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zheka24 [161]
3 years ago
14

A car starts from rest, speeds up with constant acceleration, and travels 400 meters in 10 seconds. What was the acceleration of

the car in m/s2?
Physics
1 answer:
PIT_PIT [208]3 years ago
7 0

Answer:

Acceleration of the car will be a=8m/sec^2

Explanation:

We have given that car starts from rest so initial velocity of the car u = 0 m/sec

And car traveled 400 m in 10 sec

So distance traveled by car s = 400 m

Time taken to compete this distance t = 10 sec

We have to find the acceleration of the car

From second equation of motion we know that s=ut+\frac{1}{2}at^2

So 400=0\times 10+\frac{1}{2}\times a\times 10^2

a=8m/sec^2

So acceleration of the car will be a=8m/sec^2

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A piece of thin uniform wire of mass m and length 3b is bent into an equilaeral triangle.
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Hence, moment of inertia of two rods is as follows.

      M.I of two rods = 2 \times \frac{(\frac{m}{3} \times b^{2})}{3}

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As third rod have no connection with vertices. So, moment of inertia of a rod along an axis passing through its center is as follows.

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Using parallel axis theorem moment of inertia through vertices is as follows.

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        \frac{mb^{2}}{36} + [(\frac{m}{3}) \times 3\frac{b^{2}}{4}]

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Therefore, total moment of inertia is calculated as follows.

            Total M.I = \frac{2mb^{2}}{9} + \frac{5mb^{2}}{18}

                            = \frac{mb^{2}}{2}

Thus, we can conclude that the moment of inertia of the wire triangle about an axis perpendicular to the plane of the triangle and passing through one of its vertices is \frac{mb^{2}}{2}.

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4 years ago
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