Answer:
d = 61.75 m
Explanation:
Given that,
A ball droped from a building.
We need to find how fast is it traveling after falling 3.55 s.
As it is dropped, its initial velocity is equal to 0.
Let d is the distance it covers after falling 3.55 s.
We can use second equation of motion to find d.
Here, u = 0 and a =g
So, it will cover 61.75 m after falling 3.55 seconds.
Answer:
2.5kN.m
Explanation:
Torque is directly proportional to pitch diameter
= Ta/Tb= Da/Db
=120/Tb= 0.25/0.5
Tb= 2.469kN.m approx 2.5kN.m
Answer:
L = 44,096 m
Explanation:
The speed of the sound wave is constant therefore we can use the relations of uniform kinematics
v = x / t
the speed of the wave in the bar is
v = 15 v or
v = 15 343
v = 5145 m / s
The sound at the bar goes the distance
L = v t
Sound in the air travels the same distance
L = v_air (t + 0.12)
as the two recognize the same dissonance,
v t = v_air (t +0.12)
t (v- v_air) = 0.12 v_air
t = 0.12 v_air / (v -v_air)
l
et's calculate
t = 0.12 343 / (5145 - 343)
t = 8.57 10-3 s
The length of the bar is
L = 5145 8.57 10-3
L = 44,096 m
