Answer: add 10 cm plus 200 mpa divided by 2 min then caculate that into 160 bars.
Explanation:
Answer:
Explanation:
a = F / m
where a is acceleration , F is thrust and m is mass
taking log and differentiating
da / a = dF / F - dm / m
(da / a)x 100 = (dF / F)x100 - (dm / m) x100
percentage increase in a = percentage increase in F - percentage increase in m
= percentage increase in acceleration a = 39 - 13 = 26 %
required increase = 26 %.
Answer:
Explanation:
Given:
Initial θ = 0 rad (from rest)
Final θ = 14.3 rad
Time, t = 5 s
B.
Angular velocity, w = dθ / dt
= (14.3 - 0)/5
= 2.86 rad/s
A.
Acceleration, ao = dw/dt
Initial angular velocity, wi = 0 rad/s (from rest)
Final angular velocity, wf = 2.86 rad/s
a = (2.86 - 0)/5
= 0.572 rad/s^2
Answer:

Explanation:
As we know that the frequency of the wave is given as

here we know that

also we know that

now we have


The frequency of the
scattered photon decreases or it will be lower compare to the frequency of
incident photon. An x-ray photon scatters in one direction after a collision
and some energy is transferred to the electron as it recoils in another
direction resulting to have less energy in the scattered photon. In addition, the
frequencies will also depend on the differences of the angle at which the
scattered photon leaves the collision and this incident is called Compton Effect.