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2 years ago

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Consider two celestial objects with masses m1 and m2 with a separation distance between their centers r. If the first mass m1 we

re to double and the second mass m2 were to triple, what would happen to the magnitude of the force of attraction

1 answer:

Scilla [17]2 years ago

6 0

The new magnitude of the force of attraction will be 6 times the original force of attraction

<h3>How to determine the initial force </h3>

Mass 1 = m₁
Mass 2 = m₂
Gravitational constant = G
Distance apart = r
Initial force (F₁) = ?

F = Gm₁m₂ / r²

F₁ = Gm₁m₂ / r²

<h3>How to determine the new force </h3>

Mass 1 = 2m₁
Mass 2 = 3m₂
Gravitational constant = G
Distance apart (r) = r
New force (F₂) =?

F = Gm₁m₂ / r²

F₂ = G × 2m₁ × 3m₂ / r²

F₂ = 6Gm₁m₂ / r²

But

F₁ = Gm₁m₂ / r²

Therefore

F₂ = 6Gm₁m₂ / r²

F₂ = 6F₁

Thus, the new magnitude of the force of attraction will be 6 times the original force of attraction

Learn more about gravitational force:

brainly.com/question/21500344

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A gun shoots a bullet with a velocity of 500 m/s. The gun is aimed horizontally and fired from a height of 1.5 m. How far does t

MrMuchimi

The bullet travels a horizontal distance of 276.5 m

The bullet is shot forward with a horizontal velocity $u_x$ . It takes a time <em>t</em> to fall a vertical distance <em>y</em> and at the same time travels a horizontal distance <em>x. </em>

The bullet's horizontal velocity remains constant since no force acts on the bullet in the horizontal direction.

The initial velocity of the bullet has no component in the vertical direction. As it falls through the vertical distance, it is accelerated due to the force of gravity.

Calculate the time taken for the bullet to fall through a vertical distance <em>y </em>using the equation,

$y=u_yt+\frac{1}{2} gt^2$

Substitute 0 m/s for $u_y$ , 9.81 m/s²for <em>g</em> and 1.5 m for <em>y</em>.

$y=u_yt+\frac{1}{2} gt^2\\ 1.5 m=(0m/s)t+\frac{1}{2} (9.81m/s^2)t^2\\ t=\sqrt{\frac{2(1.5m)}{9.81m/s^2} } =0.5530s$

The horizontal distance traveled by the bullet is given by,

$x=u_xt$

Substitute 500 m/s for $u_x$ and 0.5530s for t.

$x=u_xt\\ =(500m/s)(0.5530s)\\ =276.5m$

The bullet travels a distance of 276.5 m.

5 0

3 years ago

8. A meter reader determines that a business has used 5000 kW.h of energy in 4 months. If

ladessa [460]

Answer:

ENERGY AND COST. One kllowatt hour is 1,000 watts of power for one hour of time. ... Determine power: P = V XI ... Calculate the total kilowatt hours used. ... If the electric costs are 150 per kWh, how much does it cost to run the refrigerator in ... 8. A room was lighted with three 100-watt bulbs for 5 hours per day. If the cost of.

Explanation:

7 0

3 years ago

in a solar system far, far away the sun's intensity is 200 w/m2 for an inner planet located a distance r away. what is the sun's

GarryVolchara [31]

The sun's intensity for an outer planet located at a distance 6r from the sun is 5.55 W/m². The result is obtained by using the inverse square law formula.

<h3>What is the Inverse Square Law formula?</h3>

The Inverse Square Law formula describes the intensity of light is inversely proportional to the square of the distance. It can be expressed as

$\frac{I_{1} }{I_{2} } = \frac{d_{2}^{2} }{d_{1}^{2}}$

Where

I₁ = Intensity at distance 1 (W/m²)
I₂ = Intensity at distance 2 (W/m²)
d₁ = distance 1 from a light source (m)
d₂ = distance 2 from a light source (m)

Given the case the sun's intensity is 200 W/m² for an inner planet at the distance r. If an outer planet is at a distance 6r, what is the sun's intensity?

By using the inverse square law formula, the sun's intensity for an outer planet is

$\frac{I_{1} }{I_{2} } = \frac{d_{2}^{2} }{d_{1}^{2}}$

$\frac{200 }{I_{2} } = \frac{(6r)^{2} }{r^{2}}$

$\frac{200 }{I_{2} } = \frac{36r^{2} }{r^{2}}$

$I_{2} = \frac{200} {36}$

I₂ = 5.55 W/m²

Hence, the sun's intensity for a planet at a distance 6r from the sun is 5.55 W/m².

Learn more about intensity of light here:

brainly.com/question/13155277

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3 0

1 year ago

!WILL GIVE BRAINLIEST!

vazorg [7]

As per angular momentum conservation we can say

$L_i = L_f$

here we know that

$mv_1r_1 = mv_2r_2$

we know that

$v_1 = 1.1 m/s$

$r_1 = 0.6 m$

$v_2 = ?$

$r_2 = 0.15 m$

now from above equation

$2(1.1)(0.6) = 2(v_2)(0.15)$

$v_2 = \frac{(1.1)(0.6)}{0.15}$

$v_2 = 4.4 m/s$

so speed is 4.4 m/s

8 0

3 years ago

If you drive 30 miles in 40 minutes, and then 15 miles in 20 minutes, do you have a constant speed?

kompoz [17]

Answer:

A) yes

Explanation:

First section of trip: 30 miles in 40 minutes

Second section of trip: 15 miles in 20 minutes

The formula for speed is distance over time $k=\frac{d}{t}$

Calculate the speeds for each section of the trip.

First:

k = d/t

k = 30miles/40minutes <= reduce fraction by 10 (30÷10 and 40÷10)

k = 3 miles / 4 minutes

Second:

k = d/t

k = 15miles/20minutes <= reduce fraction by 5 (15÷5 and 20÷5)

k = 3 miles / 4 minutes

Therefore there is a constant speed because both sections of the trip are driving at "3 miles / 4 minutes".

3 miles / 4 minutes can be also formatted as:

0.75 miles per minute.

7 0

3 years ago