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AURORKA [14]
2 years ago
11

An average person can reach a maximum height of about 60 {\rm cm} when jumping straight upfrom a crouched position. During the j

ump itself, the person's bodyfrom the knees up typically rises a distance of around 50 {\rm cm}. To keep the calculationssimple and yet get a reasonable result, assume that the entirebody rises this much during the jump. In terms of this jumper's weight W , what force does the ground exert on him or herduring the jump?
Physics
1 answer:
Sunny_sXe [5.5K]2 years ago
3 0

Answer:

2.2 W where W is the weight of the jumper.

Explanation:

A) use the formula

V^2 = 2gh= 2*9.81*60

that gives

V = 3.43 m/s

b) W = mg

during the jump , the displacement of the body is d =0.5m

so the acceleration a = (V^2-Vo^2)/2d = 11.76 m/s^2

net force applied by ground = m(g+a) = W/g(g+a)

= 2.2 W

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Answer:

The work done by the gravel to stop the truck is 520.44 kJ

Explanation:

<u>Step 1</u>: Data given

Mass of the truck = 3047.8 kg

The ramp has an angle of 9.5 °

Velocity of  the truck = 20.68 m/s

distance = 26.6 meters

<u>Step 2:</u> Calculate initial kinetic energy

sin 9.5° = 0.165

h = ℓ*sin 9.5° = 26.6*0.165= 4.39 m

Ek = 1/2m*Vo² = 1/2*3047.8*20.68² = 651714.7 Joule = 651.7 kJ  = initial kinetic energy

<u>Step 3: </u>Calculate potential energy

Epot = U = m*g*h = 3047.8*9.81*4.39 = 131256.25 Joule = 131.26 kJ

<u>Step 4:</u>  What work is done by the truck on the gravel?  

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5 0
3 years ago
Which question could be tested in a scientific manner?
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A block of mass
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(a) The work done by the applied force is 26.65 J.

(b) The work done by the normal force exerted by the table is 0.

(c) The work done by the force of gravity is 0.

(d) The work done by the net force on the block is 26.65 J.

<h3>Work done by the applied force</h3>

W = Fdcosθ

W = 14 x 2.1 x cos25

W = 26.65 J

<h3>Work done by the normal force</h3>

W = Fₙd

W = mg cosθ x d

W = (2.5 x 9.8) x cos(90) x 2.1

W = 0 J

<h3>Work done force of gravity</h3>

The work done by force of gravity is also zero, since the weight is at 90⁰ to the displacement.

<h3> Work done by the net force on the block</h3>

∑W = 0 + 26.65 J = 26.65 J

Thus, the work done by the applied force is 26.65 J.

The work done by the normal force exerted by the table is 0.

The work done by the force of gravity is 0.

The work done by the net force on the block is 26.65 J.

Learn more about work done here: brainly.com/question/8119756

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