Answer:
Acceleration = 0.0282 m/s^2
Distance = 13.98 * 10^12 m
Explanation:
we will apply the energy theorem
work done = ΔK.E ( change in Kinetic energy ) ---- ( 1 )
<em>where :</em>
work done = p * t
= 15 * 10^6 watts * ( 1 year ) = 473040000 * 10^6 J
( note : convert 1 year to seconds )
and ΔK.E = 1/2 mVf^2 given ; m = 1200 kg and initial V = 0
<u>back to equation 1 </u>
473040000 * 10^6 = 1/2 mv^2
Vf^2 = 2(473040000 * 10^6 ) / 1200
∴ Vf = 887918.92 m/s
<u>i) Determine how fast the rocket is ( acceleration of the rocket )</u>
a = Vf / t
= 887918.92 / ( 1 year )
= 0.0282 m/s^2
<u>ii) determine distance travelled by rocket </u>
Vf^2 - Vi^2 = 2as
Vi = 0
hence ; Vf^2 = 2as
s ( distance ) = Vf^2 / ( 2a )
= ( 887918.92 )^2 / ( 2 * 0.0282 )
= 13.98 * 10^12 m
Answer:
2f
Explanation:
The formula for the object - image relationship of thin lens is given as;
1/s + 1/s' = 1/f
Where;
s is object distance from lens
s' is the image distance from the lens
f is the focal length of the lens
Total distance of the object and image from the lens is given as;
d = s + s'
We earlier said that; 1/s + 1/s' = 1/f
Making s' the subject, we have;
s' = sf/(s - f)
Since d = s + s'
Thus;
d = s + (sf/(s - f))
Expanding this, we have;
d = s²/(s - f)
The derivative of this with respect to d gives;
d(d(s))/ds = (2s/(s - f)) - s²/(s - f)²
Equating to zero, we have;
(2s/(s - f)) - s²/(s - f)² = 0
(2s/(s - f)) = s²/(s - f)²
Thus;
2s = s²/(s - f)
s² = 2s(s - f)
s² = 2s² - 2sf
2s² - s² = 2sf
s² = 2sf
s = 2f
Answer:
Check attachment for your answer
Good luck
True I believe..................
