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Andrews [41]
3 years ago
14

A 2 kg object is given a displacement ∆~s = (5 m)ˆı + (3 m) ˆ + (−4 m) ˆk along a straight line. During the displacement, a con

stant force of F~ = (3 N)ˆı + (−1 N) ˆ + (7 N) ˆk acts on the object. Find the work done by F~ for this displacement.
Physics
1 answer:
nadya68 [22]3 years ago
8 0

Answer: 15Nm

Explanation:

Work is said to be done when a force cause a body to move through a distance. Mathematically,

Work = Force (F) × distance (s)

If F = F~ = (3 N)ˆı + (−1 N) ˆj + (7 N) ˆk

S = (5 m)ˆı + (3 m) ˆ + (−4 m) ˆk

According to vector notation,

i.I = j.j =k.k = 1

Multiplication of different component will give 'zero'

F×s = (3 N)ˆi × (5 m)ˆı

Work done = 15Nm

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Answer:

The y-component of the electric force on this charge is F_y = -1.144\times 10^{-6}\ N.

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<u>Given:</u>

  • Electric field in the region, \vec E = (148.0\ \hat i-110.0\ \hat j)\ N/C.
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where, \hat i,\ \hat j are the unit vectors along the positive x and y axes respectively.

The electric field at a point is defined as the electrostatic force experienced per unit positive test charge, placed at that point, such that,

\vec E = \dfrac{\vec F}{q}\\\therefore \vec F = q\vec E\\=(10.4\times 10^{-9})\times (148.0\ \hat i-110.0\ \hat j)\\=(1.539\times 10^{-6}\ \hat i-1.144\times 10^{-6}\ \hat j)\ N.

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When a hammer thrower releases her ball, she is aiming to maximize the distance from the starting ring. Assume she releases the
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Answer:

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