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3 years ago

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A car is traveling at 19m/s. It slows to a stop at a constant rate over 4.2 seconds. How far did the car travel during the 4.2se

conds it took to stop?

1 answer:

svetlana [45]3 years ago

5 0

The distance covered by the car is 39.9 m

Explanation:

Since the car is slowing at constant rate, it means that its acceleration is constant, therefore we can solve the problem by using the following suvat equation:

$s=(\frac{u+v}{2})t$

where

s is the distance covered

u is the initial velocity

v is the final velocity

t is the time elapsed

For the car in this problem, we have

u = 19 m/s (initial velocity)

v = 0 (final velocity)

t = 4.2 s (time elapsed)

Therefore, the distance covered is

$s=(\frac{19+0}{2})(4.2)=39.9 m$

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

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Four point masses of 3.0 kg each are arranged in a square on masslessrods. The length of a side of the square is 0.50m. What is

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Answer:

Part a)

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Part b)

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Part c)

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Part a)

Moment of inertia of the system about an axis passing through B and C is given as

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Part b)

Moment of inertia of the system about an axis passing through A and C is given as

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$I = 0.75 kg m^2$

Part c)

Moment of inertia of the system about an axis passing through the center of the square and perpendicular to the plane of the square

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$I = 4m\frac{L^2}{2}$

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3 years ago

What will occur when the trough of wave A overlaps the trough of wave B?

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3 years ago

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Answer:

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Now this causes a centripetal acceleration

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If accelerating potential V is same and then kinetic energy equals the potential energy difference

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$v = \sqrt{(2Vq/m)}$ put these value in equation (ii)

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simplifying we get

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for same location r will be same in both case

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$\frac{B_{10}}{B_{7}} = \sqrt{\frac{m_{10}}{{m_7}} }$

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