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3 years ago

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WHY IS SATURN ONE OF THE COLDEST PLANETS IN OUR SOLAR SYSTEM

1 answer:

Usimov [2.4K]3 years ago

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Astronomers have observed a small, massive object at the center of our Milky Way Galaxy. A ring of material orbits this massive

Burka [1]

Answer:

$1.91773\times 10^{37}\ kg$

Explanation:

v = Orbital speed = 130 km/s

d = Diameter = 16 ly

r = Radius = $\dfrac{d}{2}=\dfrac{16}{2}=8\ ly$

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

$1\ ly=9.461\times 10^{15}\ m$

As the centripetal force balances the gravitational energy we have the following relation

$\dfrac{GMm}{r^2}=\dfrac{mv^2}{r}\\\Rightarrow M=\dfrac{v^2r}{G}\\\Rightarrow M=\dfrac{130000^2\times 8\times 9.461\times 10^{15}}{6.67\times 10^{-11}}\\\Rightarrow M=1.91773\times 10^{37}\ kg$

Mass of the the massive object at the center of the Milky Way galaxy is $1.91773\times 10^{37}\ kg$

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3 years ago

Lina20 [59]

Explanation:

Make a table, listing the x and y coordinates of each square's center of gravity and its mass. Multiply the coordinates by the mass, add the results for each x and y, then divide by the total mass.

$\left\begin{array}{ccccc}x&y&m&xm&ym\\\frac{a}{2} &\frac{a}{2} &10&5a&5a\\\frac{3a}{2}&\frac{a}{2}&70&105a&35a\\\frac{a}{2}&\frac{3a}{2}&80&40a&120a\\\frac{3a}{2}&\frac{3a}{2}&50&75a&75a\\&\sum&210&225a&235a\\&&Avg&\frac{15a}{14}&\frac{47a}{42}\end{array}\right$

The x-coordinate of the center of gravity is 15/14 a.

The y-coordinate of the center of gravity is 47/42 a.

4 0

3 years ago

A free-body diagram is a special example of the vector diagrams that will

Genrish500 [490]

From someone who has taken physics this is, True

6 0

3 years ago

Capillary waves travel what than long waves

7nadin3 [17]

Faster than. Hope this helps!!!

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2 years ago

Read 2 more answers

How much work is required to accelerate a proton from rest up to a speed of 0.993 c? Express your answer with the appropriate un

Keith_Richards [23]

Answer:

(a). The work done is 7001 MeV.

(b). The momentum of this proton is $4.20\times10^{8}\ kg-m/s$ .

Explanation:

Given that,

Speed = 0.993 c

We need to calculate the work done

Using work energy theorem

The work done is equal to the kinetic energy relative to the proton

$W=K.E$

$W=\dfrac{m_{p}c^2}{\sqrt{1-\dfrac{v^2}{c^2}}}-m_{p}c^2$

Put the value into the formula

$W=\dfrac{1.67\times10^{-27}\times(3\times10^{8})^2}{\sqrt{1-(\dfrac{0.993c}{c})^2}}-1.67\times10^{-27}\times(3\times10^{8})^2$

$W=\dfrac{1.67\times10^{-27}\times(3\times10^{8})^2}{\sqrt{1-(0.993)^2}}-1.67\times10^{-27}\times(3\times10^{8})^2$

$W=1.122\times10^{-9}\ J$

$W=7001\ MeV$

(b). We need to calculate the momentum of this proton

Using formula of momentum

$p=\dfrac{m_{0}v}{\sqrt{1-\dfrac{v^2}{c^2}}}$

Put the value into the formula

$p=\dfrac{1.67\times10^{-27}\times0.993c}{\sqrt{1-(\dfrac{0.993c}{c})^2}}$

$p=\dfrac{1.67\times10^{-27}\times0.993c}{\sqrt{1-(0.993)^2}}$

$p=1.404\times10^{-26}c$

$p=4.20\times10^{8}\ kg-m/s$

Hence, (a). The work done is 7001 MeV.

(b). The momentum of this proton is $4.20\times10^{8}\ kg-m/s$ .

4 0

3 years ago