Answer:
v ’= 21.44 m / s
Explanation:
This is a doppler effect exercise that changes the frequency of the sound due to the relative movement of the source and the observer, the expression that describes the phenomenon for body approaching s
f ’= f (v + v₀) / (v-
)
where it goes is the speed of sound 343 m / s, v_{s} the speed of the source v or the speed of the observer
in this exercise both the source and the observer are moving, we will assume that both have the same speed,
v₀ = v_{s} = v ’
we substitute
f ’= f (v + v’) / (v - v ’)
f ’/ f (v-v’) = v + v ’
v (f ’/ f -1) = v’ (1 + f ’/ f)
v ’= (f’ / f-1) / (1 + f ’/ f) v
v ’= (f’-f) / (f + f’) v
let's calculate
v ’= (3400 -3000) / (3000 +3400) 343
v ’= 400/6400 343
v ’= 21.44 m / s
Answer:
T²= 4π²R³/GM
Explanation:
First we know that
Fg= Fc
Because centripetal force must equal gravitational force
So
GMm/R² = Mv²/R
But velocity is 2πR/T
So by substitution we have
GMm/R²= M (2πR/T)/T
We have
T²= 4π²R³/GM as period
Answer:
Potential difference and charge will also increase.
Explanation:
Asking that :
What will happen to the charge and potential difference if the plate area were increased while the plate separation remains unchanged?
The charge is directly proportional to area of the plate. That is, increase in area of the plate of a capacitor will lead to the increase in the charges between the plates.
And since charge is also proportional to the magnitude of potential difference between the plates from the definition of capacitance of a capacitor which says that:
Q = CV
Therefore, increase in the area of the plate will also lead to increase in potential difference between the plates.
Therefore, if the plate area were increased while the plate separation remains unchanged, the charge and potential difference between them will also increase.
Remember that sound intensity decreases in inverse proportion to the distance squared. So, to solve this we are going to use the inverse square formula:

where

is the intensity at distance 2

is the intensity at distance 1

is distance 2

is distance 1
We can infer for our problem that

,

, and

. Lets replace those values in our formula to find

:





dB
We can conclude that the intensity of the sound when is <span>3 m from the source is
30 dB.</span>