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Margarita [4]
2 years ago
8

Suppose that air resistance cannot be ignored. For the position at which the person has jumped from the platform and the cord re

aches its maximum stretch length, how does the stretch length of the cord in the case with air resistance compare with the stretch length in the case in which air resistance is ignored? Briefly state your reasoning.
Physics
1 answer:
Kamila [148]2 years ago
6 0

Answer:

The stretch cord stores potential energy as a result of stretching but due to kinetic energy, it will move back to its original state. Since air resistance is not being ignored in this case, it will experience a slight delay in stretching at first.

Explanation:

In case, where air resistance is being ignored the stretch cord will stretch as it normally does.              

  • Air resistance is a force that any object experiences as a result of its motion through the air.  

There are various factors that affect air resistance like speed, the density of air, area, the shape of an object etc. Meanwhile, the density of air changes with temperature or altitude. <em>Hence this force is not constant but is thought to be constant during short time frames.  </em>

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A 0.200-kg mass attached to the end of a spring causes it to stretch 5.0 cm. If another 0.200-kg mass is added to the spring, th
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Answer:

A: 4 times as much

B: 200 N/m

C: 5000 N

D: 84,8 J

Explanation:

A.

In the first question, we have to caculate the constant of the spring with this equation:

m*g=k*x

Getting the k:

k=\frac{m*g}{x} =\frac{0,2[kg]*9,81[\frac{m}{s^{2} } ]}{0,05[m]} =39,24[\frac{N}{m}]

Then we can calculate how much the spring stretch whith the another mass of 0,2kg:

x=\frac{m*g}{k} =\frac{0,4[kg]*9,81[\frac{m}{s^{2} } ]}{39,24[\frac{N}{m}]} =0,1[m]\\

The energy of a spring:

E=\frac{1}{2}*k*x^{2}

For the first case:

E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,05[m])^{2} =0,049 [J]

For the second case:

E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,1[m])^{2} =0,0196 [J]

If you take the relation E2/E1 = 4.

B.

We have the next facts:

x=0,005 m

E = 0,0025 J

Using the energy equation for a spring:

E=\frac{1}{2}*k*x^{2}⇒k=\frac{E*2}{x^{2} } =\frac{0,0025[J]*2}{(0,005[m])^{2} } =200[\frac{N}{m} ]

C.

The potential energy of the diver will be equal to the kinetic energy in the moment befover hitting the watter.

E=W*h=500[N]*10[m]=5000[J]

Watch out the units in this case, the 500 N reffer to the weighs of the diver almost relative to the earth, thats equal to m*g.

D.

The work is equal to the force acting in the direction of the motion. so we have to do the diference beetwen angles to obtain the effective angle where the force is acting: 47-15=32 degree.

The force acting in the direction of the ramp will be the projection of the force in the ramp, equal to F*cos(32). The work will be:

W=F*d=F*cos(32)*d=10N*cos(32)*10m=84,8J

7 0
2 years ago
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