Answer:
A)5524J,
B) 29.2Nm
Explanation:
This question can be treated using work- energy theorem
Work= change in Kinectic energy
W= Δ KE
Work= difference between the final Kinectic energy and intial Kinectic energy.
We know that
Kinectic energy= 1/2 mv^2 .............eqn(1)
This can be written in term of angular velocity, as
KE= 1/2 I
Answer:
45 s .
Explanation:
The accelerator will first accelerate , then move with uniform velocity and at last it will decelerate to rest .
displacement s = ?
acceleration a = 1 m /s²
Final speed v = 5 m/s
initial speed u = 0
v² = u² + 2as
5² = 0 + 2 x 1 x s
s = 12.5 m
B) Let time of acceleration or deceleration be t
v = u + a t
5 = 0 + 1 t
t = 5 s
Similarly displacement during deceleration = 12.5 m
Total distance during uniform motion = 200 - ( 12.5 + 12.5 ) = 175 m .
velocity of uniform motion = 5 m /s
time during which there was uniform velocity = 175 / 5 = 35 s
Total time = 5 + 35 + 5 = 45 s .
Answer: The answer is D
Explanation: i had the same question and i just guessed and got it first try
Answer:
∈=
Explanation:
Using the Gauss Law to determine the electric field of the net flux at the surface of the nucleus
∈
The P is the charge density and 'Eo' is the constant of permittivity in free space
to find P
So replacing
∈
∈=
Y, bc the height of the bounce back is higher than x