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3 years ago

What is the free-fall acceleration at the surface of the jupiter?

Physics

1 answer:

Rufina [12.5K]3 years ago

4 0

Answer:

24.79 m/s2

Explanation:

Let us assume that there is an object with a mass of 'm' on the

Jupiter. Jupiter will attract this object:

$mg_j=G\frac{mM_j}{r_j^{2} }$

$g_j=G\frac{M_j}{r_j^{2} }$

$g_j=6.67*10^{-11} \frac{1.9*10^{27} }{71492000^{2} }$

$g_j=24.79m/s^{2}$

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Does an infrared wave or an x-ray travel faster in the vacuum of space?

Aneli [31]

Ok no se si te puedo

3 0

2 years ago

Read 2 more answers

A ball is thrown upward. At a height of 10 meters above the ground, the ball has a potential energy of 50 joules (with the poten

lions [1.4K]

Answer:

h = 20 m

Explanation:

given.

height, h = 10 m

Potential energy at 10 m = 50 J

Kinetic energy at 10 m = 50 J

maximum height the ball will reach, H = ?

Total energy of the system

T E = 50 J + 50 J

T E = 100 J

now,

A h = 10 m

P E = m g h

50 = m g x 10

mg = 5 ..............(1)

at the top most Point the only Potential energy will be acting on the body.

now, TE = Potential energy

100 = m g h

5 h = 100

h = 20 m

hence, the maximum height reached by the ball is equal to 20 m.

5 0

3 years ago

a stone is dropped from top of a tower of 50m high ,simltaneously another stone is thrown upward with a speed of 20m/s find the

aleksklad [387]

At the time that I'll call ' Q ', the height of the stone that was
dropped from the tower is

   H = 50 - (1/2 G Q²) ,

and the height of the stone that was tossed straight up
from the ground is

   H = 20Q - (1/2 G Q²) .

The stones meet when them's heights are equal,
so that's the time when

   <span>50 - (1/2 G Q²) = 20Q - (1/2 G Q²) .

This is looking like it's going to be easy.

Add </span><span>(1/2 G Q²) to each side.
Then it says
       50 = 20Q

Divide each side by 20: 2.5 = Q .

And there we are. The stones pass each other

   2.5 seconds

after they are simultaneously launched.
</span>

5 0

3 years ago

Find the position vector of a particle that has the given acceleration and the specified initial velocity and position. a(t) = 1

kondor19780726 [428]

Answer:

Explanation:

Given

Acceleration $a(t)=14t\hat{i]+\sin (t)\hat{j}+\cos (2t)\hat{k}$ [/tex]

and $v(0)=\hat{i}$

$r(0)=\hat{j}$

we know $a=\frac{\mathrm{d} v}{\mathrm{d} t}$

$\int dv=\int adt$

$v(t)=\int (14t\hat{i}+\sin (t)\hat{j}+\cos (2t)\hat{k})dt$

$v(t)=7t^2\hat{i}-\cos t\hat{j}+\frac{\sin (2t)\hat{k}}{2}+c$

at $t=0$

$v(0)=0-1\cdot \hat{j}+0+c$

$c=\hat{i}+\hat{j}$

$v(t)=(7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2}$

and $\frac{\mathrm{d} r}{\mathrm{d} t}=v(t)$

$\int dr=\int vdt$

$r(t)=\int ((7t^2+1)\hat{i}+(1-\cos t)\hat{j}+\frac{\sin (2t)\hat{k}}{2})dt$

$r(t)=(\frac{7}{2}t^3+t)\hat{i}+(t-\sin (t))\hat{j}+\frac{1}{2}\times (-\frac{1}{2}\cos 2t)\hat{k}+c_2$

at $t=0$

$r(0)=\hat{j}$

$r(t)=(\frac{7}{3}t^3+t)\hat{i}+(1+t-\sin t)\hat{j}+\frac{1}{4}(1-\cos 2t)\hat{k}$

4 0

3 years ago

A ball is connected to a light spring suspended vertically. When pulled downward from its equilibrium position and released, the

babunello [35]

Answer:

The forms of energy involved are

1. Kinetic energy

2. Potential energy

Explanation:

The system consists of a ball initially at rest. The ball is pulled down from its equilibrium position (this builds up its potential energy) and then released. The released ball oscillates due to a continuous transition between kinetic and potential energy.

5 0

3 years ago