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:
length of the ladder is 13.47 feet
base of wall to latter distance 6.10 feet
angle between ladder and the wall is 26.95°
Explanation:
given data
height h = 12 feet
angle 63°
to find out
length of the ladder ( L) and length of wall to ladder ( A) and angle between ladder and the wall
solution
we consider here angle between base of wall and floor is right angle
we apply here trigonometry rule that is
sin63 = h/L
put here value
L = 12 / sin63
L = 13.47
so length of the ladder is 13.47 feet
and
we can say
tan 63 = h / A
put here value
A = 12 / tan63
A = 6.10
so base of wall to latter distance 6.10 feet
and
we say here
tanθ = 6.10 / 12
θ = 26.95°
so angle between ladder and the wall is 26.95°
Answer:
hola me llamo bruno y tu?
Explanation:
pero yo soy de mexico
If it is completely elastic, you can calculate the velocity of the second ball from the kinetic energy
<span>v1 = velocity of #1 </span>
<span>v1' = velocity of #1 after collision </span>
<span>v2' = velocity of #2 after collision. </span>
<span>kinetic energy: v1^2 = v1' ^2 + v2' ^2 (1/2 and m cancel out) </span>
<span>5^2 = 4.35^2 + v2' ^2 </span>
<span>v2 = 2.46 m/s <--- ANSWER</span>