How many times louder is the intensity of sound at a rock concert in comparison with that of a whisper, if the two intensity lev
1 answer:
Answer:
5 × 10¹⁰ or 5 billion times louder is the intensity of sound at a rock concert in comparison with that of a whisper
Explanation:
Given that at a rock concert;
the intensity of sound is 110 dB compared to a whisper of 3 dB
Now; how many times louder will that of the whisper be compared to the sound of the rock concert
Using the formula for calculating decibels (dB);
For 110 dB
dB = 10log₁₀(W)
110 dB =
110 dB = 10¹¹
For 3dB
dB = 10log₁₀(W)
3 dB =
3 dB = 2
Now:
(110/3)dB = 10¹¹/2
(110/3)dB = 5 × 10¹⁰ or 5 billion times louder is the intensity of sound at a rock concert in comparison with that of a whisper
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For constant acceleration:
a = v,final - v,initial /t
The solutions is as follows:
a = v,final - v,initial /t
3.8 = (v - 0)/2.8 s
v = 10.64 m/s
After 2.8 seconds, the speed of the blue car is 10.64 m/s.
For constant acceleration:
a = v,final - v,initial /t
The solutions is as follows:
a = v,final - v,initial /t
3.8 = (v - 0)/2.8 s
v = 10.64 m/s
After 2.8 seconds, the speed of the blue car is 10.64 m/s.
<u>Given data:</u>
m= 50 Kg,
W= m×g = 50 × 9.81 = 490.5 N
ramp angle (α) = 30 degrees,
coefficient of friction (μs) = 0.5773,
Push at an angle (Θ) = 40 degrees,
Determine: Push to get box move up (P)=?
From the figure,
Resolving the forces along the plane
W sinα + μs.R = P cos Θ --------------------- (i)
Resolving the forces perpendicular to the inclined plane
W cosα = R+Psin Θ => R= W cosα - Psin Θ -------------- (ii)
Solving (i) and (ii) and keeping <em>μs = tan Φ, Φ = Θ </em>
<em>Pmin = W sin( α +Θ )</em>
<em> = W[ sin α.Cos Θ + cos α.sin Θ]</em>
<em> = 490.5 [ (sin 30.cos40) + (cos30.sin 40)]</em>
<em> = 460.9 N</em>
<em>Minimum push required to move the box up the ramp is 460.9 N</em>
