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Artemon [7]
3 years ago
9

A roller coaster has a vertical loop with radius 25.7 m. With what minimum speed should the roller-coaster car be moving at the

top of
the loop so that the passengers do not lose contact with the seats?
m/s
Physics
1 answer:
gogolik [260]3 years ago
7 0

Answer:

15.88m/s

Explanation:

At the top of the roller coaster you will have three forces acting on the roller-coaster. See the image below. Fc is the centripetal force (for an object in circular motion), Fg is the gravitational force, and Fn is the normal force. To achieve the minimum speed we assume the roller-coaster is barely touching the vertical loop and so the normal force is zero. This leaves two acting forces.

F_g = F_c\\mg = \frac{m\times v^2}{r}\\v = \sqrt{gr} = \sqrt{9.81 \times 25.7}  = 15.88 m/s

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a stone is thrown by a person from the top of the building, which is 200m tall. at the same time, another stone is thrown with v
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Answer:

The time after which the two stones meet is tₓ = 4 s

Explanation:

Given data,

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The velocity of the stone thrown from foot of the building, U = 50 m/s

Using the II equation of motion

                             S = ut + ½ gt²

Let tₓ be the time where the two stones  meet and x be the distance covered from the top of the building

The equation for the stone dropped from top of the building becomes

                            x = 0 + ½ gtₓ²

The equation for the stone thrown from the base becomes

                S - x = U tₓ - ½ gtₓ²  (∵ the motion of the stone is in opposite direction)

Adding these two equations,

                      x + (S - x) = U tₓ

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3 years ago
A long string is wrapped around a 6.6-cm-diameter cylinder, initially at rest, that is free to rotate on an axle. The string is
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\omega_f=571.42\ rpm

Explanation:

It is given that,

Diameter of cylinder, d = 6.6 cm

Radius of cylinder, r = 3.3 cm = 0.033 m

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Displacement, d = 1.3 m

The angular acceleration is given by :

\alpha =\dfrac{a}{r}

\alpha =\dfrac{1.5}{0.033}

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The angular displacement is given by :

\theta=\dfrac{d}{r}

\theta=\dfrac{1.3}{0.033}

\theta=39.39\ rad

Using the third equation of rotational kinematics as :

\omega_f^2-\omega_i^2=2\alpha \theta

Here, \omega_i=0

\omega_f=\sqrt{2\alpha \theta}

\omega_f=\sqrt{2\times 45.46\times 39.39}

\omega_f=59.84\ rad/s

Since, 1 rad/s = 9.54 rpm

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\omega_f=571.42\ rpm

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