Answer:
distance is 13 m for 100 dB
distance is 409 km for 10 dB
Explanation:
Given data
distance r = 2.30 m
source β = 115 dB
to find out
distance at sound level 100 dB and 10 dB
solution
first we calculate here power and intensity and with this power and intensity we will find distance
we know sound level β = 10 log(I/
) ......................a
put here value (I/) = 10^−12 W/m² and β = 115
115 = 10 log(I/10^−12)
so
I = 0.316228 W/m²
and we know power = intensity × 4π r² ...............b
power = 0.316228 × 4π (2.30)²
power = 21.021604 W
we know at 100 dB intensity is 0.01 W/m²
so by equation b
power = intensity × 4π r²
21.021604 = 0.01 × 4π r²
so by solving r
r = 12.933855 m = 13 m
distance is 13 m
and
at 10 dB intensity is 1 × 10^–11 W/m²
so by equation b
power = intensity × 4π r²
21.021604 = 1 × 10^–11 × 4π r²
by solving r we get
r = 409004.412465 m = 409 km
Answer:
Hello your question is poorly written below is the complete question
Suppose the battery in a clock wears out after moving Ten thousand coulombs of charge through the clock at a rate of 0.5 Ma how long did the clock run on does battery and how many electrons per second slowed?
answer :
a) 231.48 days
b) n = 3.125 * 10^15
Explanation:
Battery moved 10,000 coulombs
current rate = 0.5 mA
<u>A) Determine how long the clock run on the battery. use the relation below</u>
q = i * t ----- ( 1 )
q = charge , i = current , t = time
10000 = 0.5 * 10^-3 * t
hence t = 2 * 10^7 secs
hence the time = 231.48 days
<u>B) Determine how many electrons per second flowed </u>
q = n*e ------ ( 2 )
n = number of electrons
e = 1.6 * 10^-19
q = 0.5 * 10^-3 coulomb ( charge flowing per electron )
back to equation 2
n ( number of electrons ) = q / e = ( 0.5 * 10^-3 ) / ( 1.6 * 10^-19 )
hence : n = 3.125 * 10^15
Answer:
horizontal component of normal force is equal to the centripetal force on the car
Explanation:
As the car is moving with uniform speed in circle then the force required to move in the circle is towards the center of the circle
This force is due to friction force when car is moving in circle with uniform speed
Now it is given that car is moving on the ice surface such that the friction force is zero now
so here we can say that centripetal force is due to component of the normal force which is due to banked road
Now we have
so we have
so this is horizontal component of normal force is equal to the centripetal force on the car
This may helpv^2=u^2+2as. v=0 at top of flight. a=acceleration of gravity(vo^2)/2a=s.
