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tia_tia [17]
3 years ago
10

A student places her 490 g physics book on a frictionless table. She pushes the book against a spring, compressing the spring by

7.10 cm , then releases the book. What is the book's speed as it slides away? The spring constant is 1550 N/m .
Physics
1 answer:
Paul [167]3 years ago
5 0

Answer:

book speed is 3.99 m/s

Explanation:

given data

mass m = 490 g = 0.490 kg

compressing x = 7.10 cm = 0.0710 m

spring constant k = 1550 N/m

to find out

book speed

solution

we know energy is conserve so

we can say

loss in spring energy is equal to gain in kinetic energy

so

\frac{1}{2}*k*x^2 = \frac{1}{2}*m*v^2    ..................1

put here value

\frac{1}{2}*1550*0.071^2 = \frac{1}{2}*0.490*v^2

v = 3.99 m/s

so book speed is 3.99 m/s

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Answer:

I = 0.96 A

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I=\dfrac{q}{t}

q = ne (Quantization of electric charge)

I=\dfrac{ne}{t}\\\\I=\dfrac{1.8\times 10^{16}\times 1.6\times 10^{-19}}{3\times 10^{-3}}\\\\I=0.96\ A

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3 years ago
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Answer:

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Explanation:

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Answer:

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