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Kipish [7]
3 years ago
14

In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of t

he legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, g. How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)
Physics
2 answers:
dolphi86 [110]3 years ago
6 0

Answer:

H_{max} = 0.75 m

Explanation:

Acceleration achieved is given as

a = 1.25 g

So we will have

a = 12.26 m/s^2

now the velocity that he will gain till he moved 0.600 m is given as

v^2 - v_i^2 = 2 a d

v^2 - 0 = 2(12.26)(0.600)

v = 3.84 m/s

now with this speed the maximum height achieved by him

H_{max} = \frac{v_f^2 - v_i^2}{2g}

H_{max} = \frac{0 - 3.84^2}{2(-9.81)}

H_{max} = 0.75 m

muminat3 years ago
3 0
<h2><em><u>⇒</u></em>Answer:</h2>

In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, g . How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)

Step-by-Step Solution:

Solution 35PE

This question discusses about the increased range. So, we shall assume that the angle of jumping will be  as the horizontal range is maximum at this angle.

Step 1 of 3<

/p>

The legs have an extension of 0.600 m in the crouch position.

So,  m

The person is at rest initially, so the initial velocity will be zero.

The acceleration is  m/s2

Acceleration  m/s2

Let the final velocity be .

Step 2 of 3<

/p>

Substitute the above given values in the kinematic equation  ,

m/s

Therefore, the final velocity or jumping speed is  m/s

Explanation:

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Answer:

The shortest distance is  S = 24.86 ft

Explanation:

The free body diagram of this question is shown on the first uploaded image

From the question we are told that

   The speed of the bicycle is v_b = 22\ mph = 22 * \frac{5280}{3600}   =  32.26 ft/s

     The distance between the axial is  d = 42 \ in

The mass center of the cyclist and the bicycle is m_c = 26 \ in  behind the front axle

The mass center of the cyclist and the bicycle is m_h = 40 \ in above the ground

   For the bicycle not to be thrown over the

     Momentum about the back wheel must be zero so

                \sum _B = 0

=>             mg (26) = ma(40)

=>             a = \frac{26}{40} g

Here  g = 32.2 ft/s^2

     So     a =  \frac{26}{40} (32.2)

             a =  20.93 ft/s^2

Apply the equation of motion to this motion we have

       v^2 = u^2 + 2as

 Where  u = 32.26 ft /s

             and v = 0 since the bicycle is coming to a stop

        v^2 = (32.26)^2 - 2(20.93) S

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The drum of a washing machine spins at 200 rpm the drum had a radius of 0.5m what will the centripetal force experienced by a to
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The centripetal force experienced by the towel is 55 N.

The given parameters;

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  • radius of the machine' drum, r = 0.5 m
  • mass of the towel, m = 0.25 kg

The centripetal force experienced by the towel spinning along the walls of the drum is calculated as follows;

Fc = mrω²

where;

<em>Fc is the centripetal force</em>

<em>ω is angular speed in rad/s</em>

The angular speed in rad/s is calculate as;

\omega = \frac{200 \ rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 \ s} \\\\\omega = 20.95 \ rad/s

The centripetal force experienced by the towel is calculated as;

F_c = mr\omega ^2\\\\F_c = 0.25 \times 0.5 \times (20.95)^2\\\\F_c = 54.9 \ N \ \approx 55 \ N

Thus, the centripetal force experienced by the towel is 55 N.

Learn more here: brainly.com/question/20905151

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