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3 years ago

What kind of charge does an object have when it has given away electrons?

2 answers:

6 0

Electrons have a negative charge, so if you give away electrons, you give away negative charge, thus ending with a positive charge.

FrozenT [24]3 years ago

6 0

An object that loses electrons becomes Positively charged

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Hello, why do all electromagnetic waves travel at the same speed in vacuum?

Viktor [21]

Answer:

due to the magnetic field

Explanation:

magnetic field is the same in the vacuum

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3 years ago

A _______ property of gasoline is that it will _______ in air.

Misha Larkins [42]

A chemical property of gasoline is that it will burn in air. Gasoline is a substance that is used to power automobiles. Gasoline will oxidize in air which means that it reacts with oxygen in air. Hope this answers the question. Have a nice day.

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3 years ago

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Heather puts a straw into glass of water. She notices that when she looks through the glass and the water from the side, the str

Alex73 [517]

B. Refraction because objects refract in water

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3 years ago

bumper car A (281 kg) moving +2.82 m/s makes an elastic collision with bumper car B (209 kg) moving -1.72 m/s. what is the veloc

USPshnik [31]

Answer:

The final velocity of the car A is -1.053 m/s.

Explanation:

For an elastic collision both the kinetic energy and the momentum of the system are conserved.

Let us call

$m_A$ = mass of car A;

$v_{A1}$ = the initial velocity of car A;

$v_{A2}$ = the final velocity of car A;

and

$m_B$ = mass of car B;

$v_{B1}$ = the initial velocity of car B;

$v_{B2}$ = the final velocity of car B.

Then, the law of conservation of momentum demands that

$m_Av_{A1}+m_Bv_{B1} =m_Av_{A2}+m_Bv_{B2}$

And the conservation of kinetic energy says that

$\dfrac{1}{2} m_Av_{A1}^2+\dfrac{1}{2}m_Bv_{B1}^2=\dfrac{1}{2}m_Av_{A2}^2+\dfrac{1}{2}m_Bv_{B2}^2$

These two equations are solved for final velocities $v_{A2}$ and $v_{B2}$ to give

$v_{A2} =\frac{m_A-m_B}{m_A+m_B} v_{A1}+\frac{2m_B}{m_A+m_B} v_{B1}$

$v_{B2} =\frac{2m_A}{m_A+m_B} v_{A1}+\frac{m_B-m_A}{m_A+m_B} v_{B1}$

by putting in the numerical values of the variables we get

$v_{A2} =\frac{281-209}{281+209} (2.82)+\frac{2*209}{281+209} (-1.72)$

$\boxed{v_{A2} = -1.05m/s}$

and

$v_{B2} =\frac{2*281}{281+209} (2.82)+\frac{209-281}{281+209} (-1.72)$

$\boxed{v_{B2} = 3.49m/s}$

Thus, the final velocity of the car A is -1.053 m/s and of car B is 3.49 m/s.

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3 years ago

What real-world examples show no work begin done? Can you think of examples other than resisting the force of gravity?

irga5000 [103]

Oh my gosh ! Resisting the force of gravity always DOES involve doing work.
If no work is being done, then you're NOT resisting the force of gravity.

Example:

-- ball rolling on the floor . . . no work
-- ball rolling up a ramp . . . work being done
-- ball rolling down a ramp . . . work being done, BY gravity

6 0

3 years ago

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