between the capacitor plates is thus approximately 1 cm.

1) Static friction coefficient: 0.355

The crate is initially at rest. The crate remains at rest until the horizontal pushing force is less than the maximum static frictional force.

The maximum static frictional force is given by

[tex]F_s = \mu_s mg[/tex]

where

[tex]\mu_s[/tex] is the static coefficient of friction

m = 21 kg is the mass of the crate

g = 9.8 m/s^2 is the acceleration due to gravity

The horizontal force required to set the crate in motion is 73 N: this means that this is the value of the maximum static frictional force. So we have

[tex]F_s=73 N[/tex]

Using this information into the previous equation, we can find the coefficient of static friction:

[tex]\mu_s = \frac{F}{mg}=\frac{73 N}{(21 kg)(9.8 m/s^2)}=0.355[/tex]

2) Kinetic friction coefficient: 0.267

Now the crate is in motion: this means that the kinetic friction is acting on the crate, and its magnitude is

[tex]F_k = \mu_k mg[/tex] (1)

where

[tex]\mu_k[/tex] is the coefficient of kinetic friction

There is a horizontal force of

F = 55 N

pushing the crate. Moreover, the speed of the crate is constant: this means that the acceleration is zero, a = 0.

So we can write Newton's second law as

[tex]F-F_k = ma = 0[/tex]

And by substituting (1), we can find the value of the coefficient of kinetic friction:

[tex]F-\mu_k mg = 0\\\mu_k = \frac{F}{mg}=\frac{55 N}{(21 kg)(9.8 m/s^2)}=0.267[/tex]

The coefficients of static and kinetic friction between the crate and floor are;

μ_s = 0.3547

μ_k = 0.2672

1) To find the coefficient of static friction, is given by the formula;

F_s = μ_s*mg

We are given;

Mass; m = 21 kg

Horizontal force to set it to motion; F_h = 73 N

Now, in this static friction equation, we are using this force of 73 N because static friction is friction at rest and 73 N is the force to pull the object from rest. Thus;

73 = μ_s(21 × 9.8)

μ_s = 73/(21 × 9.8)

μ_s = 0.3547

2) To find the coefficient of kinetic friction, is given by the formula;

F_k = μ_k*mg

We will make use of the force of 55N because we are dealing with friction associated with motion and 55N is the force that keeps the object in motion. Thus;

μ_k = F_k/(mg)

μ_k = 55/(21 × 9.8)

μ_k = 0.2672

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