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SVETLANKA909090 [29]
3 years ago
14

An asteroid is moving along a straight line. A force acts along the displacement of the asteroid and slows it down. The asteroid

has a mass of 3.7× 104 kg, and the force causes its speed to change from 7500 to 4500m/s. (a) What is the work done by the force? (b) If the asteroid slows down over a distance of 1.7× 106 m determine the magnitude of the force.
Physics
1 answer:
Katena32 [7]3 years ago
3 0

Answer:

666000000000 J

391764.71 N

Explanation:

u = Initial velocity

v = Final velocity

s = Displacement

Change in energy

\Delta KE=\frac{1}{2}m(v^2-u^2)\\\Rightarrow \Delta KE=\frac{1}{2}3.7\times 10^4(4500^2-7500^2)\\\Rightarrow \Delta KE=-666000000000\ J

Work done be the force is 666000000000 J

W=F\times s\\\Rightarrow F=\frac{W}{s}\\\Rightarrow F=\frac{-666000000000}{1.7\times 10^6}\\\Rightarrow F=-391764.71\ N

The magnitude of force is 391764.71 N

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Explanation:

Solution:-

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( See attachments )

- The equation for the magnetic field strength - B - (magnitude) produced by a long straight current-carrying wire is given by the Biot Savart Law:

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Where,

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r : The shortest distance to the wire,

uo : The permeability of free space. = 4π * 10^-7  T. m/A

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                             B = \frac{uo*I}{4\pi }*\int\frac{dl xr}{r^2}      

Where the integral sums over,

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 3)  r^ is a unit vector in the direction of r.

   

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