Answer:
Explanation:
0.2 rev/s = 0.2 rev/s * 2π rad/rev = 0.4π rad/s
Since the angular acceleration is assumed to be constant, and the wheel's angular speed is increasing from rest (0 rad/s) to 0.4π rad/s within 23.8s. Then the angular acceleration must be
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
The work done to pull the object 7.0 m is the total area under the graph from 0.0 m to 7.0 m, determined as 245 J.
<h3>Work done by the applied force</h3>
The area under force versus displacement graph is work done.
The total work done by pulling the object 7 m, can be grouped into two areas;
- First area, A1 = area of triangle from 0 m to 2.0 m
- Second area, A2 = area of trapezium, from 2.0 m to 7.0 m
A1 = ¹/₂ bh
A1 = ¹/₂ x (2) x (20)
A1 = 20 J
A2 = ¹/₂(large base + small base) x height
A2 = ¹/₂[(7 - 2) + (7-3)] x 50
A2 = ¹/₂(5 + 4) x 50
A2 = 225 J
<h3>Total work done </h3>
W = A1 + A2
W = 20 J + 225 J
W = 245 J
Learn more about work done here: brainly.com/question/8119756
The Law of Conservation of Energy states that, in an isolated system, energy remains constant and can not be created or destroyed, only transferred from one form to another. This law was created by Julius Robert Mayer.
