According to Hooke's Law formula. The force is proportional to the displacement of the spring. I believe
Answer:
a=2.378 m/s^2
Explanation:
a=Δv/Δt------eq(1)
Δv=Vf-Vi=120 km/h-0 km/h=120 km/h
or Δv=33.3 m/sec
or time=t=14s
putting values in eq(1)
a=33.3/14
a=2.378 m/s^2
I'm not sure what "60 degree horizontal" means.
I'm going to assume that it means a direction aimed 60 degrees
above the horizon and 30 degrees below the zenith.
Now, I'll answer the question that I have invented.
When the shot is fired with speed of 'S' in that direction,
the horizontal component of its velocity is S cos(60) = 0.5 S ,
and the vertical component is S sin(60) = S√3/2 = 0.866 S . (rounded)
-- 0.75 of its kinetic energy is due to its vertical velocity.
That much of its KE gets used up by climbing against gravity.
-- 0.25 of its kinetic energy is due to its horizontal velocity.
That doesn't change.
-- So at the top of its trajectory, its KE is 0.25 of what it had originally.
That's E/4 .
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:
um d. but I am guessing this ans
