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Veseljchak [2.6K]
2 years ago
9

how does gravity affect objects of different mass close to earth, and how does that effect change as an object moves farther fro

m earth?
Physics
1 answer:
Luda [366]2 years ago
3 0

Answer:

It makes it lighter when its closer and heavier when its farther way.

Explanation:

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b.  \: light

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HELP!! ALL MY POINTS WILL BE GIVEN
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-6 m/s^2

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30 - 90 = -60

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3 years ago
An object with total mass mtotal = 16.2 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.7 k
vichka [17]

Answer:

1.) 0 kgm/s

2) 6.3 kg

3) -0.0978 m/s

4)

5)

6)

An object with total mass mtotal = 16.2 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.7 kg moves up and to the left at an angle of θ1 = 23° above the –x axis with a speed of v1 = 25.4 m/s. A second piece with mass m2 = 5.2 kg moves down and to the right an angle of θ2 = 28° to the right of the -y axis at a speed of v2 = 23.8 m/s.

2) What is the mass of the third piece?

3) What is the x-component of the velocity of the third piece?

4) What is the y-component of the velocity of the third piece?

5) What is the magnitude of the velocity of the center of mass of the pieces after the collision?

6) Calculate the increase in kinetic energy of the pieces during the explosion

Explanation:

Since explosions and collisions follow the law of conservation of Momentum.

1) Magnitude of the final momentum of the system = Magnitude of the initial momentum of the system

Since the body was initially at rest,

Magnitude of the initial momentum of the system = 0 kgm/s

Hence, Magnitude of the final momentum of the system is also equal to 0 kgm/s.

2) mass of the third piece

Sum of all the masses = 16.2 kg

4.7 + 5.2 + c = 16.2

c = 6.3 kg

3) doing an x-component balance on momentum

(4.7)×(-25.4 cos 23) + (5.2)×(23.8 cos 28) + (6.3)(v) = 0

-0.6163 + 6.3v = 0

v = -0.0978 m/s

Hope this Helps!!!

4 0
3 years ago
a 2100-kg car drives with a speed of 18 m/s onb a flat road around a curve that has a radius of curvature of 83m. The coefficien
Law Incorporation [45]

Answer:

<u>The magnitude of the friction force is 8197.60 N</u>

Explanation:

Using the definition of the centripetal force we have:

\Sigma F=ma_{c}=m\frac{v^{2}}{R}

Where:

  • m is the mass of the car
  • v is the speed
  • R is the radius of the curvature

Now, the force acting in the motion is just the friction force, so we have:

F_{f}=m\frac{v^{2}}{R}

F_{f}=2100\frac{18^{2}}{83}

F_{f}=8197.60 \: N

<u>Therefore the magnitude of the friction force is 8197.60 N</u>

I hope it helps you!

7 0
3 years ago
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