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OLEGan [10]
3 years ago
8

The magnitude of the gravitational acceleration at the surface of a planet of mass M and radius R is g. What is the magnitude of

the gravitational acceleration at the surface of a planet of mass 2M and radius 2R?​
Physics
1 answer:
maksim [4K]3 years ago
4 0
2M/4R2=.5M/R2= .0.5 g
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Oberon is a moon of Uranus. It has a mass of 3.01 x 1021 kg and is 761.4 km away from Uranus. If the force of gravity
ZanzabumX [31]

Answer:

acc to formula= F= G m1m2/rsq

mass of Uranus= F×r sq/G×m

M=3.00×10^25×761.4×761.4×10^6/6.67×10^ -11×3.01×10^21

=3×10^25-8×7614×7614/667×301×10^-2-11+21-2

=173918988×10^11/200767

=866×10^11(approx)

4 0
3 years ago
A 10 kg ball strikes a wall with a velocity of 3 m/s to the left. The ball bounces off with a velocity of 3 m/s to the right. If
lord [1]

Answer:

The force is 272.73 newtons

Explanation:

We're going to use impulse-momentum theorem that states impulse is the change on the linear momentum this is:

\overrightarrow{J}=\overrightarrow{p}_{f}-\overrightarrow{p}_{i} (1)

Impulse is also defined as average force times the time the force is applied:

\overrightarrow{J}=\overrightarrow{F}_{avg}(\varDelta t) (2)

By (2) on (1):

\overrightarrow{F}_{avg}(\varDelta t)= \overrightarrow{p}_{f}-\overrightarrow{p}_{i}

solving for \overrightarrow{F}_{avg}:

\overrightarrow{F}_{avg}=\frac{\overrightarrow{p}_{f}-\overrightarrow{p}_{i}}{\varDelta t} (3)

We already know Δt is equal to 0.22 s, all we should do now is to find \overrightarrow{p}_{f}-\overrightarrow{p}_{i} and put on (3) (\overrightarrow{p_{i}} the initial momentum and \overrightarrow{p_{f}} the final momentum). Linear momentum is defined as \overrightarrow{p}=m\overrightarrow{v} , using that on (3):

\varDelta\overrightarrow{p}=m \overrightarrow{v_{f}}-m \overrightarrow{v_{i}} (4)

Velocity (v) are vectors so direction matters, if positive direction is the right direction and negative direction left \overrightarrow{v_{i}}=+3\, \frac{m}{s} and \overrightarrow{v_{f}}=-3\, \frac{m}{s} so (4) becomes:

\varDelta\overrightarrow{p}=m(-3\frac{m}{s}- (+3\frac{m}{s}))=-(10kg)(6\frac{m}{s})

\varDelta\overrightarrow{p}=-60\, \frac{mkg}{s} (5)

Using (5) on (3):

\overrightarrow{F}_{avg}=\frac{-60\, \frac{mkg}{s}}{0.22s}

F_{avg}=272.73N

8 0
4 years ago
An artificial satellite circles Earth in a circular orbit at a location where the acceleration due to gravity is 9.00 m/s2. Dete
Ksju [112]
 <span>g = GMe/Re^2, where Re = Radius of earth (6360km), G = 6.67x10^-11 Nm^2/kg^2, and Me = Mass of earth. On the earth's surface, g = 9.81 m/s^2, so the radius of your orbit is:


R = Re * sqrt (9.81 m/s^2 / 9.00 m/s^2) = 6640km 

here, the speed of the satellite is:

v = sqrt(R*9.00m/s^2) = 7730 m/s 

  the time it would take the satellite to complete one full rotation is:

T = 2*pi*R/v = 5397 s * 1h/3600s = 1.50 h 

Hope it help i know it's long and may be confusing but if you have any more questions regarding this topic just hmu!  :)</span>
6 0
3 years ago
What would you use to to measure the weight of a truck?? I’m needing a word that is 8 letters long
myrzilka [38]

Answer:

Railroad Scale?

Explanation:

5 0
3 years ago
What is the accepted value of the speed of light through empty space?
galina1969 [7]

Answer:

299,792 kilometers per second

Explanation:

8 0
3 years ago
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