Answer:
v₀ = 9,798 ft / s
Explanation:
We can solve this problem with kinematics in one dimension, when the train stops the speed is zero, the acceleration is negative so that the train stops. Let's use the equation
v² = v₀² - 2 a d
v = 0
v₀ = √2 a d
In the problem it indicates that the acceleration is g / 2, we substitute
v₀ = √2 (g / 2) d
Let's calculate
v₀ = √ 2 32/2 3 = √32 3
v₀ = 9,798 ft / s
Power is defined as rate of work done which means it is work done in 1 second of time
Now it is given that we have a horse that will give power 745.7 W
So it will do work of 745.7 J in 1 second of time
now if we wish to find the work done by horse in 0.55 s
so we can say
So it will do total work of 410.14 J in 0.55 s of time
<span>The glowing of a neon light is caused by electrons emitting energy as they </span>move from higher to lower energy levels.
Answer:
The maximum static frictional force is 40N.
Explanation:
When an object of mass M is on a surface with a coefficient of static friction μ, there is a minimum force that you need to apply to the object in order to "break" the coefficient of static friction and be able to move the object (Called the threshold of motion, once the object is moving we have a coefficient of kinetic friction, which is smaller than the one for static friction).
This coefficient defines the maximum static friction force that we can have.
So if we apply a small force and we start to increase it, the static frictional force will be equal to our force until it reaches its maximum, and then we can move the object and now we will have frictional force.
In this case, we know that we apply a force of 40N and the object just starts to move.
Then we can assume that we are just at the point of transition between static frictional force and kinetic frictional force (the threshold of motion), thus, 40 N is the maximum of the static frictional force.
Answer:
A. Repeat the experiment to be sure the results are valid.
