A 500 kg satellite is placed in a circular orbit 6394 km above the surface of the earth. At this elevation, the acceleration of
1 answer:
Answer:
KE = 7.7160 GJ
Explanation:
given data
mass = 500 kg
radius = 6394 km
acceleration of gravity = 4.09 m/s²
orbital speed = 20,000 km/h = 20000 × = 5555.56 m/s
solution
we Kinetic energy is express as
KE = 0.5 × m × v² .................1
here m is mass and v is velocity
put here value and we get
KE = 0.5 × 500 × 5555.56²
KE = 7716061728.400
KE = 7.7160 GJ
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Answer:
The minimum frequency is 702.22 Hz
Explanation:
The two speakers are adjusted as attached in the figure. From the given data we know that
=3m
=4m
By Pythagoras theorem
Now
The intensity at O when both speakers are on is given by
Here
- I is the intensity at O when both speakers are on which is given as 6
- I1 is the intensity of one speaker on which is 6
- δ is the Path difference which is given as
- λ is wavelength which is given as
Here
v is the speed of sound which is 320 m/s.
f is the frequency of the sound which is to be calculated.
where k=0,1,2
for minimum frequency , k=1
So the minimum frequency is 702.22 Hz
