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Marysya12 [62]
3 years ago
11

In the parallel spring system, the springs are positioned so that the 37 N weight stretches each spring equally. The spring cons

tant for the left-hand spring is 2.7 N/cm and the spring constant for the righ-hand spring is 4.3 N/cm. How far down will the 37 N weight stretch the springs?
Physics
1 answer:
Reptile [31]3 years ago
3 0

Answer:

x = 5.29 m

Explanation:

given,

weight of stretch = 37 N

left-hand spring constant (k₁)= 2.7 N/cm

right hand spring constant(k₂)= 4.3 N/ cm

spring are connected in parallel

F = F₁ + F₂

F = k₁x + k₂x

F = (k₁+ k₂)x

37= (4.3+ 2.7)x

7 x = 37

x = 5.29 m

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Answer:

40N

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The minimum speed of the water must be 3.4 m/s

Explanation:

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Since the water is in circular motion, the net force must be equal to the centripetal force, so:

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When this occurs, the equation becomes:

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