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lesantik [10]
3 years ago
5

Someone please please help me with this.

Physics
1 answer:
Softa [21]3 years ago
4 0
600 g=.6 kg
mgh=pe
.6(10)(1)=6
therefore, the basketball will have roughly 6 joules of ke.
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In part one of this experiment, a 0.20 kg mass hangs vertically from a spring and an elongation below the support point of the s
statuscvo [17]

To solve this problem it is necessary to apply the concepts related to Hooke's Law as well as Newton's second law.

By definition we know that Newton's second law is defined as

F = ma

m = mass

a = Acceleration

By Hooke's law force is described as

F = k\Delta x

Here,

k = Gravitational constant

x = Displacement

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The force to keep in balance must be preserved, so the force by the weight stipulated in Newton's second law and the force by Hooke's elongation are equal, so

k\Delta x = mg

So for state 1 we have that with 0.2kg there is an elongation of 9.5cm

k (9.5-l)=0.2*g

k (9.5-l)=0.2*9.8

For state 2 we have that with 1Kg there is an elongation of 12cm

k (12-l)= 1*g

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We have two equations with two unknowns therefore solving for both,

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k = 313.6N/m

Therefore the spring constant is 313.6N/m

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3 years ago
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