All categories

2 years ago

What holds the moon in place, orbiting around earth?

Physics

2 answers:

pshichka [43]2 years ago

5 0

Answer:

Gravity

Explanation:

velikii [3]2 years ago

3 0

Answer:Gravitational force

Explanation: According to Newton's law of universal gravitation, two massive celestial bodies in space attract each other. This Force of attraction set the moon in motion.

You might be interested in

At t = 0, Ball 1 is dropped from the top of a 22 m-high building. At the same instant Ball 2 is thrown straight up from the base

Crazy boy [7]

Answer:

The two balls pass each other at a height of 5.53 m

vf1=17.97 m/s

vf2=-5.96 m/s

Explanation:

Vertical Motion

An object thrown from the ground at speed vo, is at a height y given by:

$y=vo.t-g.t^2/2$

Where t is the time and $g=9.8\ m/s^2$

Furthermore, an object dropped from a certain height h will fall a distance y, given by:

$y=g.t^2/2$

Thus, the height of this object above the ground is:

$H = h-g.t^2/2$

The question describes that ball 1 is dropped from a height of h=22 m. At the same time, ball 2 is thrown straight up with vo=12 m/s.

We want to find at what height both balls coincide. We'll do it by finding the time when it happens. We have written the equations for the height of both balls, we only have to equate them:

$vo.t-g.t^2/2=h-g.t^2/2$

Simplifying:

$vo.t=h$

Solving for t:

$t=h/vo=22/12=1.833\ s$

The height of ball 1 is:

$H = 22-9.8.(1.833)^2/2$

H = 5.53 m

The height of ball 2 is:

$y=12\cdot(1.833)-9.8\cdot(1.833)^2/2$

y=5.53 m

As required, both heights are the same.

The speed of the first ball is:

$vf1=g.t=9.8\cdot 1.833=17.97\ m/s$

vf1=17.97 m/s

The speed of the second ball is:

$vf2=vo-gt=12-9.8\cdot 1.833=-5.96\ m/s$

vf2=-5.96 m/s

This means the second ball is returning to the ground when both balls meet

3 0

3 years ago

A bicycle wheel is rotating at 49 rpm when the cyclist begins to pedal harder, giving the wheel a constant angular acceleration

raketka [301]

In 9 sec, it increases the angular velocity by (0.45 x 9) rad/s which will give us 4.05 rad/s
Now get the angular velocity and divide it 2pi = 4.05 by 2(pi) to give how many revolutions 4.05 rad is equivalent to = 0.6446 rps
Them, multiply this by 60 to get it from rps to rpm increase (0.6446 x 60)=38.676 rpm
Add this and the starting revolution frequency of 49 rpm to give: (49 rpm + 38.676 rpm) = 87.6760 rpm

6 0

4 years ago

Read 2 more answers

. Which of the following are forces that push upward on an indoor skydiver? Choose all that apply. *

SpyIntel [72]

The forces that push upward on an indoor skydiver are lift force and air resistance.

The forces that act on a skydiver moving downwards includes, gravity due to his weight, air resistance and lift force.

The downward forces on the indoor skydiver include the following;

gravity due to its weight
downward force due its acceleration

The upward force on the indoor skydiver include the following;

lift force
air resistance

Thus, we can conclude that the forces that push upward on an indoor skydiver are lift force and air resistance.

Learn more here:brainly.com/question/8947470

7 0

3 years ago

Jessica stretches her arms out 0.60 m from the center of her body while holding a 2.0 kg mass in each hand. She then spins aroun

Juliette [100K]

Answer:

a.) L = 2.64 kgm^2/s

b.) V = 4.4 m/s

Explanation: Jessica stretches her arms out 0.60 m from the center of her body. This will be considered as radius.

So,

Radius r = 0.6 m

Mass M = 2 kg

Velocity V = 1.1 m/s

Angular momentum L can be expressed as;

L = MVr

Substitute all the parameters into the formula

L = 2 × 1.1 × 0.6 = 1.32kgm^2s^-1

the combined angular momentum of the masses will be 2 × 1.32 = 2.64 kgm^2s-1

b. If she pulls her arms into 0.15 m,

New radius = 0.15 m

Using the same formula again

L = 2( MVr)

2.64 = 2( 2 × V × 0.15 )

1.32 = 0.3 V

V = 1.32/0.3

V = 4.4 m/s

Her new linear speed will be 4.4 m/s

4 0

3 years ago

The Tevatron acceleator at the Fermi National Accelerator Laboratory (Fermilab) outside Chicago boosts protons to 1 TeV (1000 Ge

Eva8 [605]

Answer:

a) $v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s$

b) $v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s$

c) $v = c \cdot 0.994 = 2.97\cdot 10^{8} m/s$

d) $v = c \cdot 0.999 = 2.997\cdot 10^{8} m/s$

e) $v = c \cdot 0.9999 = 2.999\cdot 10^{8} m/s$

Explanation:

At that energies, the speed of proton is in the relativistic theory field, so we need to use the relativistic kinetic energy equation.

$KE=mc^{2}(\gamma -1) = mc^{2}(\frac{1}{\sqrt{1-\beta^{2}}} -1)$ (1)

Here β = v/c, when v is the speed of the particle and c is the speed of light in vacuum.

Let's solve (1) for β.

$\beta = \sqrt{1-\frac{1}{\left (\frac{KE}{mc^{2}}+1 \right )^{2}}}$

We can write the mass of a proton in MeV/c².

$m_{p}=938.28 MeV/c^{2}$

Now we can calculate the speed in each stage.

a) Cockcroft-Walton (750 keV)

$\beta = \sqrt{1-\frac{1}{\left (\frac{0.75 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.04$

$v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s$

b) Linac (400 MeV)

$\beta = \sqrt{1-\frac{1}{\left (\frac{400 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.71$

$v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s$

c) Booster (8 GeV)

$\beta = \sqrt{1-\frac{1}{\left (\frac{8000 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.994$

$v = c \cdot 0.994 = 2.97\cdot 10^{8} m/s$

d) Main ring or injector (150 Gev)

$\beta = \sqrt{1-\frac{1}{\left (\frac{150000 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.999$

$v = c \cdot 0.999 = 2.997\cdot 10^{8} m/s$

e) Tevatron (1 TeV)

$\beta = \sqrt{1-\frac{1}{\left (\frac{1000000 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.9999$

$v = c \cdot 0.9999 = 2.999\cdot 10^{8} m/s$

Have a nice day!

4 0

4 years ago