Answer:
solved
Explanation:
a) F_net = (F2 - F3)i - F1 j
b) |Fnet| = sqrt( (F2 - F3)^2 + F1^2)
= sqrt( (9- 5)^2 + 1^2)
= 4.123 N
c) θ = tan^-1( (Fnet_y/Fnet_x)
= tan^-1( -1/(9-5) )
= -14.036°
Answer: 5m/L^2
Explanation:
Inertial I = mr^2 where r = distance from axis of rotation, while m is the mass of the object.
I = 2[m(1L/2)^2] + 2[m(3L/2)^2] = 2m×. 25/L^2+ 3m×2. 25/L^2= 0. 5m/l^2 +4. 5m/l^2
= 5m/l^2.
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
