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Answer:
Work done, W = 337.35 Joules
Explanation:
It is given that,
Mass of the crate, m = 21 kg
Force exerted by the worker on the crate, F = 224.9 N
The crate is inclined at an angle of 40 degrees.
Distance covered by the slide, d = 1.5 m
Let W is the work is done on the crate by the worker's applied force. it is equal to the product of force and distance. It is given by :
W = 337.35 Joules
So, the work is done on the crate by the worker's applied force is 337.35 Joules. Hence, this is the required solution.
Answer: Length axis f= 114.3 Hz, Width axis f=228.67 Hz
Explanation:
We are given that,
Length of tub= 1.5 m
Width of tub= 0.75 m
Sound speed= 343 m/s
Now, we are also given shower is closed.
So, frequency is given as:
f=
For length axis
Put v= 343 m/s, m=1 and L=1.5 m
f= 1 *
f= 114.3 Hz
For next resonant frequency, m=2
f= 2*
f= 228.67
For width axis
Put v= 343 m/s, m=1 and L= 0.75 m
f= 1*
f= 228.67 Hz
For next frequency, m=2
f= 2*
f= 457.34 Hz
Answer:
= 1.6
Explanation:
= Angle of incidence = 65.9°
= Angle of refraction = 34.8°
= Index of refraction of air = 1
= Index of refraction of glass = ?
Using Snell's law
Sin
=
(1) Sin65.9 = Sin34.8
= 1.6