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3 years ago

Increasing the two objects will cause the gravitational force between the objects to decrease.

2 answers:

8 0

Well first, I don't quite understand how you can "increase two objects".
I mean, you can increase their size, density, mass, weight, or cost,
but how do you "increase" the objects ?

If you were to, say, increase their mass ... or even the mass of only
one of them ... then the gravitational forces between the objects would
increase in strength.

So if you meant to say "increasing the mass of two objects", then
you've got yourself a nice genuine false statement there.

ELEN [110]3 years ago

4 0

This statement is false. Increasing the two objects' mass (I'm guessing) will actually increase their gravitational force. This is because of the equation:

$F_g = \frac{Gm_1m_2}{d^2}$

If the distance was increased, then the statement would be true, but since you are increasing mass, which is proportional to the Force of Gravity, you are in fact, increasing the gravitational force between the two objects.

You might be interested in

A ball thrown with 50N of force accelerates at 25 m/s2. What is the mass of the ball?

MrRissso [65]

F=ma
Force is 50N. Acceleration is 25 m/s^2.
50N=m*25 m/s^2
Divide both sides by 25.
mass=2 kg

7 0

3 years ago

Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa

anyanavicka [17]

Answer:

The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

Explanation:

Given that,

First charge $q_{1}= 2.9\mu C$

Second charge $q_{2}= 2.9\mu C$

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

$E_{1}=\dfrac{kq}{r'^2}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}$

$E_{1}=52200=52.2\times10^{3}\ N/C$

For E₂,

Using formula of electric field

$E_{1}=\dfrac{kq}{r^2}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}$

$E_{2}=104400=104.4\times10^{3}\ N/C$

We need to calculate the horizontal electric field

$E_{x}=E_{1}\cos\theta$

$E_{x}=52.2\times10^{3}\times\cos45$

$E_{x}=36910.97=36.9\times10^{3}\ N/C$

We need to calculate the vertical electric field

$E_{y}=E_{2}+E_{1}\sin\theta$

$E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45$

$E_{y}=141310.97=141.3\times10^{3}\ N/C$

We need to calculate the net electric field

$E_{net}=\sqrt{E_{x}^2+E_{y}^2}$

Put the value into the formula

$E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}$

$E_{net}=146038.69\ N/C$

$E_{net}=146.03\times10^{3}\ N/C$

We need to calculate the direction of electric field

Using formula of direction

$\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}$

$\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})$

$\theta=75.36^{\circ}$

Hence, The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

4 0

3 years ago

You plan to take your hair blower to europe, where the electrical outlets put out 240 v instead of the 120 v seen in the united

Dmitrij [34]

P = U × I
I = P / U = 1600W / 120V = 13.4A
P = 240V × 13.4A = 3216W

If your hair blower isnt rated for 220- 230V(this is the voltage in EU) you are most likely going to burn it.

8 0

3 years ago

A student pulls a 2.0-kg object to the left with a force of 30 N, while another student is pulling against the object in the opp

alex41 [277]

Answer:

5 m/s2, left

Explanation:

We can solve the problem by applying Newton's second law of motion, which states that:

$\sum F=ma$