A. Using the third equation of motion:
v2 = u2 + 2as
from the question;
the jet was initially at rest
hence u = 0
a = 1.75m/s2
s = 1500m
v2 = 02 + 2(1.75)(1500)
v2 = 5250
v = √5250
v = 72.46m/s
hence it moves with a velocity of 72.46m/s.
b. s = ut + 1/2at2
1500 = 0(t) + 1/2(1.75)t2
1500 × 2 = 2× 1/2(1.75)t2
3000 = 1.75t2
1714.29 = t2
41.4 = t
hence the time taken for the plane to down the runway is 41.4s.
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Answer:
30 m/s
Explanation:
Speed is distance over time. 60 meters / 2 seconds, = 30 m/s.
It's the only star in the sky (visible from the northern hemisphere) that never seems to move. It stays at almost exactly the same point in the sky, while the other stars all circle around it once a day.
Answer:
6 amps
Explanation:by Kirchhoff's loop rule the current at any point in the loop must be equal or charge would be building up. The current at the ammeter is equally to the total current through the sun of the paths in parallel which it is in series with
Answer:
Number of photons travel through pin hole=
Explanation:
First we will calculate the energy of single photon using below formula:
Where :
h is plank's constant with value 
c is the speed of light whch is
λ is the wave length = 532nm
E=
J
Number of photons emitted per second:
Number of photons emitted per second=
=
Where:
A-hole is area of hole
A-beam is area of beam
d-hole is diameter of hole
d-beam is diameter if beam
=
=
=
Number of photons travel through pin hole=
Number of photons travel through pin hole=
