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3 years ago

Which objects will likely have the smallest gravitational force between them?

Physics

1 answer:

DENIUS [597]3 years ago

3 0

Answer:

A. Two tennis balls that are near each other

Explanation:

The formula for gravitational force (F) between two objects is

$F = \dfrac{Gm_{1}m_{2}}{d^{2} }$

where m₁ and m₂ are the masses of the two objects, d is the distance between their centres, and G is the gravitational constant.

Thus, two objects that are far from each other will have a smaller gravitational force. We can eliminate Options C and D.

If the objects are at the same distance, those with the smaller mass will have a smaller force.

The mass of a tennis ball is 57 g.

The mass of a soccer ball is 430 g.

Two tennis balls that are near each other will have a smaller gravitational attraction.

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two bumper cars at an amusement park collide together abd get stuck together. assuming that the system of the two bumper cars is

Fiesta28 [93]

The answer to this problem is M1v1-m2v2=(m1+m2)v

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3 years ago

A 0.50-kg bomb is sliding along an icy pond (frictionless surface) with a velocity of 2.0 m/s to the west. The bomb explodes int

nevsk [136]

Answer:

2.667m/s to the north and 3.333 m/s to the west

Explanation:

According to law of momentum conservation, the total momentum should be conserved before and after the explosion.

Before the explosion, the momentum was

0.5*2 = 1 kg m/s to the west

Therefore the total momentum after the explosion should be the same horizontally and vertically.

Vertically speaking, it was 0 before the explosion. After the explosion:

0.2*4 + 0.3v = 0

0.3v = -0.8

v = -0.8/0.3 = -2.667 m/s

So the vertical component of the 0.3kg piece is 2.667m/s to the north

Horizontally speaking, since the 0.2kg-piece doesn't move west or east post-explosion:

0.2*0 + 0.3V = 1

0.3V = 1

V = 1/0.3 = 3.333 m/s

So the horizontal component of the 0.3kg piece is 3.333 m/s to the west

5 0

3 years ago

Read 2 more answers

Explain the relationship between lightning and storm clouds.

LuckyWell [14K]

Answer: Often lightning occurs between clouds or inside a cloud. But the lightning we usually care about most is the lightning that goes from clouds to ground.

Explanation: As the storm moves over the ground, the strong negative charge in the cloud attracts positive charges in the ground.

3 0

1 year ago

A force of 14 N acts on a 5 kg object for 3 seconds.

DiKsa [7]

Answer: a) 42Nm b) 8.4m/s

Explanation:

Impulse is defined as object change in momentum.

Since Force = mass × acceleration

F = ma

Acceleration is the rate of change in velocity.

F = m(v-u)/t

Cross multiply

Ft = m(v-u)

Since impulse = Ft

and Ft = m(v-u)... (1)

The object change in velocity (v-u) = Ft/m from eqn 1

Going to the question;

a) Impulse = Force (F) × time(t)

Given force = 14N and time = 3seconds

Impulse = 14×3

Impulse = 42Nm

b) The object change in velocity (v-u) = Ft/m where mass = 5kg

v-u = 14×3/5

Change in velocity = 42/5 = 8.4m/s

3 0

3 years ago

Compute the size of the charge necessary for two spheres separated by 1m to be attached with the force of 1N. How many electrons

yarga [219]

Answer:

$q\approx 6.6\cdot 10^{13}~electrons$

Explanation:

<u>Coulomb's Law</u>

The force between two charged particles of charges q1 and q2 separated by a distance d is given by the Coulomb's Law formula:

$\displaystyle F=k\frac{q_1q_2}{d^2}$

Where:

$k=9\cdot 10^9\ N.m^2/c^2$

q1, q2 = the objects' charge

d= The distance between the objects

We know both charges are identical, i.e. q1=q2=q. This reduces the formula to:

$\displaystyle F=k\frac{q^2}{d^2}$

Since we know the force F=1 N and the distance d=1 m, let's find the common charge of the spheres solving for q:

$\displaystyle q=\sqrt{\frac{F}{k}}\cdot d$

Substituting values:

$\displaystyle q=\sqrt{\frac{1}{9\cdot 10^9}}\cdot 1$

$q = 1.05\cdot 10^{-5}~c$

This charge corresponds to a number of electrons given by the elementary charge of the electron:

$q_e=1.6 \cdot 10^{-19}~c$

Thus, the charge of any of the spheres is:

$\displaystyle q = \frac{1.05\cdot 10^{-5}~c}{1.6 \cdot 10^{-19}~c}$

$\mathbf{q\approx 6.6\cdot 10^{13}~electrons}$

5 0

2 years ago