The submarine sends out a sound wave that returns in 1.08 seconds. If this sound wave has a frequency of 2.50 × 106 Hz and a wav
elength of 6.00 × 10-4 meters, how far away is the ship?
1 answer:
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Answer:
113.7
Explanation:
maximum distance (s) = 8.9 km
reference intensity (I0) = 1 x 10^{-12} W/m^{2}
power of a juvenile howler monkey (p) = 63 x 10^{-6} W
distance (r) = 210 m
intensity (I) = power/area
where we assume the area of a sphere due to the uniformity of the output in all directions
area = 4π = 4π x
= 554,176.9 m^{2}
intensity (I) =
therefore the desired ratio I/I0 = = 113.7
Answer:
The value of tension on the cable T = 1065.6 N
Explanation:
Mass = 888 kg
Initial velocity ( u )= 0.8
Final velocity ( V ) = 0
Distance traveled before come to rest = 0.2667 m
Now use third law of motion =
- 2 a s
Put all the values in above formula we get,
⇒ 0 = - 2 × a ×0.2667
⇒ a = 1.2
This is the deceleration of the box.
Tension in the cable is given by T = F = m × a
Put all the values in above formula we get,
T = 888 × 1.2
T = 1065.6 N
This is the value of tension on the cable.