<h3>
Answer:</h3>
539.56 Joules
<h3>
Explanation:</h3>
- Efficiency of a machine is the ratio of work output to work input expressed as a percentage.
- Efficiency = (work output/work input) × 100%
- Efficiency of a machine is not 100% because so energy is lost due to friction of the moving parts and also as heat.
In this case;
Efficiency = 94%
Work input = 574 Joules
Therefore, Assuming work output is x
94% = (x/574 J) × 100%
0.94 = (x/574 J)
<h3>x = 539.56 J</h3>
Thus, you get work of 539.56 J from the machine
56 Newtons bc w=F×D so if you divide by D on both side you get w/D=F
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
