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3 years ago

Which change will always result in an increase in the gravitational force between two objects?

2 answers:

8 0

The change that will always result in an increase in the gravitational force between two objects is increasing the masses of the objects and decreasing the distance between the objects.

nikdorinn [45]3 years ago

6 0

(a) and (b) can increase the gravitational attraction between two objects
IF you pick the increase and decrease just so, but not always.

(d) will always decrease the gravitational force between them.

(c) is the only choice that will always increase the gravitational force between them, no matter how much the increase and the decrease are.

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One disadvantage of using proprietary licensed software is that

Vaselesa [24]

Answer:

its b

Explanation:

5 0

3 years ago

Read 2 more answers

How much work was done by a hot air balloon to lift up a 100 Newton to a height of 300 meters?

Monica [59]

So, the work was done by that hot air-balloon is <u>30,000 J or 30 kJ</u>.

<h3>Introduction</h3>

Hi ! In this question, I will help you. <u>Work is the amount of force exerted to cause an object to move a certain distance from its starting point</u>. In physics, the amount of work will be proportional to the increase in force and increase in displacement. Amount of work can be calculated by this equation :

$\boxed{\sf{\bold{W = F \times s}}}$

With the following condition :

W = work (J)
F = force (N)
s = shift or displacement (m)

Now, the s (displacement) can be written as ∆h (altitude change) because the object move to vertical line. The formula can also be changed to:

$\boxed{\sf{\bold{W = F \times \Delta h}}}$

With the following condition :

W = work (J)
F = force (N)
$\sf{\Delta h}$ = change of altitude (m)

If an object has mass, then the object will also be affected by gravity. Always remember that F = m × g. So that :

$\sf{W = F \times \Delta h}$

$\boxed{\sf{\bold{W = m \times g \times \Delta h}}}$

With the following condition :

W = work (J)
m = mass of the object (kg)
g = acceleration of the gravity (m/s²)
$\sf{\Delta h}$ = change of altitude (m)

<h3>Problem Solving</h3>

We know that :

F = force = 100 N
$\sf{\Delta h}$ = change of altitude 300 m

What was asked :

W = work = ... J

Step by step :

$\sf{W = F \times \Delta h}$

$\sf{W = 100 \times 300}$

$\boxed{\sf{W = 30,000 \: J = 30 \: kJ}}$

<h3>Conclusion</h3>

So, the work was done by that hot air-balloon is 30,000 J or 30 kJ.

Work that he had done to lift object brainly.com/question/26341717
Converting work to potential energy brainly.com/question/26487284

5 0

2 years ago

A bowling ball is far from uniform. Lightweight bowling balls are made of a relatively low-density core surrounded by a thin she

tester [92]

Answer:

a) I = 1,75 10-² kg m² and b) I = 1.49 10⁻² kg m²

Explanation:

The expression for the moment of inertia is

I = ∫ r² dm

The moment of inertia is a scalar by which an additive magnitude, we can add the moments of inertia of each part of the system, taking into account the axis of rotation.

I = I core + I shell

The moment of inertia of a solid sphere is

I sphere = 2/5 MR²

The moment of inertia of a thin spherical shell is

I shell = 2/3 M R²

a) Let's apply to our system, first to the core of weight 1.6 kg and diameter 0.196m, the radius is half the diameter

R = d / 2

R= 0.196 m / 2 = 0.098 m

I core = 2/5 1.6 0.098²

I core = 6.147 10-3 kg m²

Let's calculate the moment of inertia of the shell of mass 1.6 kg with a diameter of 0.206 m

R = 0.206 / 2

R = 0.103 m

I shell = 2/3 1.6 0.103²

I shell = 1,132 10-2 kg m²

The moment of inertia of the ball is the sum of these moments of inertia,

I = I core + I shell

I = 6,147 10⁻³ + 1,132 10⁻² = 6,147 10⁻³ + 11.32 10⁻³

I = 17.47 10⁻³ kg m²

I = 1,747 10-² kg m²

b) Now the ball is report with mass 3.2kg and diameter 0.216 m

R = 0.216 / 2

R = 0.108 m

It is a uniform sphere

I = 2/5 M R²

I = 2/5 3.2 0.108²

I = 1.49 10⁻² kg m²

7 0

3 years ago

How would you explain force and movement to a friend who doesn’t know much about it?

kirza4 [7]

Answer:

I can explain it using the law of force

or Newton's law of motion

6 0

3 years ago

A rocket initially at rest accelerates at a rate of 99.0 meters/second2. Calculate the distance covered by the rocket if it atta

dezoksy [38]

Vi = 0m/s
A = 99m/ $s^{2}$
Vf = 445m/s
T = 4.5s
D = ?

D = Vi(t) + 1/2(a)(t²)
D = 0m/s ( 4.5s) + 1/2(99m/s²)(4.5s)²
D = 1002 m
D = 1 x 10³ m

6 0

3 years ago