Well, I'll try to write the formula in a way that's not confusing,
but I'm afraid it might be slightly confusing anyway.
When you're working with dB, the basic rule is
A change of 10 dB means either multiplying or dividing by 10 .
Multiply something by 10 ==> it increases by 10 dB.
Divide something by 10 ==> it decreases by 10 dB.
It turns out that another way to write all of this is . . .
An increase of 10 dB ===> multiply the original amount by 10¹
An increase of 20 dB ===> multiply the original amount by 10²
An increase of, say, 7 dB ===> multiply the original amount by 10⁰·⁷
A decrease of 10 dB ===> multiply the original amount by 10⁻¹
A decrease of 30 dB ===> multiply the original amount by 10⁻³
A decrease of, say, 13 dB ===> multiply the original amount by 10⁻¹·³
This question says: The sound increases by 5 dB .
That means the original 'intensity' or 'power' of the sound
is multiplied by
10⁰·⁵ = √10 = about 3.162 (rounded) .
From the choices listed, the closest one is (c).
M = 10kg
P = 40 kg.m/s
V= 4m/s
Formula v = p/m
V= p/m
=4okg.m/s / 10kg
=4m/s
1 kg=100000 cg
2 kg=200000 cq
If mass is the quantity then kg is the S.I
2 kg=2kg
Answer:
Answer:B
Explanation:
Because it all stayed consistant
The work function is what we call the minimum energy that is required by an electron to leave the metal target in the photoelectric effect.
