Answer:
Explanation:
Since the force applied is parallel to the displacement of the car, the work done on the car is simply given by:
where
F = 1210 N is the force applied on the car
d = 201 m is the displacement of the car
Substituting numbers into the equation, we find:
Answer:
120 m
Explanation:
Given:
wavelength 'λ' = 2.4m
pulse width 'τ'= 100T ('T' is the time of one oscillation)
The below inequality express the range of distances to an object that radar can detect
τc/2 < x < Tc/2 ---->eq(1)
Where, τc/2 is the shortest distance
First we'll calculate Frequency 'f' in order to determine time of one oscillation 'T'
f = c/λ (c= speed of light i.e 3 x m/s)
f= 3 x / 2.4
f=1.25 x hz.
As, T= 1/f
time of one oscillation T= 1/1.25 x
T= 8 x s
It was given that pulse width 'τ'= 100T
τ= 100 x 8 x => 800 x
s
From eq(1), we can conclude that the shortest distance to an object that this radar can detect:
= τc/2 => (800 x
x 3 x
)/2
=120m
Answer:
The final velocity of 18 kg ball is = 42.09
Explanation:
Mass of first ball = 18 kg
Mass of second ball = 47 kg
Initial velocity of 18 kg ball = 76
Initial velocity of 47 kg ball = 0
Final velocity of 18 kg ball = ??
Final velocity of 18 kg ball is given by the formula
=
Put all the values in above formula we get
= 2 × 18 ×
= 42.09
Thus, the final velocity of 18 kg ball is = 42.09