A wheel 31 cm in diameter accelerates uniformly from 250 rpm to 370 rpm in 7.0 s. How far will a point on the edge of the wheel have traveled in this time?

Answers

Answer 1

A wheel 31 cm in diameter accelerates uniformly from 250 rpm to 370 rpm in 7.0 s. A point on the edge of the wheel will have traveled approximately 196.218 cm in 7.0 seconds.

To calculate the distance traveled by a point on the edge of the wheel, we need to find the circumference of the wheel and then multiply it by the number of revolutions it completes in the given time.

The diameter of the wheel is given as 31 cm, which means the radius (r) of the wheel is half of the diameter:

r = 31 cm / 2 = 15.5 cm.

The circumference of the wheel can be calculated using the formula

C = 2πr.

Plugging in the radius value, we have:

C = 2π(15.5 cm).

Now, let's calculate the initial and final distances traveled by a point on the edge of the wheel.

Initial distance: The initial speed of the wheel is given as 250 revolutions per minute (rpm). To convert it to revolutions per second, we divide by 60:

250 rpm / 60 s = 4.17 revolutions per second.

Therefore, the initial distance traveled is:

Initial distance = 4.17 revolutions * C.

Final distance: The final speed of the wheel is given as 370 rpm. Converting it to revolutions per second:

370 rpm / 60 s = 6.17 revolutions per second.

Hence, the final distance traveled is:

Final distance = 6.17 revolutions * C.

To find the total distance traveled, we subtract the initial distance from the final distance:

Total distance = final distance - initial distance.

Now, let's calculate the values:

C = 2π(15.5 cm) = 97.4 cm (approx.)

Initial distance = 4.17 revolutions * 97.4 cm = 405.58 cm (approx.)

Final distance = 6.17 revolutions * 97.4 cm = 601.798 cm (approx.)

Total distance = 601.798 cm - 405.58 cm ≈ 196.218 cm.

Therefore, a point on the edge of the wheel will have traveled approximately 196.218 cm in 7.0 seconds.

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Related Questions

A ball dropped and accelerates downwards at a rate of 10m/s^2 for 15 seconds how much will the balls velocity increase

Answers

Answer:

v = 150 m/s

Explanation:

Given that,

The acceleration of the ball, a = 10 m/s²

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v = 150 m/s

So, the ball's final velocity is 150 m/s.

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10 POINTS and WILL MARK AS A Brainlest PLEASE

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240 cm/sec

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The speed line is closest to 240sec at 60cm.

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ears

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Can u please help mee?
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a A car of mass 1000 kg traveling at 31.6 m/s
b A train with a mass of 120 000 kg and a velocity of 40 m/s (about 90 mph)

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a

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Answer: the last one

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C

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what is mass in physics​

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Answer:

Mass, in physics, quantitative measure of inertia, a fundamental property of all matter. It is, in effect, the resistance that a body of matter offers to a change in its speed or position upon the application of a force. The greater the mass of a body, the smaller the change produced by an applied force.

Explanation:

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Answer:

Mass, in physics, quantitative measure of inertia, a fundamental property of all matter. It is, in effect, the resistance that a body of matter offers to a change in its speed or position upon the application of a force. The greater the mass of a body, the smaller the change produced by an applied force.

Explanation:

Hope that helps! :)

a effort of 100n can raise a load of 2000n in a hydraulic press. calculate the cross-sectional area of a small piston in it. The cross-sectional area of a large piston is 4m^s​

Answers

Answer:

[tex]A_{1}[/tex] = 0.2 [tex]m^{2}[/tex]

Explanation:

The pressure on the pistons is given as;

Pressure = [tex]\frac{Force}{Area}[/tex]

So that,

Pressure on the small piston = [tex]\frac{F_{1} }{A_{1} }[/tex] and Pressure on the large piston = [tex]\frac{F_{2} }{A_{2} }[/tex]

Thus,

[tex]\frac{F_{1} }{A_{1} }[/tex] = [tex]\frac{F_{2} }{A_{2} }[/tex]

Given that: [tex]F_{1}[/tex] = 100 N, [tex]F_{2}[/tex] = 2000 N, [tex]A_{2}[/tex] = 4 [tex]m^{2}[/tex].

[tex]\frac{100}{A_{1} }[/tex] = [tex]\frac{2000}{4}[/tex]

[tex]A_{1}[/tex] = [tex]\frac{100*4}{2000}[/tex]

    = [tex]\frac{400}{2000}[/tex]

    = 0.2

[tex]A_{1}[/tex] = 0.2 [tex]m^{2}[/tex]

The area of the small piston is 0.2 [tex]m^{2}[/tex].

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The answer is B. 8000.

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Answer:

12 neutrons

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shadow is a area of darkness

will mark as brainliest !! plz answer it
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False

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An earthquake is a sudden shaking movement of the surface of the earth. It is known as a quake, tremblor or tremor. Earthquakes can range in size from those that are so weak that they cannot be felt to those violent enough to toss people around and destroy whole cities. ... An earthquake is measured on Richter's scale.

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Online dictionary

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c

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Answers

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Answer:

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