An oscillator makes 360 vibrations in 3 minutes.
2 answers:
Explanation :
It is given that an oscillator makes 360 oscillations in 3 minutes.
(a) Using unitary method :
No. of vibrations in one minute is,
So, no of vibrations in one minute is 120.
(b) Similarly,
3 minutes = 180 seconds
No of vibrations in one second is
So, the no of vibrations in one second is 2.
(c) Time period of the wave is given by :
The time period of the wave is 0.5 s
(d) The no of vibation per second is called as its frequency.
Hence, this is the required solution.
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Answer:
a) 2.41 km
b) 38.8°
Questions c and d are illegible.
Explanation:
We can express the displacements as vectors with origin on the point he started (0, 0).
When he traveled south he moved to (-3, 0).
When he moved east he moved to (-3, x)
The magnitude of the total displacement is found with Pythagoras theorem:
d^2 = dx^2 + dy^2
Rearranging:
dy^2 = d^2 - dx^2
The angle of the displacement vector is:
cos(a) = dx/d
a = arccos(dx/d)
a = arccos(3/3.85) = 38.8°
Answer:
The depth is 5.15 m.
Explanation:
Lets take the depth of the pool = h m
The atmospheric pressure ,P = 101235 N/m²
The area of the top = A m²
The area of the bottom = a m²
Given that A= 1.5 a
The force on the top of the pool = P A
The total pressure on the bottom = P + ρ g h
ρ =Density of the water = 1000 kg/m³
The total pressure at the bottom of the pool = (P + ρ g h) a
The bottom and the top force is same
(P + ρ g h) a = P A
P a +ρ g h a = P A
ρ g h a = P A - P a
h=5.15 m
The depth is 5.15 m.