PHYSICS COLLEGE

A long copper cylindrical shell of inner radius 5 cm and outer radius 8 cm surrounds concentrically a charged long aluminum rod of radius 1 cm with a charge density of 7 pC/m. All charges on the aluminum rod reside at its surface. The inner surface of the copper shell has exactly opposite charge to that of the aluminum rod while the outer surface of the copper shell has the same charge as the aluminum rod. Find the magnitude and direction of the electric field at points that are at the following distances from the center of the aluminum rod: (a) 0.5 cm,
(b) 1.5 cm,
(c) 2.5 cm,
(d) 3.5 cm,
(e) 7 cm.

Answers

Answer 1

Answer:

Answer:

a. 0

b. 8.4N/C

c. 5.04N/C

d. 3.6 N/C

e. 1.8N/C

Explanation:

The following data are given

inner cylindrical radius,r=5cm

outer cylindrical radius R=8cm

Charge density,p=7pc/m

radius of rod= 1cm

a. at distance 0.5cm from the center of the rod, this point falls on the rod itself and since the charge spread out on the surface of the rod, there wont be any electric field inside the rod itself

Hence E=0 at 0.5cm

b. at 1.5cm i.e 0.015m

the electric field is expressed as

$E=(2*charge density)/(4\pi E_(0)r)\nE=(2*7*10^(-12) *9*10^(9))/(0.015)\nE=8.4N/C$

The direction of the field depends on the charge on the rod

c. at 2.5cm i.e 0.025m

the electric field is expressed as

$E=(2*charge density)/(4\pi E_(0)r)\nE=(2*7*10^(-12) *9*10^(9))/(0.025)\nE=5.04N/C$

The direction of the field depends on the charge on the rod

d. at 3.5cm i.e 0.035m this point is still within the rod and the inner cylinder

the electric field is expressed as

$E=(2*charge density)/(4\pi E_(0)r)\nE=(2*7*10^(-12) *9*10^(9))/(0.035)\nE=3.6N/C$

The direction of the field depends on the charge on the rod

e. at 7cm which is a point outside the rod and the cylinder, the electric field is

$E=(2*charge density)/(4\pi E_(0)r)\nE=(2*7*10^(-12) *9*10^(9))/(0.07)\nE=1.8N/C$

The direction of the field depends on the charge on the rod

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A screw can be considered a type of

Answers

Fastener because a fastener is something that connects to objects and usually can come apart but can also be permanent

A 0.060 ???????? tennis ball, moving with a speed of 5.28 m/???? , has a head-on collision with a 0.080 ???????? ball initially moving in the same direction at a speed of 3.00 m/ ???? . Assume that the collision is perfectly elastic. Determine the velocity (speed and direction) of both the balls after the collision.

Answers

Explanation:

It is given that,

Mass of the tennis ball, $m_1=0.06\ kg$

Initial speed of tennis ball, $u_1=5.28\ m/s$

Mass of ball, $m_2=0.08\ kg$

Initial speed of ball, $u_2=3\ m/s$

In case of elastic collision, the momentum remains conserved. The momentum equation is given by :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$v_1\ and\ v_2$ are final speed of tennis ball and the ball respectively.

$0.06* 5.28+0.08* 3=0.06v_1+0.08v_2$

$0.06v_1+0.08v_2=0.5568$ ..............(1)

We know that the coefficient of restitution is equal to 1. It is given by :

$(v_2-v_1)/(u_1-u_2)=1$

$(v_2-v_1)/(5.28-3)=1$

${v_2-v_1}=2.28$ .................(2)

On solving equation (1) and (2) to find the values of velocities after collision.

$v_1=5.28\ m/s$

$v_2=3\ m/s$

So, the speed of both balls are 5.28 m/s and 3 m/s respectively. Hence, this is the required solution.

5. A 55-kg swimmer is standing on a stationary 210-kg floating raft. The swimmer then runs off the raft horizontally with the velocity of +4.6 m/s relative to the shore. Find the recoil velocity that the raft would have if there were no friction and resistance due to the water.

Answers

Answer:

The recoil velocity of the raft is 1.205 m/s.

Explanation:

given that,

Mass of the swimmer, $m_1=55\ kg$

Mass of the raft, $m_2=210\ kg$

Velocity of the swimmer, v = +4.6 m/s

It is mentioned that the swimmer then runs off the raft, the total linear momentum of the swimmer/raft system is conserved. Let V is the recoil velocity of the raft.

$m_1v+m_2V=0$

$55* 4.6+210V=0$

V = -1.205 m/s

So, the recoil velocity of the raft is 1.205 m/s. Hence, this is the required solution.

Answer:

The recoil velocity of the raft would be $v_(r)\approx 1.2(m)/(s)$ (pointing to the left if the swimmer runs to the right)

Explanation:

The problem states thatthe swimmer has a mass of m=55 kg, and the raft has a mass of M=210 kg. Then, it says that the swimmer runs off the raft with a (final) velocity of v=4.6 m/s relative to the shore.

To analyze it, we take a system of "two particles", wich means that we will consider the swimmer and the raft as a hole system, aisolated from the rest of the world.

Then, from the shore, we can put our reference system and take the initial moment when the swimmer and the raft are stationary. This means that the initial momentum is equal to zero:

$p_(i)=0$

Besides, we can use magnitudes instead of vectors because the problem will develope in only one dimension after the initial stationary moment (x direction, positive to the side of the running swimmer, and negative to the side of the recoling raft), this means that we can write the final momentum as

$p_(f)=mv-Mv_(r)=0$

The final momentum is equal to zero due to conservation of momentum (because there are no external forces in the problem, for the system "swimmer-raft"), so the momentum is constant.

Then, from that previous relation we can clear

$v_(r)=(m)/(M)v=(55)/(210)*4.6(m)/(s)=(253)/(210)(m)/(s)\approx1.2(m)/(s)$

wich is the recoil velocity of the raft, and it is pointing to the left (we established this when we said that the raft was going to the negative side of the system of reference, and when we put a minus in the raft term inside the momentum equation).

A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to the victim with an initial velocity of 1.50 m/s and observes that it takes 1.2 s to reach the water. How high above the water was the preserver released? Note that the downdraft of the helicopter reduces the effects of air resistance on the falling life preserver, so that an acceleration equal to that of gravity is reasonable. Round your answer to the nearest whole number.

Answers

Answer:

The answer is 9 m.

Explanation:

Using the kinematic equation for an object in free fall:

$y = y_o - v_o-(1)/(2)gt^(2)$

In this case:

$v_o = \textrm{Initial velocity} = 1.5[m/s]\nt = \textrm{air time} = 1.2 [s]$ $y_o = 0$

$g = \textrm{gravity} = 9.8 [m/s^(2) ]$

Plugging those values into the previous equation:

$y = 0 - 1.5*1.2-(1)/(2)*9.8*1.2^(2) \ny = -8.85 [m] \approx -9 [m]$

The negative sign is because the reference taken. If I see everything from the rescuer point of view.

A car is making a 50 mi trip. It travels the first half of the total distance 25.0 mi at 7.00 mph and the last half of the total distance 25.0 mi at 52.00 mph. (a) What is the total time in hours of the trip? Keep two decimal places. 4.05 Correct (100,0%) (b) What is the car's average speed in mph for the entire trip? Keep two decimal places. 12 35 Correct (100,0%) Submit The car travels the same distance again, but this time, in the first half of the time its speed is 7.00 mph and in the second half of the time its speed is 52.00 mph. c) What is the total time in hours of the trip? Keep two decimal places.

Answers

Answer:

a) The total time of the trip is 4.05 h.

b) The average speed of the car is 12.35 mi/h.

c) The total time of the trip is 1.69 h.

Explanation:

Hi there!

a) The equation of traveled distance for a car traveling at constant speed is the following:

x= v · t

Where:

x = traveled distance.

v = velocity.

t = time.

Solving the equation for t, we can find the time it takes to travel a given distance "x" at a velocity "v":

x/v = t

So, the time it takes the car to travel the first half of the distance will be:

t1 = 25.0 mi / 7.00 mi/h

And for the second half of the distance:

t2= 25.0 mi / 52.00 mi / h

The total time will be:

total time = t1 + t2 = 25.0 mi / 7.00 mi/h + 25.0 mi / 52.00 mi / h

total time = 4.05 h

The total time of the trip is 4.05 h.

b) The average speed (a.s) is calculated as the traveled distance (d) divided by the time it takes to travel that distance (t). In this case, the traveled distance is 50 mi and the time is 4.05 h. Then:

a.s = d/t

a.s = 50 mi / 4.05 h

a.s = 12.35 mi/h

The average speed of the car is 12.35 mi/h

c) Let's write the equations of traveled distance for both halves of the trip:

For the first half, you traveled a distance d1 in a time t1 at 7.00 mph:

7.00 mi/h = d1/t1

Solving for d1:

7.00 mi/h · t1 = d1

For the second half, you traveled a distance d2 in a time t2 at 52.00 mph.

52.00 mi/h = d2/t2

52.00 mi/h · t2 = d2

We know that d1 + d2 = 50 mi and that t1 and t2 are equal to t/2 where t is the total time:

d1 + d2 = 50 mi

52.00 mi/h · t/2 + 7.00 mi/h · t/2 = 50 mi

Solving for t:

29.5 mi/h · t = 50 mi

t = 50 mi / 29.5 mi/h

t = 1.69 h

The total time of the trip is 1.69 h.

A volumetric flask made of Pyrex is calibrated at 20.0°C. It is filled to the 150-mL mark with 34.5°C acetone. After the flask is filled, the acetone cools and the flask warms so that the combination of acetone and flask reaches a uniform temperature of 32.0°C. The combination is then cooled back to 20.0°C. (The average volume expansion coefficient of acetone is 1.50 10-4(°C)−1.) (a) What is the volume of the acetone when it cools to 20.0°C?

Answers

Answer:149.73 ml

Explanation:

Given

$\beta \ of\ acetone=1.50* 10^(-4) ^(\circ)C^(-1)$

change in volume is given by

$\Delta V=V_(final)-V_(initial)$

$\Delta V=\nu_(initial)\beta _(acetone)\left [ T_f-T_i\right ]$

$V_(final)=\nu_(initial)+\nu_(initial)\beta _(acetone)\left [ T_f-T_i\right ]$

$V_(final)=150+150* 1.50* 10^(-4)\left [ 20-32\right ]$

$V_(final)=149.73 ml$

Final answer:

The volume of the acetone when it cools to 20.0°C is approximately 142.39 mL.

Explanation:

In order to determine the volume of the acetone when it cools to 20.0°C, we can use the equation for the volume change caused by a temperature change at constant pressure, known as Charles's law. Charles's law states that the volume of a gas is directly proportional to its temperature in Kelvin. We can use the formula V2 = V1 * (T2 / T1) to calculate the volume of the acetone at the lower temperature.

Given that the initial volume of the acetone is 150 mL at a temperature of 34.5°C, we need to convert this temperature to Kelvin by adding 273.15. Therefore, T1 = 34.5°C + 273.15 = 307.65 K.

Since the final temperature is 20.0°C, the final temperature in Kelvin will be T2 = 20.0°C + 273.15 = 293.15 K. We can now plug these values into the equation to find the volume of the acetone at the lower temperature: V2 = 150 mL * (293.15 K / 307.65 K) = 142.39 mL.

Learn more about Volume calculation here:

brainly.com/question/33318354

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