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trasher [3.6K]
2 years ago
10

Suppose two objects are gravitationally attracted to each other with some force F. If the mass of object 1 is multiplied by a fa

ctor of five and the mass of object 2 is multiplied by a factor of two, what will the new gravitational force be between the objects?
A. F2

B. 1/2F

C. 0.1F

D. 2F/5

E. 5F

J. 10F

Please help will mark brainliest

Physics
2 answers:
RoseWind [281]2 years ago
8 0

If F = Gm₁m₂/d², and we change m₁ to 5m₁ and m₂ to 2m₂, then the new magnitude of the gravitational force is

F' = G (5m₁) (2m₂) / d²

F' = 10  Gm₁m₂ / d²

but this is really just F' = 10F. So J is the correct choice.

miskamm [114]2 years ago
4 0

Hi there!

Using Newton's Law of Universal Gravitation:

\large\boxed{F_g = G\frac{m_1m_2}{d^2}}

Fg = Force of gravity (N)

G = Gravitational Constant

m₁ = mass of object 1 (kg)

m₂ = mass of object 2 (kg)

d = distance between the objects (m)

There is a direct relationship between the masses of the objects and the resulting force of gravity, so we can plug new values in:

F_g = G\frac{5m_12m_2}{d^2}} = 10G\frac{m_1m_2}{d^2}} = 10F_g

Thus, the correct answer is J. 10Fg.

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Two electrons with a charge of magnitude 1.6×10-19 C in an atom are separated by 1.5×10-10 m, the typical size of an atom. What
vesna_86 [32]

Answer:

1.02\cdot 10^{-8} N, repulsive

Explanation:

The magnitude of the electric force between two charged particles is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the two charges of the two particles

r is the separation between the two charges

The force is:

- repulsive if the two charges have  same  sign

- Attractive if the two charges have opposite signs

In this problem, we have two electrons, so:

q_1=q_2=1.6\cdot 10^{-19}C is the magnitude of the two electrons

r=1.5\cdot 10^{-10} m is their separation

Substituting into the formula, we find the electric force between them:

F=(8.99\cdot 10^9)\frac{(1.6\cdot 10^{-19})^2}{(1.5\cdot 10^{-10})^2}=1.02\cdot 10^{-8} N

And the force is repulsive, since the two electrons have same sign charge.

4 0
3 years ago
Need help will mark you the Brainliest
Hatshy [7]
The last one is correct (D)
8 0
3 years ago
Two rigid tanks of equal size and shape are filled with different gases. The tank on the left contains oxygen, and the tank on t
fredd [130]

Answer:

The number of oxygen molecules in the left container greater than the number of hydrogen molecules in the right container.

Explanation:

Given:

Molar mass of oxygen, M_O=32

Molar mass of hydrogen, M_H=2

We know ideal gas law as:

PV=nRT

where:

P = pressure of the gas

V = volume of the gas

n= no. of moles of the gas molecules

R = universal gs constant

T = temperature of the gas

∵n=\frac{m}{M}

where:

m = mass of gas in grams

M = molecular mass of the gas

∴Eq. (1) can be written as:

PV=\frac{m}{M}.RT

P=\frac{m}{V}.\frac{RT}{M}

        as: \frac{m}{V}=\rho\ (\rm density)

So,

P=\rho.\frac{RT}{M}

Now, according to given we have T,P,R same for both the gases.

P_O=P_H

\rho_O.\frac{RT}{M_O}=\rho_H.\frac{RT}{M_H}

\Rightarrow \frac{\rho_O}{32}=\frac{\rho_H}{2}

\rho_O=16\rho_H

∴The molecules of oxygen are more densely packed than the molecules of hydrogen in the same volume at the same temperature and pressure. So, <em>the number of oxygen molecules in the left container greater than the number of hydrogen molecules in the right container.</em>

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