Answer:
They both have the same angular speed.
Explanation:
The mathematical formula for angular speed is:
where 
is angular speed, 
is a constant, and 
is the period (the time it takes the marry-go-round to complete a lap).
What we can see from the formula is that, since the does not change its value, the angular speed depends only on the period T.
In this case for both the children closer to the outher edge and for the children closer to the center, the time to complete a lap is the same, because the time does not depend on where they are sitting in the marry go round. This means that the period for both is the same.
Thus, since the period for both is the same, the angular speed given by
will also be the same
Answer:
about 19.6° and 73.2°
Explanation:
The equation for ballistic motion in Cartesian coordinates for some launch angle α can be written ...
y = -4.9(x/s·sec(α))² +x·tan(α)
where s is the launch speed in meters per second.
We want y=2.44 for x=50, so this resolves to a quadratic equation in tan(α):
-13.6111·tan(α)² +50·tan(α) -16.0511 = 0
This has solutions ...
tan(α) = 0.355408 or 3.31806
The corresponding angles are ...
α = 19.5656° or 73.2282°
The elevation angle must lie between 19.6° and 73.2° for the ball to score a goal.
_____
I find it convenient to use a graphing calculator to find solutions for problems of this sort. In the attachment, we have used x as the angle in degrees, and written the function so that x-intercepts are the solutions.
Answer:
1.876 J
Explanation:
First, let’s calculate the compression of the spring from the Hooke’s law:
F=kx,
here, F=75 N is the force acted on the spring, k=1500 N⁄m is the force constant of the spring, x is the compression of the spring.
Then, we get:
x=F/k=(75 N)/(1500 N/m)=0.05 m.
Finally, we can find the potential energy stored in the spring:
PE=1/2 kx^2=1/2∙1500 N/m∙(0.05 m)^2=1.875 J.
correct my answer if it's wrong ^^
