Ask question

Ask question

All categories

2 years ago

9

how does the size of objects impact the pull of gravity between Earth and a baseball thrown into the air

2 answers:

Reil [10]2 years ago

3 0

it's how much it weighs and how much force is pushing on it like a egg if i drop it the weigh can cause it to break and how much force the gravity is pushing on it.

attashe74 [19]2 years ago

3 0

The size of an object tossed into the air doesn't matter. The gravitational force on it depends on its mass not its size. Gravity exerts more force on a small rock then on a large styrofoam wiffle ball.

You might be interested in

Why is it that an object can accelerate while

posledela

For the same reason that you can skate around a curve at constant speed but not with constant velocity.

The DIRECTION you're going is part of your velocity, but it's not part of your speed.

If the DIRECTION changes, that's a change of velocity.

The object doesn't have to change speed to have a different velocity. A change of direction is enough to do it.

And any change of velocity is called acceleration.

3 0

2 years ago

What are the possible effects of global warming?

11Alexandr11 [23.1K]

A possible effect is the risk of Icecaps melted

6 0

3 years ago

A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow

Oksana_A [137]

Answer:

a) $\omega=1.36rad/s$

b) $\omega=12.99rpm$

c) $F=705.6N$

Explanation:

a) The angular velocity is related to the centripetal acceleration by the formula $a_{cp}=\omega^2r$ , which for our purposes we will write as:

$\omega=\sqrt{\frac{a_{cp}}{r}}$

Since <em>we want this acceleration to be 1.5 times that due to gravity</em>, for our values we will have:

$\omega=\sqrt{\frac{1.5g}{r}}=\sqrt{\frac{(1.5)(9.8m/s^2)}{(8m)}}=1.36rad/s$

b) 1 rpm (revolution per minute) is equivalent to an angle of $2\pi$ radians in 60 seconds:

$1\ rpm=\frac{2\pi rad}{60s} =\frac{\pi}{30}rad/s$

Which means <em>we can use the conversion factor</em>:

$\frac{1\ rpm}{\frac{\pi}{30}rad/s}=1$

So we have (multiplying by the conversion factor, which is 1, not affecting anything but transforming our units):

$\omega=1.36rad/s=1.36rad/s(\frac{1\ rpm}{\frac{\pi}{30}rad/s})=12.99rpm$

c) The centripetal force will be given by Newton's 2nd Law F=ma, so on the centripetal direction for our values we have:

$F=ma=(48kg)(1.5)(9.8m/s^2)=705.6N$

8 0

3 years ago

A solar prominence is

8_murik_8 [283]

B. a solar storm that moves across the sun's surface

8 0

2 years ago

Read 2 more answers

CAN SOMEONE PLEASE HELP ME??!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! HURRY!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

tiny-mole [99]

Hello, this is customer service. How may I help you?

3 0

3 years ago