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sdas [7]
3 years ago
7

What keeps planets in orbit? (i need help asap please T-T 10 points im low on points sorry)

Physics
1 answer:
Arturiano [62]3 years ago
5 0

Answer:

the answer is C gravity

Explanation:

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A lunar module weighs 12 metric tons on the surface of the Earth. How much work is done in propelling the module from the surfac
zhuklara [117]

Answer:

W=76.55 miles.metric tons

Explanation:

Given that

Weight on the earth = 12 tons

So weight on the moon =12/6 = 2 tons

 ( because at moon g will become g/6)

As we know that

F=\dfrac{K}{x^2}

Here x= 1100 miles

F 2 tons

2=\dfrac{K}{1100^2}

So

K=2.4\times 10^6

We know that

Work = F. dx

W=\int_{x_1}^{x_2}F.dx

W=\int_{1100}^{1140}\dfrac{2.4\times 10^6}{x^2}.dx

W=-2.4\times 10^6\left[\dfrac{1}{x}\right]_{1100}^{1140}

W=-2.4\times 10^6\left[\dfrac{1}{1140}-\dfrac{1}{1100}\right]

W=76.55 miles.metric tons

6 0
3 years ago
Will a series circuit work if there are two batteries and one light bulb? Help asappp please!
Dmitry_Shevchenko [17]

Answer:

Doubling the voltage in this arrangement both doubles the voltage drop across the resistor and the current through it. The bulb will be much brighter.

7 0
3 years ago
For a huge luxury liner to move with constant velocity, its engines need to supply a forward thrust of 6.85 105 N. What is the m
Leviafan [203]

Answer:

The magnitude of the resistive force exerted by the water is 6.85\times 10^{5} newtons.

Explanation:

By First and Second Newton's Law, the resistive force exerted by the water on the cruise ship has the same magnitude of forward thrust, with which it is antiparallel to. The equation of equilibrium for the luxury liner is:

\Sigma F = F-R = 0 (Eq. 1)

Where:

F - Forward trust, measured in newtons.

R - Resistive force exerted by the water, measured in newtons.

From (Eq. 1), we get that: (F = 6.85\times 10^{5}\,N)

R = F

R = 6.85\times 10^{5}\,N

The magnitude of the resistive force exerted by the water is 6.85\times 10^{5} newtons.

4 0
4 years ago
One way to describe a wave crest is the amount of ___________ applied over an area.
Pani-rosa [81]
I believe it’s force but i’m not really sure.
8 0
3 years ago
Read 2 more answers
I really need help please just answer at least one
yuradex [85]

Answer:

9) a = 25 [m/s^2], t = 4 [s]

10) a = 0.0875 [m/s^2], t = 34.3 [s]

11) t = 32 [s]

Explanation:

To solve this problem we must use kinematics equations. In this way we have:

9)

a)

v_{f}^{2} = v_{i}^{2}-(2*a*x)\\

where:

Vf = final velocity = 0

Vi = initial velocity = 100 [m/s]

a = acceleration [m/s^2]

x = distance = 200 [m]

Note: the final speed is zero, as the car stops completely when it stops. The negative sign of the equation means that the car loses speed or slows down as it stops.

0 = (100)^2 - (2*a*200)

a = 25 [m/s^2]

b)

Now using the following equation:

v_{f} =v_{i} - (a*t)

0 = 100 - (25*t)

t = 4 [s]

10)

a)

To solve this problem we must use kinematics equations. In this way we have:

v_{f} ^{2} =  v_{i} ^{2} + 2*a*(x-x_{o})

Note:  The positive sign of the equation means that the car increases his speed.

5^2 = 2^2 + 2*a*(125 - 5)

25 - 4 = 2*a* (120)

a = 0.0875 [m/s^2]

b)

Now using the following equation:

v_{f}= v_{i}+a*t\\

5 = 2 + 0.0875*t

3 = 0.0875*t

t = 34.3 [s]

11)

To solve this problem we must use kinematics equations. In this way we have:

v_{f} ^{2} =  v_{i} ^{2} + 2*a*(x-x_{o})

10^2 = 2^2 + 2*a*(200 - 10)

100 - 4 = 2*a* (190)

a = 0.25 [m/s^2]

Now using the following equation:

v_{f}= v_{i}+a*t\\

10 = 2 + 0.25*t

8 = 0.25*t

t = 32 [s]

4 0
3 years ago
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