Tính lực tương tác giữa 2 điện tích q1=4.10^-6C và q2=-3.10^-6C cách nhau một 3cm đặt trong dầu hoả = 2
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Use the equation
plug in the variables for the equation. Use 0 for Vf because at the highest point the velocity would be zero. Use -9.8 for acceleration because that is the speed at which gravity pulls down.
your answer would be 2.06122 seconds.
plug in the variables for the equation. Use 0 for Vf because at the highest point the velocity would be zero. Use -9.8 for acceleration because that is the speed at which gravity pulls down.
your answer would be 2.06122 seconds.
Explanation:
Given that,
Initial speed of the rock, u = 30 m/s
The acceleration due to gravity at the surface of the moon is 1.62 m/s².
We need to find the time when the rock is ascending at a height of 180 m.
The rock is projected from the surface of the moon. The equation of motion in this case is given by :
It is a quadratic equation, after solving whose solution is given by:
t = 7.53 s
or
t = 8 seconds
(e)If it is decending, v = -20 m/s
Now t' is the time of descending. So,
Let h' is the height of the rock at this time. So,
or
h' = 155 m