Answer:
70.15 Joule
Explanation:
mass of man, m = 70 kg
intial length, l = 11 m
extension, Δl = 1.5 m
Let K is the spring constant.
In the equilibrium position
mg = K l
70 x 9.8 = K x 11
K = 62.36 N/m
Potential energy stored, U = 0.5 x K x Δl²
U = 0.5 x 62.36 x 1.5 x 1.5
U = 70.15 Joule
Answer:
The beat frequency is 0.0019 MHz.
Explanation:
Given that,
Velocity = 0.32 m/s
Frequency = 4.40 MHz
Speed of wave = 1540 m/s
We need to calculate the frequency
Case (I),
Observer is moving away from the source
Using Doppler's effect

Where, v' = speed of observer
Put the value into the formula


Case (II),
Cell is as the source of sound of frequency f' and it moving away from the observer.
Using formula of frequency



We need to calculate the beat frequency


Hence, The beat frequency is 0.0019 MHz.
Explanation:
The ball, for example, will feel gravity pulling it downward and the ground pushing it upward in the direction it is rolling. (Add this if the ball is rolling on the floor.) Friction is the force that causes the ball to slow down because it acts in the opposite direction that it is moving.
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The question is incomplete. The complete question is :
A viscoelastic polymer that can be assumed to obey the Boltzmann superposition principle is subjected to the following deformation cycle. At a time, t = 0, a tensile stress of 20 MPa is applied instantaneously and maintained for 100 s. The stress is then removed at a rate of 0.2 MPa s−1 until the polymer is unloaded. If the creep compliance of the material is given by:
J(t) = Jo (1 - exp (-t/to))
Where,
Jo= 3m^2/ GPA
to= 200s
Determine
a) the strain after 100's (before stress is reversed)
b) the residual strain when stress falls to zero.
Answer:
a)-60GPA
b) 0
Explanation:
Given t= 0,
σ = 20Mpa
Change in σ= 0.2Mpas^-1
For creep compliance material,
J(t) = Jo (1 - exp (-t/to))
J(t) = 3 (1 - exp (-0/100))= 3m^2/Gpa
a) t= 100s
E(t)= ΔσJ (t - Jo)
= 0.2 × 3 ( 100 - 200 )
= 0.6 (-100)
= - 60 GPA
Residual strain, σ= 0
E(t)= Jσ (Jo) ∫t (t - Jo) dt
3 × 0 × 200 ∫t (t - Jo) dt
E(t) = 0