initial height of the pole vaulter = 4.2 m
height of the pole vaulter just before it touch the pad = 80 cm
so the total displacement of pole vaulter just before he will touch the pad = 4.2 - 0.80 = 3.4 m
now by kinematics
now after this he will come to rest after compressing the pad by 50 cm
so again we can use kinematics to find its acceleration
so here its acceleration will be - 66.64 m/s^2
Answer:
Therefore the amplitude of the resultant wave is
Explanation:
The equation of wave:
y=A sin (kx-ωt)
For wave 1:
y₁=A sin (kx-ωt) = sin (kx-ωt)
For wave 2:
y₂=A sin (kx-ωt+Φ) = sin (kx-ωt+Φ)
Where A= amplitude=
The angular frequency
,
= wave length.
t= time
T= Time period
= phase difference =
The resultant wave will be
y = y₁ + y₂
= sin (kx-ωt) +
sin (kx-ωt+Φ)
{sin (kx-ωt) + sin (kx-ωt+Φ)}
Therefore the amplitude of the resultant wave is
Answer:
1.8m
Explanation:
Let the Elastics of the steel ASTM-36
The strain of the bar when subjected to 150 MPa is
Therefore, if the bar elongates by 1.35 mm, then the original length L would be:
or 1.8m