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Vinil7 [7]
3 years ago
6

using hooke's law, f spring = k triangle x, find the elastic constant of a spring that stretches 2 cm when a 4 newton force is a

pplied to it.
Physics
1 answer:
Ksivusya [100]3 years ago
8 0

As we know that spring force is given as

F = kx

here we know that

F = 4 N

x = 2 cm = 0.02 m

now from the above equation we will have

4 = k(0.02)

k = 200 N/m

so the elastic constant of the spring will be 200 N/m

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Two blocks are connected by a massless cable which goes through the center of a rotating turntable. The blocks have masses M1 =
Airida [17]

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If the final question is; at what velocity will the first block start to move outward in m/s?

v = 3.5596 \frac{m}{s}

Explanation:

The motion have the velocity that will make the block move using:

F_{1}*F_{r}+ F_{2}= Ec \\Ec= \frac{M*v^{2} }{r}

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μs= 0,54

Resolving:

\frac{m_{1}*v^{2}  }{r} = us*m_{1}* g + m_{2} * g

v^{2} = \frac{((us *m_{1} + m_{2})*g )* r}{m_{1} }

v^{2} = \frac{((0.54 *1.2 + 2.8)*9.8 )* 0.45}{1.2}

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3 years ago
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Answer:

v = 50.5 m/s

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F = (m)(^v/^t)

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v = 50.53834322 m/s

v = 50.5 m/s

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wlad13 [49]

Answer:

(A) Impulse = 9Ns

(B) F = 1286N

Explanation:

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F = Impulse/ t

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