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nirvana33 [79]
3 years ago
7

A man carries a load of 20N on his head over a horizontal distance of 20 m how much work is done by the man

Physics
1 answer:
sdas [7]3 years ago
4 0

Work done in horizontal direction = 0 as the potential energy is not changed and the change in kinetic energies : initial and final - is zero.


work done when the person carries the suitcase in vertical direction =

  = m g h = change in potential energy = 30 kg * 9.8 m/sec/sec * 10 m

  =  2940 Joules

=============

There is a force F1 lifting the body upwards. There is Weight W = m g pulling the body downwards.  The net force F2 on the body is upwards with acceleration a.


  F2 = m a  = F1 - W = F1 - m g

          a =  F1 / m - g


Work done by Force F1 with which the object is lifted = F1 . s = 230 Joules

     F1  = energy spent / distance = 230 J / 2 m  = 115 Newtons


Net Acceleration of the body = a = F1 / m - g = 115/10 - 10  = 1.5 m/sec/sec

==========================


 m = 2 kg          F = 8 Newtons        t = 12 sec

 a = F/m = 4 m/sec/sec

v = u + a t = 0 + 4 * 12 = 48 m/sec


kinetic energy = 1/2 m v² = 1/2 * 2 * 48 * 48 = 48² Joules



Read more on Brainly.in - https://brainly.in/question/71130#readmore


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Now, let's convert our seconds to hour:
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Answer:

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Explanation:

<h2>a) </h2>

We know that the cross product of linearly independent vectors \vec{A} and \vec{B} gives us a nonzero, orthogonal to both, vector. So, if we can find two linearly independent vectors on the plane through the points P, Q, and R, we can use the cross product to obtain the answer to point a.

Luckily for us, we know that vectors \vec{A} = \vec{P}-\vec{Q} and \vec{B} = \vec{R} - \vec{Q} are living in the plane through the points P, Q, and R, and are linearly independent.

We know that they are linearly independent, cause to have one, and only one, plane through points P Q and R, this points must be linearly independent (as the dimension of a plane subspace is 3).

If they weren't linearly independent, we will obtain vector zero as the result of the cross product.

So, for our problem:

\vec{A} = \vec{P} - \vec{Q} \\\\\vec{A} = (1,0,1) - (-2,1,4)\\\\\vec{A} = (1 +2,0-1,1-4)\\\\\vec{A} = (3,-1,-3)

\vec{B} = \vec{R} - \vec{Q} \\\\\vec{B} = (6,2,7) - (-2,1,4)\\\\\vec{B} = (6 +2,2-1,7-4)\\\\\vec{B} = (8,1,3)

\vec{A} \times  \vec{B} = (A_y B_z - B_y A_z) \  \hat{i} - ( A_x B_z-B_xA_z) \ \hat{j} + (A_x B_y - B_x A_y ) \ \hat{k}

\vec{A} \times  \vec{B} = ( (-1) * 3 - 1 * (-3) ) \  \hat{i} - ( 3 * 3 - 8 * (-3)) \ \hat{j} + (3 * 1 - 8 * (-1) ) \ \hat{k}

\vec{A} \times  \vec{B} = ( - 3 + 3 ) \  \hat{i} - ( 9 + 24 ) \ \hat{j} + (3 + 8 ) \ \hat{k}

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\sqrt{(-33)^2 + (12)^2} = 2 * area_{triangle}

\sqrt{1233} = 2 * area_{triangle}

35.114= 2 * area_{triangle}

17.55 \ units \  of \ area =  area_{triangle}

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