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meriva
3 years ago
15

A stiff wire 32.5 cm long is bent at a right angle in the middle. One section lies along the z axis and the other is along the l

ine y=2x in the xy plane. A current of 25.0 A flows in the wire-down the z axis and out the line in the xy plane. The wire passes through a uniform magnetic field given by B =(0.318i)T.
Determine the magnitude and the direction of the total force on the wire.
Physics
1 answer:
Flauer [41]3 years ago
4 0

Answer: i personally think it would be 789

Explanation:

Hopefully I am right

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A car's bumper is designed to withstand a 5.04 km/h (1.4-m/s) collision with an immovable object without damage to the body of t
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Answer:

the magnitude of the average force on the bumper is 3189.8 N

Explanation:

Given the data in the question;

In terms of force and displacement, work done is;

W =F^> × x^>

W = Fxcos\theta    ------- let this be equation 1

where F is force applied, x is displacement and θ is angle between force and displacement.

Now, since the displacement of the bumper and force acting on it is in the same direction,

hence, θ = 0°

we substitute into equation 1

W = Fxcos( 0° )

W = Fx ------- let this be equation 2

Now, using work energy theorem,

total work done on the system is equal to the change in kinetic energy of the system.

W_{net = ΔKE

= \frac{1}{2}mv² -  \frac{1}{2}mu² --------- let this be equation 3

where m is mass of object, v is final velocity, u is initial velocity.

from equation 2 and 3

Fx = \frac{1}{2}mv² -  \frac{1}{2}mu²

we make F, the subject of formula

F = \frac{m}{2x}( v² - u² )

given that mass of car m = 830 kg, x = 0.255 m, v = 0 m/s, and u = 1.4 m/s

so we substitute

F = \frac{830}{(2)(0.255)}( (0)² - (1.4)² )

F = 1627.45098 ( 0 - 1.96 )

F = 1627.45098 ( - 1.96 )

F = -3189.8 N

The negative sign indicates that the direction of the force was in opposite compare to the direction of the velocity of the car.

Therefore, the magnitude of the average force on the bumper is 3189.8 N

6 0
2 years ago
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