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:
n = 5 approx
Explanation:
If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back
= e ( coefficient of restitution ) = 
and

h₁ is height up-to which the ball bounces back after first bounce.
From the two equations we can write that


So on

= .00396
Taking log on both sides
- n / 2 = log .00396
n / 2 = 2.4
n = 5 approx
Answer:
- 0.6
Explanation:
Given that angle between normal y axis is 62° so angle between normal
and x axis will be 90- 62 = 28 °. Since incident ray is along x axis , 28 ° will be the angle between incident ray and normal ie it will be angle of incidence
Angle of incidence = 28 °
angle of reflection = 28°
Angle between incident ray and reflected ray = 28 + 28 = 56 °
Angle between x axis and reflected ray = 56 °
x component of reflected ray
= - cos 56 ( it will be towards - ve x axis. )
- 0.6
Explanation:
speed of an object is the magnitude of the rate of change of its position with time or the magnitude of the change of its position per unit of time; it is thus a scalar quantity.
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SEISMOGRAM