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Over [174]
2 years ago
9

If Earth were 10.0 times farther away from the Sun than it is now, how many times weaker would the gravitational force between t

he Sun and Earth be
Physics
1 answer:
choli [55]2 years ago
5 0

If Earth were 10.0 times farther away from the Sun than it is now, 100 times weaker would the gravitational force between the Sun and Earth.

What is Gravitational Force?

According to Newton's universal law of gravitation, The force of attraction between any two bodies is directly proportional to the product of their masses and is inversely proportional to the square of the distance between them.

What causes gravitational force?

Earth's gravity comes from all its mass. All its mass makes a combined gravitational pull on all the mass in your body. That's what gives you weight. And if you were on a planet with less mass than Earth, you would weigh less than you do here.

Learn more about gravitational force:

brainly.com/question/862529

#SPJ4

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1.) If you make a false statement or commit a forgery about your motor vehicle insurance you can be guilty of a _____ degree mis
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You would be required to have bodily injury liability insurance in Florida if you are involved in a crash where your vehicle has caused damage to the property of others.

 

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As a mass on a spring moves farther from the equilibrium position, how do the velocity, acceleration, and force change
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Refer to the diagram shown below.

m =  the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A =  the amplitude ( the maximum distance) of the mass from the equilibrium
        position

The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω =  the circular frequency of the motion
T =  the period of the motion so that ω = (2π)/T

The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)

In the equilibrium position,
x is zero;
v is maximum;
a is zero.

At the farthest distance (A) from the equilibrium position,
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6 0
3 years ago
The surface is tilted to an angle of 37 degrees from the horizontal, as shown above in Figure 3. The blocks are each given a pus
hoa [83]

Answer:

Incomplete question: "Each block has a mass of 0.2 kg"

The speed of the two-block system's center of mass just before the blocks collide is 2.9489 m/s

Explanation:

Given data:

θ = angle of the surface = 37°

m = mass of each block = 0.2 kg

v = speed = 0.35 m/s

t = time to collision = 0.5 s

Question: What is the speed of the two-block system's center of mass just before the blocks collide, vf = ?

Change in momentum:

delta(P)=F*delta(t)

P_{f} -P_{i}=F*delta(t)

2m(v_{f} -v_{i})=F*delta(t)

v_{i} =0.35-0.35=0

It is neccesary calculate the force:

F=(m+m)*g*sin\theta

Here, g = gravity = 9.8 m/s²

F=(0.2+0.2)*9.8*sin37=2.3591N

v_{f} =\frac{F*delta(t)}{2m} =\frac{2.3591*0.5}{2*0.2} =2.9489m/s

6 0
3 years ago
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