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scoundrel [369]
3 years ago
5

Anyone know a GOOD show on Netflix please kid shows pleaseeeee

Physics
2 answers:
Afina-wow [57]3 years ago
7 0

Answer:

like horror? or action haha

Explanation:

raketka [301]3 years ago
4 0

Answer:

Avatar and last kids on earth ( your missing out on a lot of great shows the there when you get older watch Arcane.)

Explanation:

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A planet has two
lozanna [386]
Kepler's third law hypothesizes that for all the small bodies in orbit around the
same central body, the ratio of (orbital period squared) / (orbital radius cubed)
is the same number.

<u>Moon #1:</u>  (1.262 days)² / (2.346 x 10^4 km)³

<u>Moon #2:</u>  (orbital period)² / (9.378 x 10^3 km)³

If Kepler knew what he was talking about ... and Newton showed that he did ...
then these two fractions are equal, and may be written as a proportion.

Cross multiply the proportion:

(orbital period)² x (2.346 x 10^4)³ = (1.262 days)² x (9.378 x 10^3)³

Divide each side by (2.346 x 10^4)³:

(Orbital period)² = (1.262 days)² x (9.378 x 10^3 km)³ / (2.346 x 10^4 km)³

               =  0.1017 day²

Orbital period = <u>0.319 Earth day</u> = about 7.6 hours.
7 0
3 years ago
A 97.0 kg ice hockey player hits a 0.150 kg puck, giving the puck a velocity of 48.0 m/s. If both are initially at rest and if t
OverLord2011 [107]

Answer:

s₁ = 0.022 m

Explanation:

From the law of conservation of momentum:

m_1u_1 + m_2u_2 = m_1v_1+m_2v_2

where,

m₁ = mass of hockey player = 97 kg

m₂ = mass of puck = 0.15 kg

u₁ = u₂ = initial velocities of puck and player = 0 m/s

v₁ = velocity of player after collision = ?

v₂ = velocity of puck after hitting = 48 m/s

Therefore,

(97\ kg)(0\ m/s)+(0.15\ kg)(0\ m/s)=(97\ kg)(v_1)+(0.15\ kg)(48\ m/s)\\\\v_1 = -\frac{(0.15\ kg)(48\ m/s)}{97\ kg} \\v_1 = - 0.074 m/s

negative sign here shows the opposite direction.

Now, we calculate the time taken by puck to move 14.5 m:

s_2 =v_2t\\\\t = \frac{s_2}{v_2} = \frac{14.5\ m}{48\ m/s} \\\\t =  0.3\ s

Now, the distance covered by the player in this time will be:

s_1 = v_1t\\s_1 = (0.074\ m/s)(0.3\ s)

<u>s₁ = 0.022 m</u>

4 0
3 years ago
A planet has a mass of 5.68 x 1026 kg and a radius of 6.03 x 107 m. What is the weight of a 65.0 kg person on the surface of thi
Paul [167]

To solve this problem, we must know the gravitational force of the planet. The equation would be,

 F=G \frac{ m_{1} m_{2}  }{ r^{2} }

This would calculate the force between two objects with masses m1 and m2 and the gravitational constant, G, is 6.67 x 10^-11 m3 s-2 kg-1 and with r as the distance between the objects.

Thus,

F = (6.67 x 10^-11 m3 s-2 kg-1) * (5.68 x 10^26 kg) * (65 kg) * ((1/6.03 x 10^7 m)^2)

F = 678 kg/s^2 or 678 N

Answer is letter B.

4 0
3 years ago
Using a simple machine, a student is able to lift a 500N weight by applying only 100N.
pishuonlain [190]
Hey there!

a \ student  \ is \ able \ to \ lift \ a \ 500(n) \\ weight \ by \ applying \ only  \ 100(n) \\ \\ so, \ this \ info \ here, \ we \ simply \ divide \ by \\ how \ much \ this \ kid \ lifted, \ by \ the \ weight \ he/she \\ \ is \ applying. \\ \\   \left[\begin{array}{ccc}\boxed{\boxed{500(n)/100(n)}}\end{array}\right] \\ \\ and \ from \ this,\ your \ answer \ would \ \\ conclude \ to \ be \ (5) \\ \\ \boxed{5} \ would \ be \ your \ answer!

Hope this helps you!
7 0
3 years ago
Mercury's surface has many craters because _____.
Alexeev081 [22]
I think it would be that it has no atmosphere
5 0
3 years ago
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