First one is D and Second one is B
<span>Question: How much power does an electric device use if the current is 36.0 amps and the resistance is 3.9 ohms? </span>
How?:
Equation: P = I^2 R
Meanings:
P = Power in Watts
I = Current in Ampere
R = Resistance in ohms.
Plugged in: P = 36^2<span> x 3.9 = 5054.4
Answer: P= </span>5100 watts.
HOPE THIS HELPS! ^_^
<span> </span>
Answer:
False
Explanation:
<em>If one of the bulbs is removed from the series, the other bulb will not come on at all.</em>
This is because the removal of one of the bulbs would interrupt the flow of current though the entire circuit.
Hence, that the other one will get brighter if one of two bulbs in a circuit is removed from its socket is not true.
Answer:
This depends on what angle they are approaching each other before they collided.The two simple cases are if they are running in the same direction or opposite direction from each other. For either case, use the conservation of momentum equation to solve: M_total*V_result = M1*V1 + M2*V2
Explanation:
Here are two possible solutions.
Head-on collision: M1=78, V1=8.5, M2=72, V2=-7.5 (that's negative because he's running the other way), M_total = 78+72 = 150, so V_result = (78*8.5 - 72*7.5)/150 = 0.82 m/s. Sanity check, they weigh about the same and so most of their velocity should cancel out.
Running the same way: change the sign of V2 to positive so V_result = (78*8.5 + 72*7.5)/150 = 8.02 m/s. Sanity check, they weigh about the same and the resultant speed is between the two starting velocities.
<em>hope it helps:)</em>
Answer:
11.1 cm
Explanation:
Let d be the depth upto which box dipped into water.
Volume of water displaced
.12\times .12\times d m^3\\
=.0144d
Weight of water displaced = upthrust
=.0144d\times 10^3\times g\\ [ density of water is 1000kg /m³]
=14.4dg This will act upwards.
Total downward force = 1.2 +.4 = 1.6 g
For equilibrium
1.6g = 14.4 gd
d = .111 m
=11.1 cm.
