Answer:
option D
Explanation:
given,
Intensity of sound = 20 dB
distance = 15 m
intensity of sound is increased to = 50 dB
distance between the sound level = ?
Using relation

L₁ = 20 dB L₂ = 50 dB r₁ = 15 m r₂ = ?





r₂ = 0.47 m
r₂ = 47 cm
hence, the correct answer is option D
Answer:
You can change the momentum of an object by giving the object more force or less force.
Explanation:
Think about a ball. It is going slow, you push it and you give it more momentum.
<h3><u>Answer;</u></h3>
= 8.55 Joules
<h3><u>Explanation;</u></h3>
Work done is the product of force and the distance moved by an object.
Work done = Force × distance
Force = 95 Newtons
Distance = X2 -X1
= 4 - (-5)
= 9 cm
Thus;
work done = 95 × 9/100
<u>= 8.55 Joules </u>
Answer:
T=0.827s
Explanation:
The period of a spring can be calculated with the equation

But we know as well that w is given by,

Replacing,

So we have that
