All categories

3 years ago

What provides the centripetal force needed to keep Earth in orbit?

Physics

2 answers:

anyanavicka [17]3 years ago

8 0

The centripetal force is the gravitational attraction
between the Earth and the Sun.

kkurt [141]3 years ago

3 0

The centripetal force needed to keep earth in orbit is gravity.

You might be interested in

The sun _____.

enyata [817]

The sun appears brighter than any other star.

(It isn't really, but it looks that way because it's much much much much much much closer to us than any other star.)

7 0

3 years ago

The power of motor is 50 w the motor drags a 2 object horizontally at a constant speed of 10m/s along a rough floor . What is th

Ilia_Sergeevich [38]

Frictional force between the object and the floor=5 N

Explanation:

power= 50 W

velocity= 10 m/s

power= force * velocity

50=F * 10

F=50/10

F=5 N

Thus the force of friction= 5 N

5 0

3 years ago

Two containers (A and B) are in thermal contact with an environment at temperature T = 280 K. The two containers are connected b

zmey [24]

Answer:

The maximum amount of work is $W = 1563.289 \ J$

Explanation:

From the question we are told that

The temperature of the environment is $T = 280\ K$

The volume of container A is $V_A = 2 m^3$

Initially the number of moles is $n = 1.2 \ moles$

The volume of container B is $V_B = 3.5 \ m^3$

At equilibrium of the gas the maximum work that can be done on the turbine is mathematically represented as

$W = P_A V_A ln[ \frac{V_B}{V_A} ]$

Now from the Ideal gas law

$P_A V_A = nRT$

So substituting for $P_A V_A$ in the equation above

$W = nRT ln [\frac{V_B}{V_A} ]$

Where R is the gas constant with a values of $R = 8.314 \ J/mol$

Substituting values we have that

$W = 1.2 * (8.314) * (280) * ln [\frac{3.5}{2} ]$

$W = 1563.289 \ J$

8 0

3 years ago

After you pick up a spare, your bowling ball rolls without slipping back toward the ball rack with a linear speed of 2.85 m/s. T

motikmotik

Answer:

V' = 0.84 m/s

Explanation:

given,

Linear speed of the ball, v = 2.85 m/s

rise of the ball, h = 0.53 m

Linear speed of the ball, v' = ?

rotation kinetic energy of the ball

$KE_r = \dfrac{1}{2}I\omega^2$

I of the moment of inertia of the sphere

$I = \dfrac{2}{5}MR^2$

v = R ω

using conservation of energy

$KE_r = \dfrac{1}{2}( \dfrac{2}{5}MR^2)(\dfrac{V}{R})2$

$KE_r = \dfrac{1}{5}MV^2$

Applying conservation of energy

Initial Linear KE + Initial roational KE = Final Linear KE + Final roational KE + Potential energy

$\dfrac{1}{2}MV^2 + \dfrac{1}{5}MV^2 = \dfrac{1}{2}MV'^2 + \dfrac{1}{5}MV'^2 + M g h$

$0.7 V^2 = 0.7 V'^2 + gh$

$0.7\times 2.85^2 = 0.7\times V'^2 +9.8\times 0.53$

V'² = 0.7025

V' = 0.84 m/s

the linear speed of the ball at the top of ramp is equal to 0.84 m/s

6 0

3 years ago

!!!HELP ASAP!!! in witch situation is work being done? theres a photo added.

Arturiano [62]

C..............................

8 0

3 years ago