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melomori [17]
3 years ago
7

A force of 5.25 newtons acts on an object of unknown mass at a distance of 6.9 × 108 meters from the center of Earth. To increas

e the force to 2.5 times its original value, how far should the object be from the center of Earth?
Physics
2 answers:
alexira [117]3 years ago
7 0

Answer:

4.36 x 10^8 m

Explanation:

Let the mass of earth is M and unknown mass is m.

According to the Newton's law of gravitation, force between two objects is given by

F = G\frac{M m}{d^{2}}

Here, F = 5.25 N, d = 6.9 x 10^8 m

5.25 = G\frac{M m}{(6.9\times 10^{8})^{2}}       .... (1)

Now, F' = 2.5 F and d be the distance

2.5\times 5.25 = G\frac{M m}{(d)^{2}}            ..... (2)

Divide equation (1) by equation by (2)

\frac{1}{2.5} = \left ( \frac{d}{6.9\times 10^{8}} \right )^{2}

d = 4.36 x 10^8 m

NemiM [27]3 years ago
5 0
The force between two objects is calculated through the equation, 
                        F = Gm₁m₂/d²
where m₁ and m₂ are the masses of the objects. In this case, an unknown mass and Earth. d is the distance between them and G is universal gravitation constant. 

In the second case, if the force is to become 2.5 times the original and all the variables are constant except d then,
                       2.5F = Gm₁m₂ / (D²)
                                D = 0.623d

Subsituting the known value of d,
                                D = 0.623(6.9 x 10^8) = <em>4.298 x 10^8 m</em>
                           
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A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.

A car experiences a deceleration (a) of -41.62 m/s² and comes to a stop (final velocity = v = 0 m/s) after 10.99 m (s).

We can calculate the initial velocity of the car (u) using the following kinematic equation.

v^{2} = u^{2} + 2as\\\\u = \sqrt[]{v^{2}-2as} = \sqrt[]{(0m/s)^{2}-2(-42.61m/s^{2} )(10.99m)} = 30.60m/s

A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.

Learn more: brainly.com/question/14851168

5 0
2 years ago
A ball is kicked from level ground at an angle 60° with initial velocity 10 m/s. The distance the ball travels, in meters, is:
RUDIKE [14]

Answer:

<em>the ball travels a distance of 8.84 m</em>

Explanation:

Range: Range is defined as the horizontal distance from the point of projection to the point where the projectile hits the projection plane again.

R = (U²sin2∅)/g.............................. Equation 1

Where R = range, U = initial velocity, ∅ = angle of projection, g = acceleration due to gravity.

<em>Given: U = 10 m/s, ∅ = 60°</em>

<em>Constant: g = 9.8 m/s²</em>

Substituting these values into equation 1

R = [10²×sin(2×60)]/9.8

R = (100sin120)/9.8

R = 100×0.8660/9.8

R = 86.60/9.8

R = 8.84 m

<em>Therefore the ball travels a distance of 8.84 m</em>

4 0
3 years ago
A planet has a gravitational acceleration on its surface of 2.2 times Earth's gravitational acceleration on its surface. The pla
lesantik [10]

Answer:

The mass of the planet is 55 times the mass of earth.

Explanation:

From the inverse-square gravitation law,

F = (GMm/r²)

If the weight of a body (the force with which the earth attracts a body to its centre) is to be calculated,

F = mg

m = mass of the body,

g = acceleration due to gravity

mg = (GMm/r²)

G = Gravitational constant

M = mass of the earth

m = mass of body

r = distance between the body and the centre of the earth = radius of the earth

The acceleration due to gravity is given by

g = (GM/r²)

Making the mass of the earth, the subject of formula

M = (gr²/G) (eqn 1)

So, the planet described,

Let the acceleration due to gravity on the planet be g₁

Mass of the planet be M₁

Radius of the planet be r₁

g₁ = 2.2g

r₁ = 5r

M₁ = ?

Note that the gravitational constant is the same for both planets.

So, we can write a similar expression for the planet's acceleration due to gravity

g₁ = (GM₁/r₁²)

Substituting all the parameters known in terms of their corresponding earth values

2.2g = [GM₁/(5r)²]

2.2g = [GM₁/25r²]

M₁ = (55gr²/G)

Recall the expression for the mass of the earth

M = (gr²/G)

M₁ = 55 M

The mass of the planet, in terms of Earth masses = 55M

The mass of the planet is 55 times the planet of earth.

Hope this Helps!!!

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