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).
Answer:
Woke done, W = 4156.92 Joules
Explanation:
The work done by the force can be calculated as :


is the angle between force and the displacement
It is assumed to find the work done for the given parameters i.e.
Force, F = 30 N
Distance travelled, s = 160 m
Angle between force and displacement, 
Work done is given by :


W = 4156.92 Joules
So, the work done by the object is 4156.92 Joules. Hence, this is the required solution.
Answer:
The crate was being lifted by a height of 1.48 meters.
Explanation:
In an attempt o move a crate;
Force applied = 2470 N
Work done by the force = 3650 J
We know that the work done is defined as the force used to move an object to a distance.
Given the Force used and the work done by that Force, we need to find out the distance the crate was lifted to.
Work done is defined as:
Work = Force*distance covered in the direction of the force
3650 = 2470*distance
distance = 3650/2470
distance = 1.48 meters
The right answer is
all of the above
good luck
Answer: 83.3 W
Explanation: I think, I’m not sure. If I’m wrong correct me ;)