Answer:
kinetic energy of the train = 2,910.6 x 10⁷ joule
Explanation:
Given:
Mass of train = 3.3 x 10⁷ kg
Speed of train = 42 m/s
Find:
kinetic energy of the train
Computation:
kinetic energy = (1/2)(m)(v²)
kinetic energy of the train = (1/2)(3.3 x 10⁷)(42²)
kinetic energy of the train = (1/2)(3.3 x 10⁷)(1,764)
kinetic energy of the train = (3.3 x 10⁷)(882)
kinetic energy of the train = 2,910.6 x 10⁷ joule
Answer:
V=14
Explanation:
PE=KE
mgh=1/2mv^2
2(9.8)10=1/2(2)v^2
(radical) 196= (radical)v
V=14
Answer with Explanation:
We are given that
Mass of bullet,
We have to find the velocity of the bullet and block B after the first impact and final velocity of the carrier.
According to law of conservation of momentum
Hence, the velocity of the bullet and block B after the first impact=4.65 m/s
According to law of conservation of momentum
Answer:
<em>Good Luck!</em>
Explanation:
Force applied by first student (F1) = 80 N
Force applied by the second student (F2) = 40 N
Displacement (d) = 10 m
Work done by first student (W1) = F1d = 80*10 = 800 J
Work done by second student (W2) = F2d = 40*10 = 400 J
Hence, the work done by the pushing student is 800 J and that by pulling student is
<h2> <u><em>400 J</em></u></h2>
Answer:
1
Explanation:
it go up than down than up
