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Bumek [7]
2 years ago
14

You are riding a rollercoaster going around a vertical loop, on the inside of the loop. If the loop has a radius of 30 meters, h

ow fast must the cart be moving in order for you to feel three times as heavy at the top of the loop
Physics
2 answers:
Brilliant_brown [7]2 years ago
8 0

Answer:

v = 24.2m/s

Explanation:

Given R = 30m

The two forces acting on you while on the roller coaster ride are the normal force and your weight.

By Newton's second law

N – W = mv²/R

For you to feel 3 time as heavy, the normal force of the seat of the roller coaster on you must be 3 times your weight. That is

N = 3×W = 3×mg

W = mg

3mg – mg = mv²/R

2mg = mv²/R

2g = v²/R

v² = 2gR

v = √2gR

v = √(2×9.8×30)

v = 24.2m/s

Debora [2.8K]2 years ago
3 0

Answer:

Speed of the cart at the top of the loop = 34.3 m/s

Explanation:

Gravitational acceleration = g = 9.81 m/s2

Your mass = m

You feel three times as heavy at the top of the loop.

Normal force on you = N = 3mg

Radius of the loop = R

Speed of the cart = V

Centripetal force required for the circular motion = Fc

F = m

The centripetal force is provided by the normal force on you which is directed downwards and your own weight which is directed downwards.

Fc = mg + N

Fc = mg + 3mg

Fc = 4mg

m12 -= 4mg R

V = 4gR

V = 4(9.81)(30)

V = 34.3 m/s

Speed of the cart at the top of the loop = 34.3 m/s

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barxatty [35]

Answer:

2.5\cdot 10^{-4}

Explanation:

The strain is defined as the ratio of change of dimension of an object under a force:

S=\frac{\Delta L}{L_0}

where

\Delta L is the change in length of the object

L_0 is the original length of the object

In this problem, we have L_0 = 4 m and \Delta L=1 mm=0.001 m, therefore the strain is

S=\frac{\Delta L}{L_0}=\frac{0.001 m}{4 m}=2.5\cdot 10^{-4}


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2 years ago
A heavy stone of mass m is hung from the ceiling by a thin 8.25-g wire that is 65.0 cm long. When you gently pluck the upper end
Triss [41]

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Explanation:

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v= \sqrt{\frac{Tension}{Linear. Mass. density} }

Here, Tension= m*g = m*9.81

and linear mass density= \frac{8.25 g}{65 cm}

Linear mass density= \frac{8.25*10^-3}{65*10^-2}

Linear mass density= 0.0127 kg/m

Velocity= 2*\frac{l}{t}

Velocity= 2 * \frac{65*10^-2}{7.84}

Velocity= 165.8 m/s

So putting all these values in equation we get

v= \sqrt{\frac{Tension}{Linear. Mass. density} }

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3 0
2 years ago
The U.S. Department of Energy had plans for a 1500-kg automobile to be powered completely by the rotational kinetic energy of a
navik [9.2K]

Answer:

230

Explanation:

\omega = Rotational speed = 3600 rad/s

I = Moment of inertia = 6 kgm²

m = Mass of flywheel = 1500 kg

v = Velocity = 15 m/s

The kinetic energy of flywheel is given by

K=\dfrac{1}{2}I\omega^2\\\Rightarrow K=\dfrac{1}{2}6\times 3600^2\\\Rightarrow K=38880000\ J

Energy used in one acceleration

K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}1500\times 15^2\\\Rightarrow K=168750\ J

Number of accelerations would be given by

n=\dfrac{38880000}{168750}\\\Rightarrow n=230.4

So the number of complete accelerations is 230

8 0
2 years ago
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