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Tcecarenko [31]

2 years ago

7

A coil of 1000 turns of wire has a radius of 12 cm and carries a counterclockwise current of 15A. If it is lying flat on the gro

und, and the Earth's magnetic field points due north, has a magnitude of 5.8 x 10-5 T, and makes a downward angle of 25Â° with the vertical, what is the torque on the loop?

1 answer:

grin007 [14]2 years ago

5 0

Answer:

torque is 1.7 * $10^{-2}$ Nm

Explanation:

Given data

turns n = 1000 turns

radius r = 12 cm

current I = 15A

magnitude B = 5.8 x 10^-5 T

angle θ = 25°

to find out

the torque on the loop

solution

we know that torque on the loop is

torque = N* I* A*B* sinθ

here area A = πr² = π(0.12)²

put all value

torque = N* I* A*B* sinθ

torque = 1000* 15* π(0.12)² *5.8 x 10-5 * sin25

torque = 0.0166 N m

torque is 1.7 * $10^{-2}$ Nm

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In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co

labwork [276]

Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

$m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m$

a)

The final velocity of cart 1 after collision is given as:

$v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\ Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s$

The final velocity of cart 2 after collision is given as:

$v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\ Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s$

b) Using the law of conservation of energy:

$\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm$

7 0

2 years ago

Four students measured the acceleration of gravity. The accepted value for their location is 9.78m/s2. Which student’s measureme

Lisa [10]

On comparing values , we see that student which has the largest percent error is <u>A. Student 4: 9.61 m/s2 .</u>

<u>Explanation:</u>

Here, we have Four students measured the acceleration of gravity. The accepted value for their location is 9.78m/s2. Let's calculate which student’s measurement has the largest percent error :

<u>A. Student 4: 9.61 m/s2 </u>

Percentage of error = $\frac{9.78-9.61}{9.78} (100) = 1.738%$ %.

<u>B. Student 3: 9.88 m/s2 </u>

Percentage of error = $\frac{9.88-9.78}{9.78} (100) = 1.022%$ %.

<u>C. Student 2: 9.79 m/s2 </u>

Percentage of error = $\frac{9.79-9.78}{9.78} (100) = 0.1022%$ % .

<u>D. Student 1: 9.78 m/s2</u>

Percentage of error = $\frac{9.78-9.61}{9.78} (100) = 0%$ % .

On comparing values , we see that student which has the largest percent error is <u>A. Student 4: 9.61 m/s2 .</u>

4 0

2 years ago

A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 4 ft/s along

babymother [125]

Answer:

Explanation:

height of pole = 15 ft

height of man = 6 ft

Let the length of shadow is y .

According to the diagram

Let at any time the distance of man is x.

The two triangles are similar

$\frac{y-x}{y}=\frac{6}{15}$

15 y - 15 x = 6 y

9 y = 15 x

$y=\frac{5}{3}x$

Differentiate with respect to time.

$\frac{dy}{dt}=\frac{5}{3}\frac{dx}{dt}$

As given, dx/dt = 4 ft/s

$\frac{dy}{dt}=\frac{5}{3}\times 4$

$\frac{dy}{dt}=\frac{20}{3}$ ft/s

6 0

2 years ago

A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then

Elan Coil [88]

Answer:

A) $W_{ff} =-744.12J$

B) $F_f=-W_{ff}*sin\theta /hy = 112.75N$

C) $F_{f2}=207.58N$

Explanation:

This question is incomplete. The full question was:

<em>A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.2 m/s. </em>

<em>Part (a) Write an expression for the work, Wf, done by the friction force between the ramp and the skateboarder in terms of the variables given in the problem statement. </em>

<em>Part (b) The ramp makes an angle θ with the ground, where θ = 30°. Write an expression for the magnitude of the friction force, fr, between the ramp and the skateboarder. </em>

<em>Part (c) When the skateboarder reaches the bottom of the ramp, he continues moving with the speed vf onto a flat surface covered with grass. The friction between the grass and the skateboarder brings him to a complete stop after 5.00 m. Calculate the magnitude of the friction force, Fgrass in newtons, between the skateboarder and the grass.</em>

For part A), we make a balance of energy to calculate the work done by the friction force:

$W_{ff}=\Delta E$

$W_{ff}=1/2*m*vf^2-m*g*hy$

$W_{ff}=-744.12J$

For part B), we use our previous value for the work:

$W_{ff}=-F_f*(hy/sin\theta)$ Solving for friction force:

$F_f=-W_{ff}*sin\theta /hy$

$F_f=112.75N$

For part C), we first calculate the acceleration by kinematics and then calculate the module of friction force by dynamics:

$Vf^2=Vo^2+2*a*d$

Solving for a:

$a=-3.844m/s^2$

Now, by dynamics:

$|F_f|=|m*a|$

$|F_f|=207.58N$

8 0

2 years ago

A small pebble and one large boulder start at the same height and begin rolling down the side of a mountain. Which object would

vitfil [10]

Answer:

The small pebble

Explanation:

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Since h and g are constant

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So, the kinetic energy of the object is directly proportional to its mass. Thus, the object with the smaller mass has the lesser kinetic energy.

Since the object with the smaller mass is the small pebble, so the small pebble would have less kinetic energy as it crashes on the road at the bottom of the mountain.

8 0

2 years ago