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3 years ago

8

In the amusement park ride known as Magic Mountain Superman, powerful magnets accelerate a car and its riders from rest to 43.4

m/s in a time of 8.59 s. The mass of the car and riders is 3.00 × 10^3 kg. Find the average net force exerted on the car and riders by the magnets.

1 answer:

Anton [14]3 years ago

5 0

Answer:

Average net force, F = 15157.15 N

Explanation:

It is given that,

The mass of the car and riders is, $m=3\times 10^3\ kg$

Initial speed of the car, u = 0

Final speed of the car, v = 43.4 m/s

Time, t = 8.59 seconds

We need to find the average net force exerted on the car and riders by the magnets. It can be calculated using second law of motion as :

F = m a

$F=m(\dfrac{v-u}{t})$

$F=3\times 10^3\ kg\times (\dfrac{43.4\ m/s-0}{8.59\ s})$

F = 15157.15 N

So, the average net force exerted on the car and riders by the magnets. Hence, this is the required solution.

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When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius

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To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

$\tau = I\alpha$

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I = Moment of inertia $\rightarrow \frac{1}{2}mr^2$ For a disk

$\alpha =$ Angular acceleration

The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

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

Where

$\omega_{f,i} =$ Final and Initial Angular velocity

$\alpha =$ Angular acceleration

$\theta =$ Angular displacement

Our values are given as

$\omega_i = 0 rad/s$

$\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega_f = 47.12rad/s$

$\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad$

$r = 7cm = 7*10^{-2}m$

$m = 17g = 17*10^{-3}kg$

Using the expression of angular acceleration we can find the to then find the torque, that is,

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

$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}$

$\alpha = \frac{47.12^2-0^2}{2*6\pi}$

$\alpha = 58.89rad/s^2$

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

$\tau = I\alpha$

$\tau = (\frac{1}{2}mr^2)\alpha$

$\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)$

$\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m$

Therefore the torque exerted on it is $2.45*10^{-3}N\cdot m$

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