Two particles experience an attractive electric force of 5 N. If one of the charges triples and
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Answer:
4.44 i + 11.11 j , 11.96 m/s
Explanation:
m1 = 0.2 kg, v1 = 10 m/s along X axis, v1 = 10 i
m2 = 0.25 kg, v2 = 20 m/s along north, v2 = 20 j
After collision they stick together and let they move with velocity v.
Use the conservation of momentum
m1 x v1 + m2 x v2 = (m1 + m2) x v
0.2 x (10 i) + 0.25 x (20 j) = (0.2 + 0.25) x v
2i + 5j = 0.45 v
v = 4.44 i + 11.11 j
magnitude of velocity = = 11.96 m/s
Thus, the velocity of 0.250 kg hockey puck is 4.44 i + 11.11 j metre per second and the magnitude of velocity is 11.96 m/s.
Given data:
* The mass of the Ceres is,
* The mass of the astronaut is,
* The radius of the Ceres is,
Solution:
The gravitational force acting on the astronaut due to the Ceres is,
where G is the gravitational force constant,
Substituting the known values,
The weight of the astronaut on the Ceres is equal to the gravitational force acting on the astronaut.
Thus, the weight of the astronaut on the Ceres is 17.7 N.