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AleksAgata [21]
3 years ago
13

According to the graph above, how large of a force is needed in order to stretch the string 1.00 meters?

Physics
2 answers:
gulaghasi [49]3 years ago
7 0

My guess for this one would be; 400 N

My reasoning would be; it starts at 0 on both X and Y, if you need to get to 1.00 meters thats 4/4. 1/4 of 1.00 is .25, and on .25 its on 100 so multiply it by 4 to make 1.00 and you get 400 N

MrMuchimi3 years ago
6 0

Answer:

The needed force is 400 N.

Explanation:

We need to calculate the force for 1.00 m string

According to graph,

The force applied for stretch from 0.10 to 0.15 is 20 N.

0.10-0.15 = 0.05 m

So, the force applied is 20 N for 0.05 m.

For stretch 0.05 m = 20 N

For stretch 1 m = \dfrac{20}{0.05}= 400\ N

Hence, The needed force is 400 N.

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4 0
3 years ago
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b) v_{min}=4.4m/s

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The net force on the car must produce the centripetal acceleration necessary to make this circle, which is a_{cp}=\frac{v^2}{R}. At the top of the circle, the normal force and the weight point downwards (like the centripetal force should), while at the bottom the normal force points upwards (like the centripetal force should) and the weight downwards, so we have (taking the upwards direction as positive):

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N_{top}=m\frac{v^2}{R}-mg=(1.2kg)\frac{(6m/s)^2}{2m}-(1.2kg)(9.8m/s^2)=9.8N\\N_{bottom}=m\frac{v^2}{R}+mg=(1.2kg)\frac{(6m/s)^2}{2m}+(1.2kg)(9.8m/s^2)=33.4N

The limit for falling off would be N_{top}=0, so the minimum speed would be:

0=m\frac{v_{min}^2}{R}-mg\\v_{min}=\sqrt{Rg}=\sqrt{(2m)(9.8m/s^2)}=4.4m/s

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