During a golf drive, the angular velocity of the driver is 20 rad/s just before impact with the golf ball. If the distance from
1 answer:
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F=nmv
where;
n=no. of bullets = 1
m=mass of bullets=2g *10^-3
V=velocity of bullets200m/sec
F=1
loss in Kinetic energy=gain in heat energy
1/2MV^2=MS∆t
let M council M
=1/2V^2=S∆t
M=2g
K.E=MV^2/2
=(2*10^-3)(200)^2/2
2 councils 2
2*10^-3*4*10/2
K.E=40Js
H=mv∆t
(40/4.2)
40Js=40/4.2=mc∆t
40/4.2=2*0.03*∆t
=158.73°C
Answer:
5.024 years
Explanation:
T1 = 1 year
r1 = 150 million km
r2 = 440 million km
let the period of asteroid orbit is T2.
Use Kepler's third law
T² ∝ r³
So,
T2 = 5.024 years
Thus, the period of the asteroid's orbit is 5.024 years.