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Lesechka [4]
3 years ago
11

1 2 3 4 5 6 7 8 9 10 TIME REMAINING 59:49 Which changes would result in a decrease in the gravitational force between two object

s? Check all that apply.
Physics
1 answer:
Masteriza [31]3 years ago
3 0

Answer:

<em>increasing the distance between the objects</em>

<em>decreasing the mass of one of the objects</em>

<em>decreasing the mass of both objects </em>

Explanation:

The complete question would be:

<em>Which changes would result in a decrease in the gravitational force between two objects? Check all that apply.</em>

<em>increasing the distance between the objects</em>

<em>decreasing the distance between the objects</em>

<em>increasing the mass of one of the objects</em>

<em>increasing the mass of both objects</em>

<em>decreasing the mass of one of the objects</em>

<em>decreasing the mass of both objects </em>

---------------------------------------------------------------------------------------

Gravitational force between two objects can be computed using the formula:

Fg = G\dfrac{Mm}{r^2}

Where:

Fg is the gravitational force (Newtons or N)

G is the gravitational constant (6.674⋅10⁻¹¹N(m/kg)²)

M is the mass of one object (Kilograms or Kg)

m is the mass of the other object (Kilograms or Kg)

r is the distance between the two objects (m)

Now notice the relationship between the mass and gravitational force. Their relationship is direct, meaning as the value of one goes up, the other goes up and the reverse holds true. If the mass of one or both of the objects decreases, the gravitational force also decreases.

Now look at the relationship between the distance between the objects and the gravitational force-- they are indirect. This means that as one increases, the other decreases. So if we increase the distance between the objects, the gravitational force between them decreases as well.

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What part of the gun position equipment supports all of the elevating parts of the gun?
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A player kicks a soccer ball in a high arc toward the opponent's goal. At the highest point in its trajectory
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At the highest point in its trajectory, the ball's acceleration is zero but its velocity is not zero.

<h3>What's the velocity of the ball at the highest point of the trajectory?</h3>
  • At the highest point, the ball doesn't go more high. So its vertical velocity is zero.
  • However, the ball moves horizontal, so its horizontal component of velocity is non - zero i.e. u×cosθ.
  • u= initial velocity, θ= angle of projection

<h3>What's the acceleration of the ball at the highest point of projectile?</h3>
  • During the whole projectile motion, the earth exerts the gravitational force with a acceleration of gravity along vertical direction.
  • But as there's no acceleration along vertical direction, so the acceleration along vertical direction is zero.

Thus, we can conclude that the acceleration is zero and velocity is non-zero at the highest point projectile motion.

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: Player kicks a soccer ball in a high arc toward the opponent's goal. At the highest point in its trajectory

A- neither the ball's velocity nor its acceleration are zero.

B- the ball's acceleration points upward.

C- the ball's acceleration is zero but its velocity is not zero.

D- the ball's velocity points downward.

Learn more about the projectile motion here:

brainly.com/question/24216590

#SPJ1

7 0
2 years ago
The weight of the atmosphere above 1 m- of
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Answer:

1.09 kg.m

Explanation:

no need

4 0
3 years ago
A uranium nucleus is traveling at 0.94 c in the positive direction relative to the laboratory when it suddenly splits into two p
Anettt [7]

Answer:

A   u = 0.36c      B u = 0.961c

Explanation:

In special relativity the transformation of velocities is carried out using the Lorentz equations, if the movement in the x direction remains

     u ’= (u-v) / (1- uv / c²)

Where u’ is the speed with respect to the mobile system, in this case the initial nucleus of uranium, u the speed with respect to the fixed system (the observer in the laboratory) and v the speed of the mobile system with respect to the laboratory

The data give is u ’= 0.43c and the initial core velocity v = 0.94c

Let's clear the speed with respect to the observer (u)

      u’ (1- u v / c²) = u -v

      u + u ’uv / c² = v - u’

      u (1 + u ’v / c²) = v - u’

      u = (v-u ’) / (1+ u’ v / c²)

Let's calculate

      u = (0.94 c - 0.43c) / (1+ 0.43c 0.94 c / c²)

      u = 0.51c / (1 + 0.4042)

      u = 0.36c

We repeat the calculation for the other piece

In this case u ’= - 0.35c

We calculate

       u = (0.94c + 0.35c) / (1 - 0.35c 0.94c / c²)

       u = 1.29c / (1- 0.329)

       u = 0.961c

6 0
3 years ago
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