Answer:
The time taken for the commercial Jet liner to reach the end of its runway is 10.18 s.
Explanation:
Given;
average acceleration of the commercial Jet liner, a = 3g = 3 x 9.8 m/s² = 29.4 m/s²
distance traveled by the commercial Jet liner, s = 1542 m
The time taken for the commercial Jet liner to reach the end of its runway is calculated as follows;
s = ut + ¹/₂at²
where;
u is the initial velocity of the commercial Jet liner = 0
s = 0 + ¹/₂at²
s = ¹/₂at²
2s = at²
Therefore, the time taken for the commercial Jet liner to reach the end of its runway is 10.18 s.
Answer:
The law of conservation of energy
Explanation:
Answer:
Explanation:
As we know that the acceleration due to gravity decreases with height.
At certain height it will get to the half of its value on the surface of the earth.
As we know that the weight on the surface of the earth is given as:
where:
m = mass of the object
g = acceleration due to gravity of the substance
Since mass of the substance is constant so the variation is weight is possible only due to change in the acceleration due to gravity.
<u>We know that the variation of the acceleration due gravity with height is given as:</u>
where:
value to acceleration due to gravity at height h
g = acceleration due to gravity at the earth's surface
h = height of the object
R = radius of the earth = 
according to question the weight becomes half, so,:
is the height a rocket has to go above Earth's surface before its weight is half of what it is on Earth.
No the vector can never be shorter than one of its components
Let
A = the amplitude of vibration
k = the spring constant
m = the mass of the object
The displacement at time, t, is of the form
x(t) = A cos(ωt)
where
ω = the circular frequency.
The velocity is
v(t) = -ωA sin(ωt)
The maximum velocity occurs when the sin function is either 1 or -1.
Therefore
Therefore
The KE (kinetic energy) is given by
The PE (potential energy) is given by
When the KE and PE are equal, then
For the oscillating spring,
Therefore
Answer:
