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2 years ago

If a Ball Falls from a girls hand, How does its speed change?

Physics

2 answers:

Lapatulllka [165]2 years ago

7 0

Answer:

When an object is in free fall, gravity increases its velocity by 9.8 m/s with every passing second.

Explanation:

Hope Helps :)

Keith_Richards [23]2 years ago

3 0

Answer:

Yes

Explanation:

At initial condition the ball is in rest because girl hold the ball.But after when girl release the ball the due to gravity it speed goes to increase and become zero again when it touch the floor.

As we know that gravity provides acceleration due to which velocity of ball increases.

Lets take the height of ball from the bottom was h ,so the velocity of ball just before when ball touch the bottom surface after falling from the hand of girl is V.

$V=\sqrt{2gh}$

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A merry-go-round is shaped like a uniform disk and has moment of inertia of 50,000 kg m 2 . It is rotating so that it has an ang

lapo4ka [179]

Answer:

r = 20 m

Explanation:

The formula for the angular momentum of a rotating body is given as:

L = mvr

where,

L = Angular Momentum = 10000 kgm²/s

m = mass

v = speed = 2 m/s

r = radius of merry-go-round

Therefore,

10000 kg.m²/s = mr(2 m/s)

m r = (10000 kg.m²/s)/(2 m/s)

m r = 5000 kg.m ------------- equation 1

Now, the moment of inertia of a solid uniform disc about its axis through its center is given as:

I = (1/2) m r²

where,

I = moment of inertia = 50000 kg.m²

Therefore,

50000 kg.m² = (1/2)(m r)(r)

using equation 1, we get:

50000 kg.m² = (1/2)(5000 kg.m)(r)

(50000 kg.m²)/(2500 kg.m) = r

5 0

2 years ago

5,6,1,9,21,37,68 ? find missing

maria [59]

Answer:

Missing numbers are 3,11,13, 15,17,19, 23,27,29 and wrong are 6,68

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2 years ago

On a hot summer day in the state of Washington while kayaking, I saw several swimmers jump from a railroad bridge into the Snoho

Svetlanka [38]

Answer: part a: 19.62m

part b: 19.62 m/s

part a: 2.83 secs

Explanation:If the air resistance is ignored then the swimmer experience free fall under gravity hence

u=0

a=9.81 m/s2

t=2 secs

$s=ut+0.5at^2$

s=h

$h=0*2+0.5*9.81*2^2\\h=19.62 meters$

Part b

$v=u+at\\v=0+9.81*2\\v=19.62m/s$

Part c

now we have h=2*19.62=39.24

$39.24=0+0.5*9.81*t^2\\t^2=8\\t=2.83 secs$

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2 years ago

How does Mr. Anderson define solubility in the video?

VikaD [51]

Answer:

7.3

Explanation:

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2 years ago

What is the mass of a man who accelerates 4 m/s2 under the action of a 200 N net force?

Over [174]

Answer:

$\huge \boxed{ \boxed{50 \: kg }}$

Explanation:

The mass of the man can be found by using the formula

$m = \frac{f}{a} \\$

f is the force

a is the acceleration

From the question we have

$m = \frac{200}{4} \\$

We have the final answer as

Hope this helps you

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2 years ago