The kinematic equations of motion that apply here are<span>y(t)=votsin(θ)−12gt2</span>and<span>x(t)=votcos(θ)</span>Setting y(t)=0 yields <span>0=votsin(θ)−12gt2</span>. If we solve for t, we obtain, by factoring,<span>t=<span>2vsin(θ)g</span></span>Substitute this into our equation for x(t). This yields<span>x(t)=<span><span>2v2cos(θ)sin(θ)</span>g</span></span><span>This is equal to x=<span><span>v^2sin(2θ)</span>g</span></span>Hence the angles that have identical projectiles are have the same range via substitution in the last equation is C. <span> 60.23°, 29.77° </span>
Answer:
μ = mg/kx
Explanation:
Since the bock does not slip, the frictional force equals the weight of the block. So, F = mg. Now, the frictional force, F = μN where μ = coefficient of static friction and N = Normal force.
Now, the normal force equals the spring force F' = kx where k = spring constant and x = compression of spring.
N = F' = kx
So, F = μN = μkx
μkx = mg
So, μ = mg/kx
The child's linear speed is
<em> (pi / 5) x (the child's distance from the center of the ride, in feet)</em>
feet per second.
