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k0ka [10]
3 years ago
11

True or false gravitational pull decreases with an increase of distance between two objects​

Physics
1 answer:
ohaa [14]3 years ago
5 0

Answer:

true! : )

(i underlined the place where the answer is the other information is just as important but if you do not want to read it you do not have to)

Explanation:

Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces. So as two objects are separated from each other, the force of gravitational attraction between them also decreases. the greater the mass, the greater the gravitational pull. <u>gravitational pull decreases with an increase in the distance between two objects.</u> Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces. So as two objects are separated from each other, the force of gravitational attraction between them also decreases.

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Determine qual é a carga elétrica de um corpo que possui 1 milhão de partículas?
MaRussiya [10]

Answer:

This can be translated to:

"find the electrical charge of a body that has 1 million of particles".

First, it will depend on the charge of the particles.

If all the particles have 1 electron more than protons, we will have that the charge of each particle is q = -e = -1.6*10^-19 C

Then the total charge of the body will be:

Q = 1,000,000*-1.6*10^-19 C = -1.6*10^-13 C

If we have the inverse case, where we in each particle we have one more proton than the number of electrons, the total charge will be the opposite of the one of before (because the charge of a proton is equal in magnitude but different in sign than the charge of an electron)

Q = 1.6*10^-13 C

But commonly, we will have a spectrum with the particles, where some of them have a positive charge and some of them will have a negative charge, so we will have a probability of charge that is peaked at Q = 0, this means that, in average, the charge of the particles is canceled by the interaction between them.

7 0
3 years ago
If the velocity of a pitched ball has a magnitude of 44.5 m/sm/s and the batted ball's velocity is 55.5 m/sm/s in the opposite d
Yuliya22 [10]

Incomplete question as the mass of baseball is missing.I have assume 0.2kg mass of baseball.So complete question is:

A baseball has mass 0.2 kg.If the velocity of a pitched ball has a magnitude of 44.5 m/sm/s and the batted ball's velocity is 55.5 m/sm/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.

Answer:

ΔP=20 kg.m/s

Explanation:

Given data

Mass m=0.2 kg

Initial speed Vi=-44.5m/s

Final speed Vf=55.5 m/s

Required

Change in momentum ΔP

Solution

First we take the batted balls velocity as the final velocity and its direction is the positive direction and we take the pitched balls velocity as the initial velocity and so its direction will be negative direction.So we have:

v_{i}=-44.5m/s\\v_{f}=55.5m/s

Now we need to find the initial momentum

So

P_{1}=m*v_{i}

Substitute the given values

P_{1}=(0.2kg)(-44.5m/s)\\P_{1}=-8.9kg.m/s

Now for final momentum

P_{2}=mv_{f}\\P_{2}=(0.2kg)(55.5m/s)\\P_{2}=11.1kg.m/s

So the change in momentum is given as:

ΔP=P₂-P₁

=[(11.1kg.m/s)-(-8.9kg.m/s)]\\=20kg.m/s

ΔP=20 kg.m/s

3 0
3 years ago
A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical ener
defon

Complete question:

A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical energy of the spring–load system is 2.00 J. Find

(a) the force constant of the spring and (b) the amplitude of the motion.

Answer:

(a) the force constant of the spring = 47 N/m

(b) the amplitude of the motion = 0.292 m

Explanation:

Given;

mass of the spring, m = 200g = 0.2 kg

period of oscillation, T = 0.410 s

total mechanical energy of the spring, E = 2 J

The angular speed is calculated as follows;

\omega = \frac{2\pi}{T} \\\\\omega = \frac{2\pi}{0.41} \\\\\omega = 15.33 \ rad/s

(a) the force constant of the spring

\omega = \sqrt{\frac{k}{m} } \\\\\omega^2 = \frac{k}{m} \\\\k = m \omega^2\\\\k = (0.2)(15.33)^2\\\\k = 47 \ N/m

(b) the amplitude of the motion

E = ¹/₂kA²

2E = kA²

A² = 2E/k

A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2\times 2}{47} }\\\\A = 0.292 \ m

7 0
3 years ago
Two ice skaters, Lilly and John, face each other while at rest, and then push against each other's hands. The mass of John is tw
seropon [69]

Answer:

Lilly's speed is two times John's speed.

Explanation:

m = Mass

a = Acceleration

t = Time taken

u = Initial velocity

v = Final velocity

The force they apply on each other will be equal

F=ma\\\Rightarrow a_l=\frac{F}{m_l}

F=ma\\\Rightarrow a_j=\frac{F}{2m_l}\\\Rightarrow a_j=\frac{1}{2}a_l

v=u+at\\\Rightarrow v_l=0+\frac{F}{m_l}\times t\\\Rightarrow v_l=a_lt

v=u+at\\\Rightarrow v_l=0+\frac{F}{2m_l}\times t\\\Rightarrow v_j=\frac{1}{2}a_lt\\\Rightarrow v_j=\frac{1}{2}v_l\\\Rightarrow v_l=2v_j

Hence, Lilly's speed is two times John's speed.

4 0
3 years ago
Explain why an experiment should test only one variable at a time
padilas [110]
We use only one variable at a time to find the accurate result. We want to see how the result of experiment changes everytime with a single variable.
4 0
2 years ago
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