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fenix001 [56]
1 year ago
11

Chris and Jamie are carrying Wayne on a horizontal stretcher. The uniform stretcher is 2.00 m long and weighs 100 N. Wayne weigh

s 800 N. Wayne's center of gravity is 75.0 cm from Chris. Chris and Jamie are at the ends of the stretcher. The upward force that Jamie is exerting to support the stretcher, with Wayne on it, is:______
Physics
1 answer:
PIT_PIT [208]1 year ago
3 0

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The value is F_j  =  550\ N

Explanation:

From the question we are told that

   The length of the stretcher is  d =  2.0 \  m

    The weight of the stretcher is W  =  100 \  N

    The weight for Wayne is  W_w =  800 \ N

     The distance of  center of gravity for Wayne from Chris is c_w = 75 cm  =  0.75 \ m

Generally taking moment about the first end where Chris is

         F_j *  d              => upward moment

Here F_j is the force applied by Jamie

Generally  taking moment about the second end where Jamie is

      W *  ( \frac{d}{2} ) +  W_w * (d - c_w)      => downward moment

Generally at equilibrium , the upward moment is equal to the downward moment

     F_j *  d = W *  ( \frac{d}{2} ) +  W_w * (d - c_w)

=>   F_j *  2  = 100 *  ( \frac{ 2}{2} ) +  800 * (2 - 0.75)

=>    F_j  =  550\ N

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Answer:

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Explanation:

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Let's calculate

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4 0
2 years ago
The block in the diagram below is AT REST. However, the tension in the cable is not the only thing holding the block back. Stati
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Answer:

The  tension in the rope is 229.37 N.

Explanation:

Given:

Mass of the block is, m=33.2\ kg

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Angle of inclination is, \theta = 31.5°

Draw a free body diagram of the block.

From the free body diagram, consider the forces in the vertical direction perpendicular to inclined plane.

Forces acting are mg\cos \theta and normal N. Now, there is no motion in the direction perpendicular to the inclined plane. So,

N=mg\cos \theta\\N=(33.2)(9.8)\cos (31.5)\\N=277.415\ N

Consider the direction along the inclined plane.

The forces acting along the plane are mg\sin \theta and frictional force, f, down the plane and tension, T, up the plane.

Now, as the block is at rest, so net force along the plane is also zero.

T=mg\sin \theta+f\\T=mg\sin \theta +\mu N\\T= (33.2)(9.8)(\sin (31.5)+(0.214\times 277.415)\\T= 170+59.37\\T=229.37\ N

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1 year ago
A ball is dropped from the top of a building.After 2 seconds, it’s velocity is measured to be 19.6 m/s. Calculate the accelerati
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Answer:

acceleration, a = 9.8 m/s²

Explanation:

'A ball is dropped from the top of a building' indicates that the initial velocity of the ball is zero.

u = 0 m/s

After 2 seconds, velocity of the ball is 19.6 m/s.

t = 2s, v = 19.6 m/s

Using

v = u + at

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a = 9.8 m/s²

6 0
2 years ago
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A jetboat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m ​ 5, point, 0, start fraction, start text, m, end tex
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The question is incomplete. Here is the entire question.

A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?

Answer: Δx = - 42m

Explanation: The jetboat is moving with an acceleration during the time interval, so it is a <u>linear</u> <u>motion</u> <u>with</u> <u>constant</u> <u>acceleration</u>.

For this "type" of motion, displacement (Δx) can be determined by:

\Delta x = v_{i}.t + \frac{a}{2}.t^{2}

v_{i} is the initial velocity

a is acceleration and can be positive or negative, according to the referential.

For Referential, let's assume rightward is positive.

Calculating displacement:

\Delta x = 5(6) - \frac{4}{2}.6^{2}

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\Delta x = - 42

Displacement of the boat for t=6.0s interval is \Delta x = - 42m, i.e., 42 m to the left.

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Answer:

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c_w = Specific heat of water = 4186 J/kg⋅°C

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The final equilibrium temperature is 9.50022°C

4 0
2 years ago
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