The distance between tow objects is increased by 3 times the original distance. How will this change the force of attraction bet
1 answer:
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Answer:
The flea will reach to a height of 5.2 cm.
Explanation:
Let us assume it is given that, a flea reaches a takeoff speed of 1.0 m/s over a distance of 0.50 mm.
Initial speed, u = 1 m/s
Initial distance, x = 0.5 mm
We need to find the height reached by the flea. Using third equation of motion to find it.
At maximum height, its final speed, v = 0
Here, a = -g
or
h = 5.2 cm
So, the flea will reach to a height of 5.2 cm.
Answer: after 1.75 seconds
Explanation:
The only force acting on the ball is the gravitational force, so the acceleration will be:
a = -9.8 m/s^2
the velocity can be obtained by integrating over time:
v = -9.8m/s^2*t + v0
where v0 is the initial velocity; v0 = -7.95 m/s.
v = -9.8m/s^2*t - 7.95 m/s.
For the position we integrate again:
p = -4.9m/s^2*t^2 - 7.95 m/s*t + p0
where p0 is the initial position: p0 = 29m
p = -4.9m/s^2*t^2 - 7.95 m/s*t + 29m
Now we want to find the time such that the position is equal to zero:
0 = -4.9m/s^2*t^2 - 7.95 m/s*t + 29m
Then we solve the Bhaskara's equation:
Then the solutions are:
t = (7.95 + 25.1)/(-9.8) = -3.37s
t = (7.95 - 25.1)/(-9.8) = 1.75s
We need the positive time, then the correct answer is 1.75s