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avanturin [10]
2 years ago
14

A baseball pitcher brings his arm forward during a pitch, rotating the forearm about the elbow. If the velocity of the ball in t

he pitcher's hand is 35.0 m/s and the ball is 0.300 m from the elbow joint, what is the angular velocity of the forearm?
Group of answer choices
A) 10.5 rad/s
B) 0.0952 rad/s
C) 0.00857 rad/s
D) 117 rad/s
Physics
1 answer:
stiv31 [10]2 years ago
3 0

Answer:

D) 117 rad/s

Explanation:

We can treat this system as a circular motion where the origin is the elbow joint, the ball rotation velocity v is 35 m/s, the rotation radius is r = 0.3m.

As the ball is leaving the pitcher hand at such speed and rotation radius. Its angular velocity is:

\omega = \frac{v}{r} = \frac{35}{0.3} = 117 rad/s

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You are standing at the midpoint between two speakers, a distance D away from each. The speakers are playing the exact same soun
Rzqust [24]

Answer:

Explanation:

wave length of sound waves = velocity / frequency

= 340 / 170

λ = 2 m.

When the position of man is exactly at the meddle point between the speakers , sound waves from the speakers reaching man are in same phases ( path difference is zero. ) so intensity of sound is maximum .

Now , the man starts moving towards one of the speakers , his distance from one speaker becomes closer than the other creating path difference for the sound waves reaching his ears.

If he walks a distance of .5 m towards one speaker , path difference created

= .5 x 2 = 1 m

So , path difference = λ /2 ,

there will be destructive interference so minimum sound will be heard there.

When he walks a distance of 1 m , path difference created = 2m

path difference =  λ

so there is constructive interference and maximum  sound will be heard there.

Again we he walks a distance of 1.5 m , path difference created = 3 m

path difference = 3 λ /2

So there will be destructive interference so minimum sound will be heard there.

In this way we see that man starts  from a point of maximum sound intensity , reaches a point of minimum sound intensity , then reaches a point of maximum sound intensity . At last he reaches a position of minimum sound intensity.

3 0
2 years ago
The acceleration due to gravity at the north pole of Neptune is approximately 11.2 m/s2. Neptune has mass 1.02×1026 kg and radiu
larisa [96]

Answer: (a) The gravitational force on the object at the North Pole of Neptune is 51.7N

(b) The apparent weight of the object at Neptune's equator is 50.4N

Explanation: Please see the attachments below

5 0
2 years ago
A boy throws a 15 kg ball at 4.7 m/s to a 65 kg girl who is stationary and standing on a skateboard. After catching the ball, th
jek_recluse [69]

Answer:

a)v_{f}=0.88m/s

Explanation:

To solve this problem we use the Momentum's conservation Law, before and after the girl catch the ball:

\\ p_{1}=p_{2}\\m_{ball}*v_{o.ball}+m_{girl}*v_{o.girl} = m_{ball}*v_{f.ball} + m_{girl}*v_{f.girl}        (1)

At the beginning the girl is  stationary:

v_{o.girl}=0m/s       (2)

If the girl catch the ball, both have the same speed:

v_{f.girl}=v_{f.ball}=v_{f}       (3)

We replace (2) and (3) in (1):

m_{ball}*v_{o.ball} = (m_{ball}+m_{girl})*v_{f} \\

We can now solve the equation for v_{f}:

v_{f}=\frac{m_{ball}*v_{o.ball}}{(m_{ball}+m_{girl})}=\frac{15*4.7}{15+65}=0.88m/s

4 0
2 years ago
The boom hoisting sheave must have pitch diameters of no less than _______times the nominal diameter of the rope used.
alexira [117]

Answer:

18 times

Explanation:

According to the security purposes which is set under the rules and regulation OSHA, which describes all the rights to the worker.

In the boom hoist receiving system all the sheaves which are used should have a pitch diameter of rope not less than 18 times the diameter of the nominal rope which is used.

7 0
2 years ago
When the balloon hits the ground, it rebounds slightly. What is the source of the energy for this rebound? A. When the balloon h
nevsk [136]

Answer:

The correct answer is c.    When the balloon hits the ground, the rubber envelope stretches, storing elastic potential energy; this elastic potential energy is converted to the gravitational potentiaL

Explanation:

Let's analyze the situation of the globe

When it touches the ground, the part that is in contact decreases its velocity to zero, but the upper part of the ball continues to move, which creates that the molecules approach slightly, if we approximate the spring links, a repulsive force is created that after all the particles reach zero speed. The force of the springs moves the ball up until the force decreases to zero.

We can relate this force of Hooke with an elastic energy

This energy can be stored in the deformation of the system, as elastic potential energy, which is subsequently transformed into gravitational potential energy when the balloon is lifted.

The correct answer is c

7 0
2 years ago
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