It accelerates from rest at 1.1 m/s^2 for 20s.
<span>s = at^2/2 </span>
<span>s = (.5)(1.1)(20)^2= 220m </span>
<span>then it travels 1100m </span>
<span>Then it slows down at 2.2m/s^2 until it stops at the station. </span>
<span>Calculate the train's speed: </span>
<span>v = at </span>
<span>v = (1.1)(20) = 22m/s
</span>
<span>Now, calculate the time to stop: </span>
<span>v = at </span>
<span>t = 22/2.2 = 10s </span>
<span>s = t(u+v)/2 </span>
<span>s = 10(0+22)/2 = 110m </span>
<span>(A) Total distance = 220 +1100+110 = 1430m </span>
<span>(B) Time = 20 + 1100/22 + 10 = 80s</span>
I think the answer is potential
Answer:
The magnitude of the applied torque is 
(e) is correct option.
Explanation:
Given that,
Mass of object = 3 kg
Radius of gyration = 0.2 m
Angular acceleration = 0.5 rad/s²
We need to calculate the applied torque
Using formula of torque
Here, I = mk²
Put the value into the formula
Hence, The magnitude of the applied torque is
Electromagnetic waves travel as what type of waves is transverse wave
