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Nata [24]
2 years ago
7

Moment of inertia describes Select one: a. How the mass of an extended object is distributed about a rotation axis. b. How a for

ce can rotate an object. c. The average position of the mass in an extended object. d. The tendency of an object to move in a circular path. e. The tendency of an object to move in a straight line.
Physics
1 answer:
olga nikolaevna [1]2 years ago
7 0

Answer: a. How the mass of an extended object is distributed about a rotation axis

Explanation: Moment of inertia is defined as the measure of the rotational inertia of a solid object, it is a quantity that defines the torque needed to reach a desired angular acceleration around a given rotation axis, and it depends mainly on the mass distribution of the object, so the correct answer is: "how the mass of an extended object is distributed about a rotation axis"

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A string has its 4th harmonic at 31.5 Hz. What is the fundamental frequency?
seropon [69]

Given data

*The given 4th harmonic frequency is 31.5 Hz

The fundamental frequency is calculated as

\begin{gathered} f_n=\frac{31.5}{4} \\ =7.875\text{ Hz} \end{gathered}

Hence, the fundamental frequency is 7.875 Hz

5 0
1 year ago
At height h above the surface of Earth, the gravitational acceleration is What is h? Note: The radius of Earth is 6380 km.
rusak2 [61]

The acceleration of gravity is inversely proportional to
the square of the distance from Earth's center.

The acceleration of gravity is 9.8 m/s² on the Earth's surface ...
6380 km from the center.

If the acceleration of gravity at 'h' is 4.9 m/s² ... 1/2 of what it is
on the surface, then the distance from the center is

                 (6380 x √2) =  9,023 km  (rounded) ,

and 'h' is the distance above the surface

                     = (9,023 - 6,380) =  2,643 km  (rounded) .

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3 years ago
According to the humanistic perspective, what motivates individuals?
Paraphin [41]
The desire for positive reinforcement.
8 0
3 years ago
Read 2 more answers
In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400
liq [111]

Complete Question:

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

Answer:

a) Speed of car A at the start of sliding = 4.23 m/s

b) speed of car B at the start of sliding = 3.957 m/s

c) Speed of car B before the collision = 7.28 m/s

Explanation:

NB: The figure is not provided but all the parameters needed to solve the question have been given.

Let the frictional force acting on car A, f_{ra} = \mu mg\\............(1)

Since frictional force is a type of force, we are safe to say f_{ra} = ma.......(2)

Equating (1) and (2)

ma = \mu mg\\a = \mu g\\\mu = 0.15\\a = 0.15 * 9.8 = 1.47 m/s^{2}

a) Speed of A at the start of the sliding

d_{A} = 6.1 m\\Speed of A at the start of sliding, v_{A} = \sqrt{2ad_{A} }\\ v_{A} = \sqrt{2*1.47*6.1 } \\v_{A} = \sqrt{17.934 } \\v_{A} = 4.23 m/s

b) Speed of B at the start of the sliding

d_{A} = 4.4 m\\Speed of A at the start of sliding, v_{B} = \sqrt{2ad_{B} }\\ v_{B} = \sqrt{2*1.47*4.4 } \\v_{B} = \sqrt{12.936 } \\v_{B} = 3.957 m/s

Let the speed of car B before collision = v_{B1}

Momentum of car B before collision = m_{B} v_{B1}

Momentum after collision = m_{A} v_{A} + m_{B} v_{B2}

Applying the law of conservation of momentum:

m_{B} v_{B1}  = m_{A} v_{A} +m_{B} v_{B2}

m_{A} = 1100 kg\\m_{B} = 1400 kg

(1400*v_{B1} ) = (1100 * 4.23) + ( 1400 * 3.957)\\(1400*v_{B1} ) = 10192.8\\v_{B1} = 10192.8/1400\\v_{B1 = 7.28 m/s

3 0
3 years ago
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Answer:

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p = 1500 \frac{kg*m}{s}

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