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3 years ago

15

Dis hard for me plz plz answer...

1 answer:

Ann [662]3 years ago

8 0

The correct answer would be it would decrease.

P.S. If you need any technical support with Windows 10, just send a message to me on Brainly and I'll try to figure it out.

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Consider as a system the Sun with Saturn in a circular orbit around it. Find the magnitude of the change in the velocity of the

Doss [256]

Answer:

$v_{su} = 19.44 m/s$

Explanation:

$m_{su}=5.68x10^{29}kg\\m_{sa}=5.68x10^{26}kg$

$T=9.29x10^8\\r_{o}=1.43x10^{12}$

If the sun considered as x=0 on the axis to put the center of the mass as a:

$m_{su}*r_{o}=(m_{sa}+m_{su})*r_{1}$

solve to r1

$r_1=\frac{m_{sa}*r_{o}}{m_{sa}+m_{su}}=\frac{5.68x10^{26}*1.43x10^{12}}{5.68x10^{26}+5.68x10^{26}}$

$r_1=1.428x10^9m$

Now convert to coordinates centered on the center of mass. call the new coordinates x' and y' (we won't need y'). Now since in the sun centered coordinates the angular momentum was

$L = \frac{m_{sa}*2*pi*r_1^2}{T}$

where T = orbital period

then L'(x',y') = L(x) by conservation of angular momentum. So that means

$L_{sun}=\frac{m_{sa}*2*\pi *( 2r_{o}*r_1 -r_1^2)}{T}$

Since

$L_{su}= m_{su}*v_{su}*r_1$

then

$v_{su}=\frac{m_{sa}*2*pi*(2r_{o}*r_{1}-r_{1}^2)}{T*m_{sa}*r_1}$

$v_{su} = 19.44 m/s$

7 0

3 years ago

An observer on Earth sees rocket 1 leave Earth and travel toward Planet X at 0.3c. The observer on Earth also sees that Planet X

Verizon [17]

Answer:

0.625 c

Explanation:

Relative speed of a body may be defined as the speed of one body with respect to some other or the speed of one body in comparison to the speed of second body.

In the context,

The relative speed of body 2 with respect to body 1 can be expressed as :

$u'=\frac{u-v}{1-\frac{uv}{c^2}}$

Speed of rocket 1 with respect to rocket 2 :

$u' = \frac{0.4 c- (-0.3 c)}{1-\frac{(0.4 c)(-0.3 c)}{c^2}}$

$u' = \frac{0.7 c}{1.12}$

$u'=0.625 c$

Therefore, the speed of rocket 1 according to an observer on rocket 2 is 0.625 c

5 0

3 years ago

Starting with a new moon, name the phases the moon will pass through before the next new moon.

ipn [44]

Hope this helps! Please mark as brainliest!

8 0

3 years ago

Which of the following is typically NOT a method scientist use to determine astronomic distances

tangare [24]

I think the answer is A

5 0

3 years ago

A uniform, spherical, 1900.0 kg shell has a radius of 5.00 m. Find the gravitational force this shell exerts on a 1.80 kg point

Mandarinka [93]

Answer:

$F=9.09\times 10^{-9} N$

Explanation:

We are given that

Mass of spherical shell, $m_1$ =1900 kg

Mass= $m_2=1.80 kg$

Radius of shell=r=5 m

Distance between two masses=r=5.01 m

Because distance measure from center .

Gravitational force

$F=G\frac{m_1m_2}{r^2}$

$G=6.67\times 10^{-11} Nm^2/kg^2$

Using the formula

$F=6.67\times 10^{-11}\times \frac{1900\times 1.80}{(5.01)^2}$

$F=9.09\times 10^{-9} N$

Hence,the gravitational force = $F=9.09\times 10^{-9} N$

6 0

3 years ago