<span>work =V*Q
=12*50*10^-6
The total work done will be equal to
work = V.Q
which means
w= 12 . 50.10^-6
Hence,
w= 0.0006 J</span>
S = ut + 1/2 at^2
a = 3.2 m/s^2
s = 15m
Find t
15 = 1/2(3.2)t^2
15 = 3.2t^2/2
30 = 3.2t^2
30/ 3.2 = 9.38
Square root of 9.38 = 3.06
It takes 3.06 seconds
Answer:
a) In order to catch the ball at the level at which it is thrown in the direction of motion.
b)Speed of the receiver will be 7.52m/s
Explanation:
Calculating range,R= Vo^2Sin2theta/g
R= (20^2×Sin(2×30)/9.8 = 35.35m
Let receiver be(R-20) = 35.35-20= 15.35m
The horizontal component of the ball is:
Vox= Vocostheta= 20× cos30°
Vox= 17.32m/s
Time taken to coverR=35.35m with 17.32m/s will be:
t=R/Vox= 35.35/17.32
t= 2.04seconds
b)Speed required to cover 15.35m at 2.04seconds
Vxreciever= d/t = 15.35/2.04 = 7.52m/s
Answer:
250 J
Explanation:
Apply the formula:
K = 1/2.m.v²
K = 1/2.5.10²
K = 1/2.5.100
K = 5.50
K = 250 Joules
But, remember that if the speed is accelerating or not, if it is, then we need to know the point in time that the question is asking for the Kinetic Energy.
In this case, I think it is just a constant speed.
After 20 s, the motorcycle attains a speed of
and it continues at this speed for the next 40 s. So at 45 s, its speed is 80 m/s.