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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:

a. 4d.

If the car travels at a velocity of 2v, the minimum stopping distance will be 4d.

Explanation:

Hi there!

The equations of distance and velocity of the car are the following:

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v = v0 + a · t

Where:

x =  position of the car at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

v = velocity of the car at time t.

Let´s find the time it takes the car to stop using the equation of velocity. When the car stops, its velocity is zero. Then:

velocity = v0 + a · t      v0 = v

0 = v + a · t

Solving for t:

-v/a = t

Since the acceleration is negative because the car is stopping:

v/a = t

Now replacing t = v/a in the equation of position:

x = x0 + v0 · t + 1/2 · a · t²     (let´s consider x0 = 0)

x = v · (v/a) + 1/2 · (-a) (v/a)²    

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x = 1/2 v²/a

At a velocity of v, the stopping distance is 1/2 v²/a = d

Now, let´s do the same calculations with an initial velocity v0 = 2v:

Using the equation of velocity:

velocity = v0 + a · t

0 = 2v - a · t

-2v/-a = t

t = 2v/a

Replacing in the equation of position:

x1 = x0 + v0 · t + 1/2 · a · t²  

x1 = 2v · (2v/a) + 1/2 · (-a) · (2v/a)²

x1 = 4v²/a - 2v²/a

x1 = 2v²/a

x1 = 4(1/2 v²/a)

x1 = 4x

x1 = 4d

If the car travels at a velocity 2v, the minimum stopping distance will be 4d.

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Answer:

I. The horizontal distance traveled by the bullet is greater for the Moon.

II. The flight time is less for the bullet on the Earth.

Explanation:

Horizontal distance depends on the initial speed, height and gravity. Bullets have the same initial speed and are shot from the same height. In these conditions horizontal distance only depends on gravity, which is inversely proportional. Therefore, the less gravity the greater the horizontal distance. Gravity slows bullet and causes its impact on the ground. Since gravity is greater in Earth, the bullet hits faster on the earth.

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You are working as an assistant to an air-traffic controller at the local airport, from which small airplanes take off and land.
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Answer:

d = 2021.6 km

Explanation:

We can solve this distance exercise with vectors, the easiest method s to find the components of the position of each plane and then use the Pythagorean theorem to find distance between them

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Angle θ = 25°

           cos 25 = x / r

           sin 25 = z / r

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           z₁ = r sin 25

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Height   y₂ = 1100 m

Angle θ = 20°

          x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m

          z₂ = 20 103 without 25 = 8.452 103 m = 8452 m

The distance between the planes using the Pythagorean Theorem is

         d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2

Let's calculate

        d² = (18126-16314)²  + (1100-800)² + (8452-7607)²

        d² = 3,283 106 +9 104 + 7,140 105

        d² = (328.3 + 9 + 71.40) 10⁴

        d = √(408.7 10⁴)

        d = 20,216 10² m

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