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2 years ago

Fan object moves in uniform circular motion in a circle of radius R=200 meters, and the objectes 5.00 seconds to

Physics

1 answer:

Leviafan [203]2 years ago

6 0

Answer:

The centripetal acceleration of the object is $31550.72\ m/s^2$ .

Explanation:

We have,

Radius of a circular path is 200 m

It takes 5 seconds to complete 10 revolutions. The angular velocity of the object is given by the rate of change of angular displacement per unit time :

$\omega=\dfrac{\Delta \theta}{\Delta t}\\\\\omega=\dfrac{2\pi \times 10}{5}\\\\\omega=12.56\ rad/s$

The centripetal acceleration of the object is given by :

$a=\omega^2 r\\\\a=(12.56)^2 \times 200\\\\a=31550.72\ m/s^2$

So, the centripetal acceleration of the object is $31550.72\ m/s^2$ .

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Answer:

According to Newton's third law, for every action force there is an equal (in size) and opposite (in direction) reaction force. Together, these two forces exerted upon two different objects form the action-reaction force pair.

Explanation:

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2 years ago

You are designing a 108 cm3 right circular cylindrical can whose manufacture will take waste into account. There is no waste in

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Explanation:

It is given that,

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$V=\pi r^2 h$

$108=\pi r^2 h$

$h=\dfrac{108}{\pi r^2}$ ............(1)

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$A=32r^2+2\pi r\times \dfrac{108}{\pi r^2}$

$A=32r^2+\dfrac{216}{r}$

For maximum area, differentiate above equation wrt r such that, we get :

$\dfrac{dA}{dr}=64r-\dfrac{216}{r^2}$

$64r-\dfrac{216}{r^2}=0$

$r^3=\dfrac{216}{64}$

r = 1.83 m

Dividing equation (1) with r such that,

$\dfrac{h}{r}=\dfrac{108}{\pi r}$

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2 years ago

A car maintains a constant speed v as it traverses the hill and valley as shown below. Both the hill and valley have a radius of

Zigmanuir [339]

Answer:

As given that the car maintains a constant speed v as it traverses the hill and valley where both the valley and hill have a radius of curvature R.

(i) At point C, the normal force acting on the car is largest because the centripetal force is up. gravity is down and normal force is up. net force is up so magnitude of normal force must be greater than the car's weight.

(ii) At point A, the normal force acting on the car is smallest because the centripetal force is down. gravity is down and normal force is up. net force is up so magnitude of normal force must be less than car's weight.

(iii) At point C, the driver will feel heaviest because the driver's apparent weight is the normal force on her body.

(iv) At point A, the driver will feel the lightest.

(v)The car can go that much fast without losing contact with the road at A can be determined as follow:

Fn=0 - lose contact with road

Fg= mv²/r

mg=mv²/r

v=sqrt (gr)

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2 years ago

When the velocity of an object Changes it is acted upon by a

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When the velocity of an object changes, it is acted upon by a force

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2 years ago

Light incident on a glass sheet is partly reflected and partly refracted. How is the reflected ray different from the refracted

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Answer:

E. The refracted ray is vertically polarized whereas the reflected ray is horizontally polarized.

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2 years ago