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Sergio039 [100]
2 years ago
8

Use a variation model to solve for the unknown value. Use as the constant of variation. The stopping distance of a car is direct

ly proportional to the square of the speed of the car. (a) If a car travelling has a stopping distance of , find the stopping distance of a car that is travelling . (b) If it takes for a car to stop, how fast was it travelling before the brakes were applied
Physics
1 answer:
svlad2 [7]2 years ago
6 0

Complete question is;

Use a variation model to solve for the unknown value.

The stopping distance of a car is directly proportional to the square of the speed of the car.

a. If a car traveling 50 mph has a stopping distance of 170 ft, find the stopping distance of a car that is traveling 70 mph.

b. If it takes 244.8 ft for a car to stop, how fast was it traveling before the brakes were applied?

Answer:

A) d = 333.2 ft

B) 60 mph

Explanation:

Let the stopping distance be d

Let the speed of the car be v

We are told that the stopping distance is directly proportional to the square of the speed of the car. Thus;

d ∝ v²

Therefore, d = kv²

Where k is constant of variation.

A) Speed is 50 mph and stopping distance of 170 ft.

v = 50 mph

d = 170 ft = 0.032197 miles

Thus,from d = kv², we have;

0.032197 = k(50²)

0.032197 = 2500k

k = 0.032197/2500

k = 0.0000128788

If the car is now travelling at 70 mph, then;

d = 0.0000128788 × 70²

d = 0.06310612 miles

Converting to ft gives;

d = 333.2 ft

B) stopping distance is now 244.8 ft

Converting to miles = 0.046363636 miles

Thus from d = kv², we have;

0.046363636 = 0.0000128788(v²)

v² = 0.046363636/0.0000128788

v² = 3599.99658

v = √3599.99658

v ≈ 60 mph

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How to measure the volume of a baseball bat ( need answers ASAP )
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<em>Measure the amount of water it displaces.</em>

This won't be easy, because the bat floats in water.  But I think you can get around that little problem like this:

-- Get some kind of a tank or tub that's big enough to hold the whole bat under water.

-- Get a heavy weight, like a big wrench or a small rock.  

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-- Slip the heavy weight into the tub, slowly.  Some water will run over the top and out of the tub.  That's OK ... it's exactly what you want.  If NO water runs over the top, pour some more in, until it runs out and then stops.  You want the tub full to the brimmy rim with the rock at the bottom of it.

-- Take the heavy weight out of the tub.

-- Now set the tub into a bigger tub or a deep pan.  The next time it overflows and some water runs out of it, you'll need to catch that water and measure it.

-- Get a short piece of heavy string.  Tie the heavy weight to somewhere near the middle of the bat.

-- Slowly slide the bat into the water, with the rock tied to it.  The bat needs to go complete underwater.

-- Some more water will run over the top and out of the tub, and INTO the lower tub.  Wait until the overflow stops and everything settles down again.

-- Take the bat (tied to the weight) out of the tub.  Slowly and carefully, so that your hand or your arm doesn't make any MORE water run over and out.

-- Lift the upper tub out of the lower tub.

-- Take the lower tub, with the overflow water in it.  Using a kitchen measuring cup, or a saucepan or a bottle, or anything else with liquid amounts marked on it, measure how much water overflowed into the lower tub.

THAT amount is the volume of the bat.

You may have to do some units conversions.  Like if you need the volume of the bat in cm³ and you used measuring vessels marked in fluid ounces.  But you can find all those conversion factors with a search on Floogle.

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3 years ago
Assume it takes 10 J to stretch a spring 10 cm beyond its natural length. Find the work required (in Joules) to stretch the spri
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Answer:

W = 30 J

Explanation:

given,

Work done = 10 J

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We know,

dW = F .dx

we know, F = k x

\int dW = \int_0^{0.1} k.x dx

W = \int_0^{0.1} k.x dx

W = k[\dfrac{x^2}{2}]_0^{0.1}

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k = 2000

now, calculating Work done by the spring when it stretched to 0.2 m from 0.1 m.

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W = 2000 [\dfrac{x^2}{2}]_{0.1}^{0.2} dx

W = 1000 x 0.03

W = 30 J

Hence, work done is equal to 30 J.

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