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nexus9112 [7]
3 years ago
14

A 0.250 kg block on a vertical spring with spring constant of 4.45 ✕ 103 N/m is pushed downward, compressing the spring 0.080 m.

When released, the block leaves the spring and travels upward vertically. How high does it rise above the point of release?
Physics
1 answer:
Elza [17]3 years ago
6 0

Answer:h=5.81 m

Explanation:

Given

Mass of block(m)=0.250 kg

Spring Constant k=4.45\times 10^3 N/m

Initial elongation =0.080 m=8 cm

Thus Initial Potential Energy stored =Final Potential Energy stored in Block

P_i=\frac{kx^2}{2}

P_i=\frac{4.45\times 10^3\times 64\times 10^{-4}}{2}=14.24 J

P_f=mgh=0.25\times 9.8\times h

P_i=P_f

14.24 =0.25\times 9.8\times h

h=\frac{14.24}{0.25\times 9.8}=5.81 m

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the resistance of the cable is 582.9 ohms

we are given the length of the cable which is  3 km,  of  1.5 mm in, the diameter  and resistivity of copper which is 1.72 m

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R = 19( ρl/A)

Here  ρ is the resisitivity =  1.72 , l is the length  = 3(1+0.05)*10³3= 3150 m

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To know more about resistance refer to the linkhttps://brainly.com/question/14547003?referrer=searchResults.

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