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MakcuM [25]
3 years ago
15

When landing after a spectacular somersault, a 35.0 kg gymnast decelerates by pushing straight down on the mat. calculate the fo

rce she must exert if her deceleration is 9.00 times the acceleration of gravity?
Physics
1 answer:
Sliva [168]3 years ago
6 0

The deceleration experienced by the gymnast is the 9 times of the acceleration due to gravity.

Now from Newton`s  first law, the net force on gymnast,

F_{net} =F-W=ma

Here, W is the weight of the gymnast and a is the acceleration experienced by the gymnast (9\times g acceleration due to gravity)  

Therefore,

F= ma+W OR F=ma+mg=m(g+a)

Given m = 30 kg anda=9\times g=9\times 9.8 m/s^{2} =88.2 m/s^{2}

Substituting these values in above formula and calculate the force exerted by the gymnast,  

F=(40 kg) (88.2 m/s^{2} +9.8 m/s^{2} )

F=3.537\times10^{3}N

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Knowing this, let's begin with the answers:

<h2 /><h2>Answer A:</h2>

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Solving (1) with these values:

D=\frac{0.155g}{0.235mL} (2)

D=0.660g/mL (3)

<h2>Answer B:</h2>

In this case the mass of a sample is m=4.71g and its density is D=3.63g/mL.

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V=\frac{m}{D} (4)

V=\frac{4.71g}{3.63g/mL} (5)

V=1.297mL (5)

<h2>Answer C:</h2>

In this case the volume of a sample is V=0.293mL and its density is D=0.930g/mL.

Isolating m from (1):

m=D.V (6)

m=(0.930g/mL)(0.293mL) (7)

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