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makvit [3.9K]
4 years ago
6

What would happen if the moon was hit by an asteroid

Physics
1 answer:
Semmy [17]4 years ago
8 0
If the moon was hit by an asteroid there would be a crater mark and possible movement.
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What's the kinetic energy of an object that has a mass of 30 kilograms and moves with a velocity of 20m/s?
icang [17]

Ek = 6KJ.

In physics, the kinetic energy of a body or object is the one that owns due to its movement and is given by the equation E_{k} = \frac{1}{2} mv^{2}, where m is the mass of the object in kilograms and v is the velocity in m/s.

An object that it has a mass of 30 kilograms and moves with a velocity of 20m/s, its kinetic energy is given by:

E_{k} = \frac{1}{2} (30kg)(20m/s)^{2}=6000J=6KJ

4 0
3 years ago
What is the speed of a transverse wave in a rope of length 2.00 m and mass 60.0 g under a tension of 500 n?
drek231 [11]

The formula we can use in this case would be:

v = sqrt (T / (m / l))

Where,

v = is the velocity of the transverse wave = unknown (?)

T = is the tension on the rope = 500 N

m = is the mass of the rope = 60.0 g = 0.06 kg

 l = is the length of the rope = 2.00 m

Substituting the given values into the equation to search for the speed v:
v = sqrt (500 N/(0.06 kg /2 m)) 
v = sqrt (500 * 2 / 0.06) 
v = sqrt (16,666.67) 
<span>v = 129.10 m/s</span>

7 0
3 years ago
A car of mass 1200Kilograms moving at 15 m/s the driver applies the brakes for 0.08 seconds and the castles down to 10 meter per
denpristay [2]
  • Initial velocity=u=15m/s
  • Final velocity=v=10m/s
  • Time=0.08s

\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}

\\ \sf\longmapsto Acceleration=\dfrac{10-15}{0.08}

\\ \sf\longmapsto Acceleration=\dfrac{-5}{0.08}

\\ \sf\longmapsto Acceleration=-62.5m/s^2

5 0
3 years ago
A 2kg 2 kg object is moving in along the x x -axis under a single conservative force. When the object is at position x=0 x = 0 ,
Luden [163]

Answer:

The particle comes to rest before reaching the position x=4m.

Explanation:

When the object is at x=0m, all of its energy is kinetic energy. Using the equation for kinetic energy yields KE=1/2mv^2=(12)(2)(3)^2=9J. Using the given equation for potential energy when the object is at x=4m yields U=4x^2=4(4)^2=64J. Since the system only has 9J of energy, the object comes to rest before reaching x=4m.

7 0
3 years ago
Planets beyond the solar system.On October 15, 2001, a planet was discovered orbiting around the star HD68988. Its orbital dista
Evgesh-ka [11]

Answer:

Part a)

M = 2.31 \times 10^{30} kg

Part b)

M = 1.15 M_s

Explanation:

Part a)

As we know that the period of revolution of the planet is given by

T = 2\pi\sqrt{\frac{r^3}{GM}}

now we know that

r = 10.5 \times 10^6 km = 1.05 \times 10^{10} m

T = 6.3 days

T = 5.44 \times 10^5 s

now we have

5.44 \times 10^5 = 2\pi\sqrt{\frac{(1.05\times 10^{10})^3}{(6.67 \times 10^{-11})M}}

M = 2.31 \times 10^{30} kg

Part b)

As we know that mass of sun is given as

M_s = 2.0 \times 10^{30} kg

now we have

\frac{M}{M_s} = \frac{2.31 \times 10^{30}}{2 \times 10^{30}}

M = 1.15 M_s

6 0
4 years ago
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