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8090 [49]
3 years ago
7

A motor keep a Ferris wheel (with moment of inertia 6.97 × 107 kg · m2 ) rotating at 8.5 rev/hr. When the motor is turned off, t

he wheel slows down (because of friction) to 7.5 rev/hr in 15 s. What was the power of the motor that kept the wheel rotating at 8.5 rev/hr despite friction?
Physics
1 answer:
Talja [164]3 years ago
8 0

Answer:

P = 133.13 Watt

Explanation:

Initial angular speed of the ferris wheel is given as

\omega_i = 2\pi f

\omega_i = 2\pi(8.5/3600)

\omega_i = 0.015 rad/s

final angular speed after friction is given as

\omega_f = 2\pi f

\omega_f = 2\pi(7.5/3600)

\omega_f = 0.013 rad/s

now angular acceleration is given as

\alpha = \frac{\omega_f - \omega_i}{\Delta t}

\alpha = \frac{0.015 - 0.013}{15}

\alpha = 1.27 \times 10^{-4} rad/s^2

now torque due to friction on the wheel is given as

\tau = I \alpha

\tau = (6.97 \times 10^7)(1.27 \times 10^{-4})

\tau = 8875.3 N m

Now the power required to rotate it with initial given speed is

P = \tau \omega

P = 8875.3 \times 0.015

P = 133.13 Watt

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A car moving with an initial speed v collides with a second stationary car that is one-half as massive. After the collision the
Mashutka [201]

Answer:

4v/3

Explanation:

Assume elastic collision by the law of momentum conservation:

m_1v = m_1v_1 + m_2v_2

where v is the original speed of car 1, v1 is the final speed of car 1 and v2 is final speed of car 2. m1 and m2 are masses of car 1 and car 2, respectively

Substitute m_2 = m_1/2 \& v_1 = v/3

m_1v = \frac{m_1v}{3} + \frac{m_1v_2}{2}

Divide both side by m_1, then multiply by 6 we have

6v = 2v + 3v_2

3v_2 = 4v

v_2 = \frac{4v}{3}

So the final speed of the second car is 4/3 of the first car original speed

5 0
3 years ago
A 100-kg running back runs at 5 m/s into a stationary linebacker. It takes 0.5 s for the running back to be completely stopped.
Elza [17]

Answer:

1000 N

Explanation:

First, we need to find the deceleration of the running back, which is given by:

a=\frac{v-u}{t}

where

v = 0 is his final velocity

u = 5 m/s is his initial velocity

t = 0.5 s is the time taken

Substituting, we have

a=\frac{0-5 m/s}{0.5 s}=-10 m/s^2

And now we can calculate the force exerted on the running back, by using Newton's second law:

F=ma=(100 kg)(-10 m/s^2)=-1000 N

so, the magnitude of the force is 1000 N.

6 0
3 years ago
Read 2 more answers
A piece of wood is floating in a bathtub. A second piece of wood sits on top of the first piece, and does not touch the water. I
Fofino [41]

Answer:

B

Explanation:

Water level remains unchanged

The first thing to realize is that the buoyancy force is the same as, or equal to the weight of the wood, this same force is also the same as or equal to the weight of the water displaced by the wood. In the two cases, the weight of the wood will be unaffected nonetheless, and thus the water level will remain the same.

Therefore, the answer is B, the water level remains unchanged.

6 0
3 years ago
In order to identify them, animal fossils and other fossils nearby can be: A. compared B. piled C. seen
tigry1 [53]

Answer:

A.compared

Explanation:

Fossils help figure out the time that organisms lived. If you know one of the fossils, it can be used as a reference for others around.

5 0
3 years ago
A 60 kg acrobat is in the middle of a 10 m long tightrope. The center of the rope dropped 30 cm in relation to the ends that are
Zigmanuir [339]

Answer:

The tension in each half of the rope, is approximately 4,908.8 N

Explanation:

The mass of the acrobat, m = 60 kg

The length of the rope, l = 10 m

The extent by which the center dropped = 30 cm = 0.3 m

Let, 'T' represent the tension in each half of the rope

Weight, W = Mass, m × The acceleration due to gravity, g

∴ W = m × g

The acceleration due to gravity, g ≈ 9.8 m/s²

∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N

The angle the dropped rope makes with the horizontal, θ is given as follows;

θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°

At equilibrium, the sum of vertical forces, \Sigma F_y = 0

The vertical component of the tension, T_y, in each half of the rope is given as follows;

T_y = T × sin(θ)

∴ \Sigma F_y = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)

Plugging in the values, with θ = arctan(0.06) for accuracy, we get;

588 N + 2 × T × sin(arctan(0.06) = 0

∴ 2 × -T × sin(arctan(0.06) = 588 N

-T= 588 N/(2 × sin(arctan(0.06)) = 4,908.81208 N ≈ 4,908.8 N

The tension in each half of the rope, T ≈ 4,908.8 N.

4 0
2 years ago
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