The voltage from one side of the battery all the way around to the other side of the battery is 12v .
If 4 of those volts show up across the circle-thing, then the rest of the 12v ... 8v ... Must show up across the set of parallel rectangles.
To get that answer, I subtracted the 4 from the 12.
Just like it says in choice-C.
By Newton's 2nd law of motion, F = ma, where F is force, m is mass, and a is acceleration.
Rearranging this equation to find acceleration would give us:
a = F/m
The horizontal force to the right is 10N, because the box is pushed to the right with a force of 20N, and the friction force of 10N opposes that, so:
20N - 10N = 10N
The mass is 2kg.
Putting these values into the equation gives us:
a = F/m
= 10/2
= 5ms^-2
The acceleration of the box is 5ms^-2
Answer:
The second one is correct.
Explanation:
=
Answer:
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Explanation:
Answer:
a) When R is very small R << r, therefore the term R+ r will equal r and the current becomes
b) When R is very large, R >> r, therefore the term R+ r will equal R and the current becomes
Explanation:
<u>Solution :</u>
(a) We want to get the consumed power P when R is very small. The resistor in the circuit consumed the power from this battery. In this case, the current I is leaving the source at the higher-potential terminal and the energy is being delivered to the external circuit where the rate (power) of this transfer is given by equation in the next form
P=∈*I-I^2*r (1)
Where the term ∈*I is the rate at which work is done by the battery and the term I^2*r is the rate at which electrical energy is dissipated in the internal resistance of the battery. The current in the circuit depends on the internal resistance r and we can apply equation to get the current by
I=∈/R+r (2)
When R is very small R << r, therefore the term R+ r will equal r and the current becomes
I= ∈/r
Now let us plug this expression of I into equation (1) to get the consumed power
P=∈*I-I^2*r
=I(∈-I*r)
=0
The consumed power when R is very small is zero
(b) When R is very large, R >> r, therefore the term R+ r will equal R and the current becomes
I=∈/R
The dissipated power due toll could be calculated by using equation.
P=I^2*r (3)
Now let us plug the expression of I into equation (3) to get P
P=I^2*R=(∈/R)^2*R
=∈^2/R
