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svetoff [14.1K]
1 year ago
10

A positive charge and a negative charge held a certain distance apart are released. as they move, the force on each particle:___

__.
Physics
1 answer:
iVinArrow [24]1 year ago
7 0

A positive charge and a negative charge held a certain distance apart are released. as they move, the force on each particle increases

The most common charge carriers are the positively charged proton and the negatively charged electron. The movement of any of these charged particles constitutes an electric current

<h3>What is a Charge ?</h3>

When there are more or fewer protons in an atom than electrons, the substance has an electric charge. Protons have a positive charge, while electrons have a negative charge. If a substance has more protons than electrons, it is positively charged; if it has more electrons, it is negatively charged.

  • The SI units for charge are ampere-second or coulomb. When one ampere of electric current goes through the conductor for one second, one coulomb of charge passes through it. Charge is denoted by the formula Q = I t.

Learn more about Charge here:

brainly.com/question/18102056

#SPJ4

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The blades in a blender rotate at a rate of 6800 rpm . When the motor is turned off during operation, the blades slow to rest in
tangare [24]
The angular acceleration of the blade when it's switched off is (-6800 rev/min) divided by (2.8 sec) = -2,428.6 rev/(min-sec) = -40.5 rev/sec^2 .
5 0
3 years ago
Two balls with masses of 2.0 kg and 6.0 kg travel toward each other at speeds of 12 m/s and 4.0 m/s, respectively. If the balls
Alina [70]

Answer:

The kinetic energy lost in the collision is 48 J

Explanation:

Given;

mass of the first ball, m₁ = 2.0 kg

mass of the second ball, m₂ = 6.0 kg

initial speed of the first ball, u₁ = 12 m/s

initial speed of the second ball, u₂ = 4 m/s

let v be the final velocity of the two balls after the inelastic collision

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

2 x 12 + 6 x 4 = v(2 + 6)

48 =  8v

48 / 8 = v

v = 6 m/s

The initial kinetic energy of the balls is calculated as;

K.E₁ = ¹/₂m₁u₁² + ¹/₂m₂u₂²

K.E₁ = ¹/₂(2)(12²) + ¹/₂(6)(4)²

K.E₁ = 144 + 48

K.E₁ = 192 J

The final kinetic of the balls is calculated as;

K.E₂ = ¹/₂(m₁ + m₂)(v²)

K.E₂ = ¹/₂(2 + 6)(6²)

K.E₂ = ¹/₂(8)(6²)

K.E₂ = 144 J

The lost in kinetic energy of the balls is K.E₂ - K.E₁ = 144 J - 192 J = -48 J

Therefore, the kinetic energy lost in the collision is 48 J

5 0
3 years ago
Answers are - <br><br>A 235 N<br><br>B 376 N<br><br>C 271 N<br><br>D 188 N<br><br>E 470 N
amm1812

Answer:

C

Explanation:

8 0
3 years ago
I need help please it’s for a lab
krok68 [10]

Answer:

idk

Explanation:

5 0
3 years ago
Why should it take significantly more energy to move a beam of alpha particles than a beam of beta minus (β–) particles?
a_sh-v [17]
An 'alpha particle' is the same thing as the nucleus of a helium atom ... 
a little bundle made of 2 protons and 2 neutrons.

A 'beta' particle is an electron.

The mass of an alpha particle is more than 7,000 times the mass of 
an electron, so it certainly takes more energy to get it moving.
4 0
3 years ago
Read 2 more answers
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