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3 years ago

6

A bus starting from rest moves with a uniform acceleration for 2 minutes. If the velocity at the end of 2 minutes is 12 m/s, its

acceleration is *
(a) 6 m/s²
(b) 0.1 m/s²
(c) 1 m/s²
(d) 0.6 m/s²

1 answer:

Andreyy893 years ago

7 0

Answer:

0.1m/s²

Explanation:

as we know that

acceleration=velocity/time taken

a=12/120

a=0.1m/s²

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2 years ago

a body of a mass 1kg suspended from a spring is found to stretch the spring by 10cm. (A) what is the spring constant? what is pe

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Answer:

(A) The spring constant is <u>98 N/m.</u>

(B) The period of oscillation is <u>0.635 s.</u>

Explanation:

Given:

Mass of the body is, $m=1\ kg$

Extension length of the spring is, $x=10\ cm=0.1\ m$

Now, let 'k' be the spring constant.

The force acting on the body is due to gravity only and is equal to its weight.

So, weight of the body is given as:

$F_g=mg\\F_g=1\times 9.8\\F_g=9.8\ N$

Now, we know that for a spring-mass system, the net force acting on the body is equal to the product of the spring constant and extension length. Therefore,

$F_g=kx\\9.8=k(0.1)\\k=\frac{9.8}{0.1}=98\ N/m$

Hence, the spring constant is 98 N/m.

(B)

Period of oscillation of a body of spring-mass system in SHM is given as:

$T=2\pi\sqrt{\frac{m}{k}}$

Plug in the given values and solve for period 'T'. This gives,

$T=2\pi\sqrt{\frac{1}{98}}\\T=2\pi\times 0.101\\T=0.635\ s$

Therefore, the period of oscillation is 0.635 s.

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3 years ago