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Zanzabum
3 years ago
5

a 65 kg skater at rest on a frictionless rink throws a 2 kg ball, giving the ball a velocity of 7 m/s. What is the velocity of t

he skater immediately after
Physics
1 answer:
gayaneshka [121]3 years ago
8 0
The answer to your question is 33
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A light year is approximately 9.5 million km long. 'Barnard's Star' is 6 light years away from Earth. Calculate how many million
hammer [34]

Answer:

6 light years = 57 million km

Explanation:

Given;

A light year = 9.5 million km

To calculate how far is 6 light years;

6 light years = 6 × 1 light year = 6 × 9.5 million km

6 light years = 57 million km

7 0
3 years ago
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Your car's speedometer works in much the same way as its odometer, except that it converts the angular speed of the wheels to a
brilliants [131]

Answer:

Speed will be 30810 rpm      

Explanation:

We have given diameter of the tire d = 24 inch

So radius r=\frac{d}{2}=\frac{24}{2}=12imch

We have given linear velocity v = 35 mph

We know that linear velocity is given by v=\omega r

35=\omega \times 12

\omega =\frac{35}{12}\times \frac{63360}{60}=3080rad/min

As we know that 1 mile = 63360 inch and 1 hour = 60 min

3 0
3 years ago
All of the fallowing are possible sources of error in a scientific investigation exept for
Whitepunk [10]

c adding research resources during an investigation

6 0
3 years ago
Where is the magnetic field of a horseshoe magnet the strongest?<br> pls hellp omg
Nuetrik [128]
The answer is slightly left and slightly right of the curved end of the horseshoe.
7 0
3 years ago
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A flat uniform circular disk (r= 2.00m,
dusya [7]

Incomplete question.The Complete question is here

A flat uniform circular disk (radius = 2.00 m, mass = 1.00 ✕ 102 kg) is initially stationary. The disk is free to rotate in the horizontal plane about a friction less axis perpendicular to the center of the disk. A 40.0-kg person, standing 1.25 m from the axis, begins to run on the disk in a circular path and has a tangential speed of 2.00 m/s relative to the ground.

a.) Find the resulting angular speed of the disk (in rad/s) and describe the direction of the rotation.

b.) Determine the time it takes for a spot marking the starting point to pass again beneath the runner's feet.

Answer:

(a)ω = 1 rad/s

(b)t = 2.41 s

Explanation:

(a) initial angular momentum = final angular momentum  

0 = L for disk + L............... for runner

0 = Iω² - mv²r ...................they're opposite in direction

0 = (MR²/2)(ω²) - mv²r ................where is ω is angular speed which is required in part (a) of question

0 = [(1.00×10²kg)(2.00 m)² / 2](ω²) - (40.0 kg)(2.00 m/s)²(1.25 m)

0=200ω²-200

200=200ω²

ω = 1 rad/s

b.)

lets assume the "starting point" is a point marked on the disk.

The person's angular speed is  

v/r = (2.00 m/s) / (1.25 m) = 1.6 rad/s

As the person and the disk are moving in opposite directions, the person will run part of a revolution and the turning disk would complete the whole revolution.

(angle) + (angle disk turns) = 2π

(1.6 rad/s)(t) + ωt = 2π

t[1.6 rad/s + 1 rad/s] = 2π

t = 2.41 s

6 0
3 years ago
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