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Anna11 [10]
3 years ago
13

A roller coaster car is loaded with passengers and has a mass of 500 kg along with a speed of 18 meters/second at the dip. The r

adius of curvature of the track at the bottom point of the dip is 12 meters (gravity = 9.8 meters/second2). What force is exerted on the roller coaster car by the track at the bottom of the dip?
Physics
2 answers:
GenaCL600 [577]3 years ago
7 0
The roller coaster is moving in a circular path, so the force that must be computed is the centripetal force. The centripetal force is the force acting on an object undergoing circular motion and is directed towards the center of the circular path. This is computed using:
F = mv²/r
F = (500 * 18²) / 12
F = 13,500 N

Now, at the bottom of the track, the track is also supporting the weight of the car and its passengers, which is:
W = mg
W = 500 * 9.81
W = 4,905 N

The total reactive force exerted by the track to counter the centripetal force and the weight of the car is:
F = 13,500 + 4,905
F = 18,405 Newtons
Pachacha [2.7K]3 years ago
6 0

the answer is D on Plato.

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ValentinkaMS [17]

A) 2.03 m/s^2

Let's start by writing the equation of the forces along the directions parallel and perpendicular to the incline:

Parallel:

mg sin \theta - \mu_k R = ma (1)

where

m is the mass

g = 9.8 m/s^2 the acceleration of gravity

\theta=22.5^{\circ}

\mu_k = 0.19 is the coefficient of friction

R is the normal reaction

a is the acceleration

Perpendicular:

R-mg cos \theta =0 (2)

From (2) we find

R=mg cos \theta

And substituting into (1)

mg sin \theta - \mu_k mg cos \theta = ma

Solving for a,

a=g sin \theta - \mu_k g cos \theta = (9.8)(sin 22.5)-(0.19)(9.8)(cos 22.5)=2.03 m/s^2

B) 5.94 m/s

We can solve this part by using the suvat equation

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

Here we have

v = ?

u = 0 (it starts from rest)

a=2.03 m/s^2

s = 8.70 m

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0+2(2.03)(8.70)}=5.94 m/s

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Answer:

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Explanation:

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6 0
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A 50-kg man is driving his snowmobile at a constant speed. The engine applies a force of 360 N and the coefficient of kinetic fr
S_A_V [24]

Answer:

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Explanation:

The snowmobile is moving at constant speed, so the acceleration is 0.

Sum of the forces in the x direction:

∑F = ma

F − 360 = 0

F = 360

Friction force equals the normal force times the coefficient of friction:

Nμ = 360

On level ground, the normal force equals the combined weight:

(m + 50)g μ = 360

m + 50 = 360 / (gμ)

m = -50 + 360 / (gμ)

m = -50 + 360 / (9.8 × 0.2)

m ≈ 134 kg

Round as needed.

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Answer: friction

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