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fenix001 [56]
2 years 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]2 years 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:

 v = 13.19 m / s

Explanation:

This problem must be solved using Newton's second law, we create a reference system where the x-axis is perpendicular to the cylinder and the Y-axis is vertical

 

X axis

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Centripetal acceleration is

       a = v² / r

Y Axis

      fr -W = 0

      fr = W

The force of friction is

     fr = μ N

Let's calculate

    μ (m v² / r) = mg

   μ v² / r = g

   v² = g r / μ

   v = √ (g r /μ)

   v = √ (9.8 11 / 0.62)

   v = 13.19 m / s

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2 years ago
A toy plane has a mass of 2.5 kg and is 18 m above the ground. It is moving at 4.5 m/s. How much mechanical energy does the toy
sleet_krkn [62]
Mechanical energy is the sum of kinetic energy and potential energy, or E=Ek+Ep. So Ek=(1/2)*m*v² where m is the mass of the object and v is it's velocity. Mp=m*g*h where m is the mass, g=9.81 m/s² and h is the height of the object. So after we input the numbers the total mechanical energy is
E=(1/2)*2.5*(4.5²) + 2.5*9.81*18 = 25.3125 J + 441.45 J = 466.7625 J. The correct answer is E= 466 J. 
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2 years ago
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Traffic officials indicate, it takes longer to ______ when you drive fast.
nignag [31]
The answer in the blank is that it is difficult to accelerate at decelerate the vehicle when it is on a fast speed because having a fast speed makes it difficult to adjust the meter as well as if you try to decelerate the vehicle, it could burn out the tires and engine as it is in the fast speed, in accelerating it, it could also be complicated because it would only make the car faster enough that you may no longer control of how to stop it.
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2 years ago
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A fighter jet is catapulted off an aircraft carrier from rest to 75 m/s. If the aircraft carrier deck is 100 m long, what is the
egoroff_w [7]

The acceleration of the jet is 28.1 m/s^2

Explanation:

Since the motion of the jet is a uniformly accelerated motion, we can use the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

For the jet in this problem, we have

u = 0

v = 75 m/s

s = 100 m

Solving for a, we find the acceleration:

a=\frac{v^2-u^2}{2s}=\frac{75^2-0}{2(100)}=28.1 m/s^2

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4 0
2 years ago
Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to thenext handhold. A 9.3kggibbon has
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Upward force provided by the branch is 260 N

<u>Explanation:</u>

Given -

Mass of Gibbon, m = 9.3 kg

Length of the branch, r = 0.6 m

Speed of the movement, v = 3.3 m/s

Upward force, T = ?

The tension force in the rod must be greater than the weight at the bottom of the swing in order to provide an upward centripetal acceleration.

Therefore,

F net = T - mg

F net = ma = mv²/r

Thus,

T = mv²/r + mg

T = m ( v²/r + g)

T = 9.3 [ ( 3.3)² / 0.6 + 9.8]

T = 259.9 N ≈ 260 N

Therefore, upward force provided by the branch is 260 N

5 0
2 years ago
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