Answer:
The current is reduced to half of its original value.
Explanation:
- Assuming we can apply Ohm's Law to the circuit, as the internal resistance and the load resistor are in series, we can find the current I₁ as follows:

- where Rint = r and RL = r
- Replacing these values in I₁, we have:

- When the battery ages, if the internal resistance triples, the new current can be found using Ohm's Law again:

- We can find the relationship between I₂, and I₁, dividing both sides, as follows:

- The current when the internal resistance triples, is half of the original value, when the internal resistance was r, equal to the resistance of the load.
Answer:
B
Explanation:
because kinetic energy is directly proportional to temperature so the hottor the object, the more kinetic energy.
Answer:
Mass is constant everywhere,
But weight is different,
If earth g = 10 then moon's is 1.6666667
Now billie's weight in moon is 41.6667
As we know,

so, let's solve for charge (q) :
time = 5 minutes = 5 × 60 seconds = 300 seconds.
hence, the charge = 60 coulombs (C)
R is proportional to the length of the wire:
R ∝ length
R is also proportional to the inverse square of the diameter:
R ∝ 1/diameter²
The resistance of a wire 2700ft long with a diameter of 0.26in is 9850Ω. Now let's change the shape of the wire, adding and subtracting material as we go along, such that the wire is now 2800ft and has a diameter of 0.1in.
Calculate the scale factor due to the changed length:
k₁ = 2800/2700 = 1.037
Scale factor due to changed diameter:
k₂ = 1/(0.1/0.26)² = 6.76
Multiply the original resistance by these factors to get the new resistance:
R = R₀k₁k₂
R₀ = 9850Ω, k₁ = 1.037, k₂ = 6.76
R = 9850(1.037)(6.76)
R = 69049.682Ω
Round to the nearest hundredth:
R = 69049.68Ω