Answer:
increase the time the force acts or you could increase the number of temptations.
hope this helped!
I will assume that you want to know if the driver can stop before hitting the obstacle or not.
First step is to use the reaction time in order to know how far can she go before her motion starts to slow:
distance = velocity x time = 20 x 0.5 = 10 m
Thus, the driver has 50 - 10 = 40 m to stop before colliding
Second step is to calculate the distance that the driver requires to stop using the rules of velocity and distance:
1- velocity = acceleration x time
20 = 6t ..............> thus, time = 3.334 seconds
2- distance = 0.5 x a x t^2 = 0.5 x 6 x (3.334)^2 = 33.3466 m
From the previous calculations, we can see that the driver has 40 m to stop and she needs only 33.3466 m to stop based on the given parameters. This means that she can stop before collision.
Answer:
The centripetal force is 663.95 N.
Explanation:
Given that,
Mass of the child, m = 19 kg
Angular velocity of the merry-go-round,
Distance from the center, r = 2 m
We need to find the centripetal force acting on the child. The centripetal force is given by the formula as :
F = 663.95 N
So, the centripetal force is 663.95 N. Hence, this is the required solution.
From the calculation, the force constant is 192 N. Also, friction would decrease the extension.
<h3>What is the force constant?</h3>
We know that the force constant can be obtained by the use of the relation;
F = Ke
F = applied force
K = force constant
e = extension
We know from Hooks law that the force applied is directly proportional to the extension.
We can write;
F = mgcosθ
F = 43 Kg * 9.8 m/s^2 * sin31°
F = 217 N
K = 217 N/1.13 m
K = 192 N/m
If there is friction between the incline and the crate, it will stretch less because some work will be lost due to friction causing only some fraction of the elastic potential energy to be converted to kinetic energy.
Learn more about Hooke's law:brainly.com/question/14140269
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