Answer:
22 m/s
Explanation:
PEf +KEf =PE0 +KE0 →PE0 −PEf =KEf
−mgΔy= 1 mv2 →v= −2gΔy = −2(9.8 m/s2)(−25 m)=22 m/s
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Answer:
A) 1.88 * 10^17 m
B) 1.22 * 10^34 J
C) 1.95 * 10^34 J
Explanation:
Parameters given:
Mass of planet = 7.00 * 10^25 kg
Radius of orbit = 6.00 * 10^11 m
Force exerted on planet = 6.51 * 10^22 N
Velocity of planet = 2.36 * 10^4 m/s
A) The distance traveled by the planet is half of the circumference of the orbit (which is circular).
The circumference of the orbit is
C = 2 * pi * R
R = radius of orbit
C = 2 * 3.142 * 6.0 * 10¹¹
C = 3.77 * 10¹² m
Hence, distance traveled will be:
D = 0.5 * 3.77 * 10¹²
D = 1.88 * 10 ¹² m/s
B) Work done is given as:
W = F * D
W = 652 * 10²² * 1.88 * 10¹¹
W = 1.22 * 10³⁴ J
C) Change in Kinetic energy is given as:
K. E. = 0.5 * m * v²
K. E. = 0.5 * 7 * 10^25 * (2.36 * 10^4)²
K. E. = 1.95 * 10³⁴ J
Answer:
Explanation:
Apply the law of conservation of energy
from the law of conservation of the linear momentum
Therefore,
Substitute the values in the above result
B) the speed of the sphere with mass 107.0 kg is
C) the magnitude of the relative velocity with which one sphere is
D) the distance of the centre is proportional to the acceleration
Thus,
and
When the sphere make contact with eachother
Therefore,
And
The point of contact of the sphere is