It’s either 0.05 or 20. Assuming that the coefficient friction is a damping factor, I feel like 0.05 would be correct m
Answer:
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Answer:
Distance for which the balloon has been pulled by the child = 12 meter
Force applied by the child to pull the balloon = 1.0 Newton
The angle below horizontal = 60 degree
These information's are already given in the question. Based on the above information's the answer to the question can be easily determined.
We know that
Work done = Force * Distance * Cos(60)
= 1 * 12 * (1/2)
= 1 * 6
= 6 Joules
So the work done by the child is 6 joules. I hope the procedure is clear enough for you to understand. Based on the above mentioned procedure, you can always solve problems of similar kind without taking any outside help.
Explanation:
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Explanation:
Answer:
0.8 seconds
Explanation:
F=ma
Let x be the seconds the force is applied.
m = 20kg
F = 50 Newtons (kg*m/sec^2)
acceleration, a, is provided for x seconds to increase the speed from 1 m/s to 3 m/s, an increase of 2m/s
Let's calculate the acceleration of the cart:
F=ma
(50 kg*m/s^2) = (20kg)*a
a = 2.5 m/s^2
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The acceleration is 2.5 m/s^2. The cart increases speed by 2.5 m/s every second.
We want the number of seconds it takes to add 2.0 m/sec to the speed:
(2.5 m/s^2)*x = 2.0 m/s
x = (2.0/2.5) sec
x = 0.8 seconds