Two resistors, A and B, are connected in a series circuit with a battery. The resistance of A is twice that of B. Which resistor
1 answer:
Answer:
A dissipates more power.
Explanation:
- In a series circuit, the current is the same at any point in it.
- The power dissipated in any resistor follows Joule's law, as follows:
- So, for a given current, the power is directly proportional to the resistance of the resistor.
- In this case, as resistor A has twice the resistance of resistor B, A dissipates twice more power than B.
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Answer: 539.4 N
Explanation:
Let's begin by explaining that Coulomb's Law establishes the following:
"The electrostatic force between two point charges
and
is proportional to the product of the charges and inversely proportional to the square of the distance
that separates them, and has the direction of the line that joins them"
What is written above is expressed mathematically as follows:
(1)
Where:
is the electrostatic force
is the Coulomb's constant
and
are the electric charges
is the separation distance between the charges
Then:
(2)
Isolating and
:
(3)
Now, if we keep the same charges but we decrease the distance to , (1) is rewritten as:
(4)
Then, the new electrostatic force will be:
(5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.
The momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.
<h3>What is the law of conservation of momentum?</h3>According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.
The given data in the problem is;
(m₁) is the mass of hockey player 1= 91.0-kg
(m₂) is the mass of hockey player 2= 91.0-kg
(u₁) is the velocity before collision of hockey player 1 = 5.50 m/s.
(u₂) is the velocity before the collision of hockey player 2=?
a)
Momentum before the collision;
Momentum before the collision = 1237.6 kg m/s².
b)
The velocity of the two hockey players after the collision from the law of conservation of the momentum as:
Momentum before collision = Momentum after the collision
1237.6 kg m/s² = (m₁+m₂)V
1237.6 kg m/s² =(2 ×91.0-kg )V
V=6.8 m/sec.
Hence, momentum before the collision velocity after the collision will be 1237.6 kg m/s² and 6.8 m/sec.
To learn more about the law of conservation of momentum refer;
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