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3 years ago

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Stuntman's Freefall, a ride at Six Flags Great Adventure in New Jersey, stands 39.6 meters high. Ignoring the force of friction,

what is the minimum power rating of the motor that raises the 1.20×105 kg ride from the ground to the top in 10.0 seconds at a constant velocity?

1 answer:

Artyom0805 [142]3 years ago

4 0

Answer:

Power, $P=4.65\times 10^6\ watts$

Explanation:

It is given that,

Mass, $m=1.2\times 10^5\ kg$

Height of the free fall, h = 39.6 m

Time taken, t = 10 s

Let P is the power rating of the motor. The energy per unit time is called power delivered to an object. Its formula is given by :

$P=\dfrac{W}{t}$

$P=\dfrac{mgh}{t}$

$P=\dfrac{1.2\times 10^5\ kg\times 9.8\ m/s^2\times 39.6\ m}{10\ s}$

P = 4656960 Watts

$P=4.65\times 10^6\ watts$

So, the power rating of the motor is $4.65\times 10^6\ watts$ . Hence, this is the required solution.

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Answer:

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Explanation:

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A block of amber is placed in water and a laser beam travels from the water through the amber. The angle of incidence is 35 degr

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Answer: 1.88

Explanation

Applying Snell’s Law, sin(1)/sin(2) = n(2)/n(1), where n is the index of refraction and sin 1 and 2 being of incidence and refracted respectively.

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2) 1.41 = n(2)/1.33
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How are volcanoes distributed

Colt1911 [192]

Answer:

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3 years ago

A car moving with a speed of 35 m/s sees a child standing in the

zlopas [31]

Answer:

2100 N

Explanation:

$v = u + at \\ 0 = 35 + a(5) \\ a = - 7m {s}^{ - 2} \\ \\ net \: force \: = ma \\ = (300)(7) \\ = 2100 \: newtons$

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3 years ago

Sort the forces as producing a torque of positive, negative, or zero magnitude about the rotational axis identified in part

Fantom [35]

a) Angular acceleration: $17.0 rad/s^2$

b) Weight: conterclockwise torque, reaction force: zero torque

Explanation:

a)

In this problem, you are holding the pencil at its end: this means that the pencil will rotate about this point.

The only force producing a torque on the pencil is the weight of the pencil, of magnitude

$W=mg$

where m is the mass of the pencil and g the acceleration of gravity.

However, when the pencil is rotating around its end, only the component of the weight tangential to its circular trajectory will cause an angular acceleration. This component of the weight is:

$W_p =mg sin \theta$

where $\theta$ is the angle of the rod with respect to the vertical.

The weight act at the center of mass of the pencil, which is located at the middle of the pencil. So the torque produced is

$\tau = W_p \frac{L}{2}=mg\frac{L}{2} cos \theta$

where L is the length of the pencil.

The relationship between torque and angular acceleration $\alpha$ is

$\tau = I \alpha$ (1)

where

$I=\frac{1}{3}mL^2$

is the moment of inertia of the pencil with respect to its end.

Substituting into (1) and solving for $\alpha$ , we find:

$\alpha = \frac{\tau}{I}=\frac{mg\frac{L}{2}sin \theta}{\frac{1}{3}mL^2}=\frac{3 g sin \theta}{2L}$

And assuming that the length of the pencil is L = 15 cm = 0.15 m, the angular acceleration when $\theta=10^{\circ}$ is

$\alpha = \frac{3(9.8)(sin 10^{\circ})}{2(0.15)}=17.0 rad/s^2$

b)

There are only two forces acting on the pencil here:

- The weight of the pencil, of magnitude $mg$

- The normal reaction of the hand on the pencil, R

The torque exerted by each force is given by

$\tau = Fd$

where F is the magnitude of the force and d the distance between the force and the pivot point.

For the weight, we saw in part a) that the torque is

$\tau =mg\frac{L}{2} cos \theta$

For the reaction force, the torque is zero: this is because the reaction force is applied exctly at the pivot point, so d = 0, and therefore the torque is zero.

Therefore:

- Weight: counterclockwise torque (I have assumed that the pencil is held at its right end)

- Reaction force: zero torque

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