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1 year ago

2 Jupiter orbits the sun in a nearly circular path with radius 7.8x10^11m. The orbital period of Jupiter is 12 years.

Physics

1 answer:

bazaltina [42]1 year ago

4 0

Mass of Jupiter=1.9×10

㎏=M

Mass of Sun=1.99×10

㎏=M

Mean distance of Jupiter from Sun=7.8×10

m=r

G=6.67×10

−11

N㎡㎏

−2

Gravitational Force, F=

(7.8×10

)

6.67×10

−11

×1.9×10

×1.99×10

F=4.16×10

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hichkok12 [17]

Answer:

$v = 3.5 \times 10^5 m/s$

Explanation:

At some distance from the Earth the force of attraction due to moon is balanced by the force due to Moon

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now we have

$\frac{d - r}{r} = \sqrt{\frac{M_m}{M_e}}$

$\frac{3.844\times 10^8 - r}{r} = \sqrt{\frac{7.36 \times 10^{22}}{5.9742\times 10^{24}}}$

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Now by energy conservation

$-\frac{GM_e}{R_e} - \frac{GM_m}{d - (R_e + R_m)} + \frac{1}{2}v^2 = -\frac{GM_e}{r} - \frac{GM_m}{d - r}$

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2 years ago

An airplane flies with a constant speed of 840km/h. how far can it travel in 1 hour?

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In this example, the speed of the airplane = 840km. The time (that the question is asking)is how far can it travel in 1 hour.

So just plug in your numbers.

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2 years ago

A horse is pulling 53.0 kg plow forward while the ground exerts a 275 N force, and the plow accelerates at 0.222 m/s^2. what is

JulsSmile [24]

Answer:
Magnitude of force the ground exerts on the plow = 263.234 N

Explanation:
Magnitude of force the ground exerts on the plow = Fground - Fplow
We are given that:
Fgound = 275 N
We will now calculate Fplow as follows:
Fplow = mass of horse * acceleration of plow
Fplow = 53 * 0.222
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Now, substitute in the above equation to get magnitude of force the ground exerts on the plow as follows:
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