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3 years ago

Missy's favorite ride at the Topsfield Fair is the rotor, which has a radius of 4.0 m. The ride

Physics

1 answer:

dolphi86 [110]3 years ago

8 0

Answer:

a. 12.57 m/s b. 39.5 m/s² c. Her centripetal force is four times her weight.

Explanation:

a. What is Missy's linear speed on the rotor?

Missy's linear speed v = 2πr/T where r = radius = 4.0 m and T = time it takes to complete one revolution = 2.0 s

So, v = 2πr/T

= 2π(4.0 m)/2.0 s

= 4π m/s

= 12.57 m/s

b. What is Missy's centripetal acceleration on the rotor?

Missy's centripetal acceleration, a = v²/r where v = linear velocity = 12.57 m/s and r = radius = 4.0 m

a = v²/r

= (12.57 m/s)²/4.0 m

= 158.01 m²/s² ÷ 4.0 m

= 39.5 m/s²

c. If her mass is 50-Kg, how is the centripetal force compare to her weight?

Her centripetal force F = ma where m = mass = 50 kg and a = centripetal acceleration = 39.5 m/s².

Her weight W = mg where m = mass = 50 kg and g = acceleration due to gravity = 9.8 m/s².

So, comparing her centripetal force to her weight, we have

F/W = ma/mg

= a/g

= 39.5 m/s² ÷ 9.8 m/s²

= 4.03

≅ 4

So her centripetal force is four times her weight.

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3 years ago

A 20-foot ladder is leaning against the wall. If the base of the ladder is sliding away from the wall at the rate of 3 feet per

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Differentiating the above equation with respect to time, 't'. This gives,

$\frac{d}{dt}(h^2+b^2)=\frac{d}{dt}(400)\\\\\frac{d}{dt}(h^2)+\frac{d}{dt}(b^2)=0\\\\2h\frac{dh}{dt}+2b\frac{db}{dt}=0\\\\h\frac{dh}{dt}+b\frac{db}{dt}=0--------2$

In the above equation the term $\frac{dh}{dt}$ is the rate at which top of ladder slides down and $\frac{db}{dt}$ is the rate at which bottom of ladder slides away.

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Therefore, the rate at which the top of ladder slide down is 6.87 ft/s. The negative sign implies that the height is reducing with time which is true because it is sliding down.

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4 years ago

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