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krek1111 [17]
3 years ago
6

It took 1700 J of work to stretch a spring from its natural length of 2 m to a length of 6 m. Find the spring constant k.

Physics
1 answer:
algol133 years ago
4 0

Answer:

212.5 N/m

Explanation:

Work, W = 1700 J

y1 = 2 m

y2 = 6 m

y = y2 - y1 = 6 - 2 = 4 m

Let K be the spring constant.

Work done is given by

W = 1/2 Ky²

1700 = 0.5 x K x 4 x 4

K = 212.5 N/m

Thus, the spring constant is given by 212.5 N/m.

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Answer:

12 kgm²

Explanation:

here angular acceleration = 10rad/sec²

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moment of inertia=?

we know,

torque= angular acceleration× moment of Inertia

or, moment of inertia = torque/angular acceleration

= 120/10

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6 0
3 years ago
Three people pull simultaneously on a stubborn donkey. jack pulls eastward with a force of 92.5 n, jill pulls with 89.9 n in the
alekssr [168]
Jack------------ force of 92.5 n   eastward-------Fjack(X)=92.5 n   Fjack(Y)=0

<span>jill ------------------------------- force of 89.9 n   northeast
Fjill(X)=cos45*89.9=63.57 n
</span>Fjill(Y)=sin45*89.9=63.57 n<span>

</span>jane -----------------------------force of 163 n   southeast
Fjane(X)=cos45*163=115.26 n
Fjane(X)=-sin45*163=-115.26 n

Ftotal (X)=92.5+63.57+115.26=271.33 n
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the magnitude of the net force the people exert on the donkey. is 294.80 n southeast
7 0
3 years ago
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