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Gnom [1K]
2 years ago
10

G. Example: A satellite of mass 100kg is in orbit around the Moon at an

Physics
1 answer:
Aloiza [94]2 years ago
8 0

Answer:

ther are many satellites out on earth only some are important

Explanation:

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A body of mass 20kg initially at rest is subjected to a force of 40m for 15sec. Calculate the change in k.e
Ede4ka [16]

Explanation:

mass(m)=20kg

velocity(v)=d/t=2.67

k.e=?

now,

k.e=1/2mv^2

=1/2*20*(2.67)^2

=71J

3 0
2 years ago
Read 2 more answers
The fragment of an asteroid or any interplanetary material is known as a: A) limestone dignitary satellite. B) moon. C) shower m
ElenaW [278]
The fragment of an asteroid or any interplanetary material is known as a a : D. Meteroid

Human came in contact with this material mostly because it penetrate the atmosphere and fall within the earth surface

hope this helps
6 0
2 years ago
How much power does it take to do 500 J of work in 10 seconds?
Dmitry_Shevchenko [17]
Power = work/time
  
          = 500/10
 
          = 50J/s or 50 watt 


7 0
3 years ago
A 3.0-kg block starts at rest at the top of a 37° incline, which is 5.0 m long. Its speed when it reaches the bottom is 2.0 m/s.
Mama L [17]

Answer: f_{r} = 16.49N

Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,

W_{x} = horizontal component of the weight = mgsinФ

W_{y} = vertical component of weight = mgcosФ

Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.

by using newton's law of motion, we have that

mgsinФ - f_{r} = ma

where m = mass of object=5kg

a = acceleration= unknown

Ф = angle of inclination = 37°

g = acceleration due to gravity = 9.8m/s^{2}

f_{r} = frictional force = unknown

we need to first get the acceleration before the frictional force which is gotten by using the equation below

v^{2} = u^{2} + 2aS

where v = final velocity = 2m/s

u = initial velocity = 0m/s (because the object started from rest)

a= unknown

S= distance covered = length of plane = 5m

2^{2} = 0^{2} + 2*a*5\\\\4= 10 *a\\\\a = \frac{4}{10} \\a = 0.4m/s^{2}

we slot in a into the equation below to get frictional force

mgsinФ - f_{r} = ma

3 * 9.8 * sin 37 - f_{r} = 3* 0.4

17.9633 - f_{r} =  1.2

f_{r} = 17.9633 - 1.2

f_{r} = 16.49N

4 0
3 years ago
Doe anyone get this ​
8_murik_8 [283]

Answer:

we know a = F/ M

Explanation:

  • 2 m/s²
  • 0.19 m/s²
  • 9.25 m/ s²
  • 0.04 m/s²
  • 100.39 m/s²

4 0
2 years ago
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