I think that the answer is friction
Answer:
D. Friction
Explanation:
Friction is a force that opposes motion. So a perpetual motion machine can never be built because it is impossible to eliminate frictional force. It can only be reduced
Answer:
yes
Explanation:
this is simple
the horizontal line is adjacent
the vertical line is opposite
recall that cos x=adj/hyp
adj=hyp(cos x)
while opp=hyp(sin x)
<u>Answer :</u>
(a) d = 0.25 m
(b) d = 0.5 m
<u>Explanation :</u>
It is given that,
Frequency of sound waves, f = 686 Hz
Speed of sound wave at
is, v = 343 m/s
(1) Perfectly destructive interference occurs when the path difference is half integral multiple of wavelength i.e.
........(1)
Velocity of sound wave is given by :




Hence, when the speakers are in phase the smallest distance between the speakers for which the interference of the sound waves is perfectly destructive is 0.25 m.
(2) For constructive interference, the path difference is integral multiple of wavelengths i.e.
( n = integers )
Let n = 1
So, 


Hence, the smallest distance between the speakers for which the interference of the sound waves is maximum constructive is 0.5 m.
Answer:

Explanation:
The angular momentum of the pulsar is given by:

where
is the mass of the pulsar
is the radius
is the angular speed
Given the period of the pulsar,
, the angular speed is given by

And so, the angular momentum is
