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2 years ago

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The square loop shown in the figure moves into a 0.80 T magnetic field at a constant speed of 10 m/s. The loop has a resistance

of 0.10 Ω, and it enters the field at t = 0 s.

1 answer:

mr Goodwill [35]2 years ago

8 0

Is there more information that goes with the question

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What type of weather modification involves the use of large fans to mix surface air with air aloft?

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Which best contrasts an electric generator and an electric motor? In a generator, slip rings function as a solenoid, and in a mo

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Sophia is planning on going down an 8-m water slide. Her weight is 50 N. She knows that she has gravitational potential energy (

Pepsi [2]

Answer:The higher up an object is the greater its gravitational potential energy. The larger the distance something falls through the greater the amount of GPE the object loses as it falls. As most of this GPE gets changed into kinetic energy, the higher up the object starts from the faster it will be falling when it hits the ground. So a change in gravitational potential energy depends on the height an object moves through.

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2 years ago

a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expres

shusha [124]

Answer:

Part a) When collision is perfectly inelastic

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b) When collision is perfectly elastic

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

Explanation:

Part a)

As we know that collision is perfectly inelastic

so here we will have

$mv_m = (m + M)v$

so we have

$v = \frac{mv_m}{m + M}$

now we know that in order to complete the circle we will have

$v = \sqrt{5Rg}$

$\frac{mv_m}{m + M} = \sqrt{5Rg}$

now we have

$v_m = \frac{m + M}{m} \sqrt{5Rg}$

Part b)

Now we know that collision is perfectly elastic

so we will have

$v = \frac{2mv_m}{m + M}$

now we have

$\sqrt{5Rg} = \frac{2mv_m}{m + M}$

$v_m = \frac{m + M}{2m}\sqrt{5Rg}$

6 0

2 years ago

Two wires are used to suspend a sign that weighs 500 N. The two wires make an angle of 100° between each other. If each wire is

vaieri [72.5K]

1) draw a diagram.
2) label diagram. (split the 100 degrees into 50, (which is right down the middle) to make a right angle triangle.)
3) since its a free body diagram, the forces known must be labelled. (force of gravity). this shows that the straight vertical line of the right angle triangle is Fg (force gravity). label it.
4) use trigonometry. rearrange the equation to solve for what needs to be known.

angles known: 50 (split 100 in half to make a right angle triangle), 90 (since its right angle), and 40 (180-90-50 = 40)

sides known: vertical lined up with the 90 degree angle. Fg. --> fg=mg=500N x 9.81m/s^2 = 4905N

use formula: sin or cos

i used sin. sin(40) = 4905 / ?
- times '?' on both sides. : sin(40) x '?' = 4905
-divide both sides by sin(40): '?' = 4905/ sin(40)
--> Solve.

8 0

2 years ago