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
So, C = kE°A/d
putting the values,
C
= 3.8 × 8.85×10^(-12) × 3.14×1.5×1.5 × 10^(-6)/0.43 × 10^(-3)
so, 1.02 × 10^(-13)
so the most appropriate answer is 2 ...that is
1.4 × 10^(-13) ....answer !!
Answer:
96 m
Explanation:
Given,
Initial velocity ( u ) = 4 m/s
Final velocity ( v ) = 20 m/s
Time ( t ) = 8 s
Let Acceleration be " a ".
Formula : -
a = ( v - u ) / t
a = ( 20 - 4 ) / 8
= 16 / 8
a = 2 m/s²
Let displacement be " s ".
Formula : -
s = ut + at² / 2
s = ( 4 ) ( 8 ) + ( 2 ) ( 8² ) / 2
= 32 + ( 2 ) ( 64 ) / 2
= 32 + ( 2 ) ( 32 )
= 32 + 64
s = 96 m
Therefore, it travels 96 m in time 8 s.
"<span>a layer in the earth's stratosphere at an altitude of about 6.2
miles (10 km) containing a high concentration of ozone, which absorbs
most of the ultraviolet radiation reaching the earth from the sun."
Hope this helps!
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