A ball is ejected to the right with an unknown horizontal velocity from the top of a pillar that is 50 meters in height. At the
2 answers:
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Answer:
lowest frequency = 535.93 Hz
distance between adjacent anti nodes is 4.25 cm
Explanation:
given data
length L = 32 cm = 0.32 m
to find out
frequency and distance between adjacent anti nodes
solution
we consider here speed of sound through air at room temperature 20 degree is approximately v = 343 m/s
so
lowest frequency will be = ..............1
put here value in equation 1
lowest frequency will be =
lowest frequency = 535.93 Hz
and
we have given highest frequency f = 4000Hz
so
wavelength = ..............2
put here value
wavelength =
wavelength = 0.08575 m
so distance = ..............3
distance =
distance = 0.0425 m
so distance between adjacent anti nodes is 4.25 cm
Answer:
Torque,
Explanation:
It is given that,
Length of the wrench, l = 0.5 m
Force acting on the wrench, F = 80 N
The force is acting upward at an angle of 60.0° with respect to a line from the bolt through the end of the wrench. We need to find the torque is applied to the nut. We know that torque acting on an object is equal to the cross product of force and distance. It is given by :
So, the torque is applied to the nut is 34.6 N.m. Hence, this is the required solution.