Quadrupling the power output from a speaker emitting a single frequency will result in what increase in loudness (in units of dB
1 answer:
Answer:
<em>6.02 dB increase </em>
<em></em>
Explanation:
Let us take the initial power from the speaker P' = P Watt
then, the final power P = 4P Watt
for a given unit area, initial intensity (power per unit area) will be
I' = P Watt/m^2
and the final quadrupled sound will produce a sound intensity of
I = 4P Watt/m^2
Increase in loudness is gotten from the relation
ΔL =
where
I = final sound intensity
I' = initial sound intensity
imputing values of the intensity into the equation, we have
==> =
= <em>6.02 dB increase </em>
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Answer:
10.09 N
Explanation:
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Torque is the vector product of the position vector of the point at which the force is applied by the force vector:
Since the effective lever arm is perpendicular to the force, the angle between them is . The magnitud of this vector product is defined as:
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Missing question:
"Determine (a) the astronaut’s orbital speed v and (b) the period of the orbit"
Solution
part a) The center of the orbit of the third astronaut is located at the center of the moon. This means that the radius of the orbit is the sum of the Moon's radius r0 and the altitude (
) of the orbit:
This is a circular motion, where the centripetal acceleration is equal to the gravitational acceleration g at this altitude. The problem says that at this altitude,
. So we can write
where
is the centripetal acceleration and v is the speed of the astronaut. Re-arranging it we can find v:
part b) The orbit has a circumference of
, and the astronaut is covering it at a speed equal to v. Therefore, the period of the orbit is
So, the period of the orbit is 2.45 hours.
"Determine (a) the astronaut’s orbital speed v and (b) the period of the orbit"
Solution
part a) The center of the orbit of the third astronaut is located at the center of the moon. This means that the radius of the orbit is the sum of the Moon's radius r0 and the altitude (
This is a circular motion, where the centripetal acceleration is equal to the gravitational acceleration g at this altitude. The problem says that at this altitude,
where
part b) The orbit has a circumference of
So, the period of the orbit is 2.45 hours.