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VLD [36.1K]
3 years ago
11

A moon with a mass of 6.0 x 10 kg orbits about a planet with a mass of 5.0x 10^24 kg. They are 4.0 x 10^8 meters apart What is t

he gravitational force between them? The answer is 1.25 x 10^20 N, it won’t let me answer my own question so someone put that as the answer!
Physics
1 answer:
allochka39001 [22]3 years ago
4 0

The gravitational force between them is <em>1.25 x 10^20 Newtons.</em>

(I think you must have typed the mass of the moon wrong.  

It must be 6.0 x 10^22 kg.)

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Com que rapidez estará sendo feito trabalho elétrico num resistor de 10 ohms, percorrido por uma corrente elétrica de 5 A? Qual
34kurt

Poder = (resistencia) x (corrente)²

Poder = (10 ohms) x (5 A)²

<em>Poder = 250 watts </em>(250 Joule por segundo)

2 horas = 7,200 segundos

Energia = (250 joule/seg) x (7,200 seg)

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7 0
3 years ago
Newton's Law of inertia is sufficient to cause a planet to orbit the sun.<br> O True<br> O False
mario62 [17]
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7 0
3 years ago
(D.) Define the following Moment (.) Moment of a force (.) couple of force (.) torque<br>​
erik [133]
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4 0
3 years ago
Pls help A car starts from rest and gains a velocity of 20m/s in 10 seconds calculate its acceleration and the distance covered
Soloha48 [4]

Answer:

\boxed{\sf Acceleration \ (a) = 2 \ m/s^{2}}

\boxed{\sf Distance \ covered \ (s) = 100 \ m}

Given:

Initial velocity (u) = 0 m/s

Final velocity (v) = 20 m/s

Time taken (t) = 10 sec

To Find:

(i) Acceleration (a)

(ii) Distance covered (s)

Explanation:

\sf (i) \ From \ 1^{st} \ equation \ of \ motion:

\sf \implies v = u + at

\sf \implies 20 = 0 + a(10)

\sf \implies 10a = 20

\sf \implies \frac{10a}{10}  =  \frac{20}{10}

\sf \implies a = 2 \: m/ {s}^{2}

\sf (ii) \ From \ 2^{nd} \ equation \ of \ motion:

\sf \implies s = ut +  \frac{1}{2} a {t}^{2}

\sf \implies s = (0)(10) +  \frac{1}{2}  \times 2 \times  {(10)}^{2}

\sf \implies s =  \frac{1}{ \cancel{2}}  \times  \cancel{2} \times  {(10)}^{2}

\sf \implies s =  {10}^{2}

\sf \implies s = 100 \: m

6 0
3 years ago
30 POINTS!!! CAN U AWNSER IT?? :)
solniwko [45]

Answer:

5235.84 kg

Explanation:

There is one theorem - whose proof I will never remember without having to drag calculus in there - that says that the variation of momentum is equal to the force applied times the time the application last.

F\Delta t = m \Delta v As long as the engine isn't ejecting mass - at this point it's a whole new can of worm - we know the force, we know the variation in speed, time to find the mass. But first, let's convert the variation of speed in meters per second. The ship gains 250 kmh, \Delta v = 69.4 m/s;

45 450 \cdot 8 = 69.4 m \rightarrow m = \frac{45450\cdot 8}{69.4} = 5235.84 kg

7 0
2 years ago
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