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timama [110]
3 years ago
8

A car starts to move from rest and covers a distance of 360m in one minute. Calculate the acceleration of the car.

Physics
1 answer:
Romashka-Z-Leto [24]3 years ago
4 0
<h2>The acceleration of car is 0.2 ms⁻²</h2>

Explanation:

When the car moves , the distance covered is calculated by the relation

S = u t + \frac{1}{2} a t²

In this question u = 0 , because car was at rest initially

Thus S =  \frac{1}{2} a t²

here S is displacement and a is the acceleration of car

Therefore  360 =  \frac{1}{2} a ( 60 )²

Because time taken is one minute or 60 seconds

Therefore a = \frac{360x2}{3600}

or a = 0.2 m s⁻²

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A man drops a penny V=0 off the top of the Golden Gate Bridge how fast will the penny be moving when it hits the ground? The gol
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Answer:

Vf = 73.4 m/s

Explanation:

This is the case of vertical motion where we have to find the final velocity of the penny when it hits the ground. We can use 3rd equation of motion to find the final velocity:

2gh = Vf² - Vi²

where,

g = 9.8 m/s²

h = height = 275 m

Vf = Final Velocity = ?

Vi = Initial Velocity = 0 m/s

Therefore,

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<u>Vf = 73.4 m/s</u>

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What is the mass of an object that requires a force of 30 N to accelerate at a rate of 5 m/sec2 ?
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3 years ago
A 4.4 kg mess kit sliding on a frictionless surface explodes into two 2.2 kg parts, one moving at 2.9 m/s, due north, and the ot
sp2606 [1]

Answer:

Original speed of the mess kit = 4.43 m/s at 50.67° north of east.

Explanation:

Let north represent positive y axis and east represent positive x axis.

Here momentum is conserved.

Let the initial velocity be v.

Initial momentum = 4.4 x v = 4.4v

Velocity of 2.2 kg moving at 2.9 m/s, due north = 2.9 j m/s

Velocity of 2.2 kg moving at 6.8 m/s, 35° north of east = 6.9 ( cos 35i + sin35 j ) = 5.62 i + 3.96 j m/s

Final momentum = 2.2 x 2.9 j + 2.2 x (5.62 i + 3.96 j) = 12.364 i + 15.092 j kgm/s

We have

         Initial momentum = Final momentum

         4.4v = 12.364 i + 15.092 j

         v =2.81 i + 3.43 j

Magnitude

        v=\sqrt{2.81^2+3.43^2}=4.43m/s

Direction

       \theta =tan^{-1}\left ( \frac{3.43}{2.81}\right )=50.67^0

       50.67° north of east.

Original speed of the mess kit = 4.43 m/s at 50.67° north of east.

6 0
3 years ago
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