Answer:
81.6 m
Explanation:
Answer: 81.6 m.
The time it takes gravity to slow 40 m/s to zero when it teaches maximum height is
-v(initial) / -g = t
-40 m/s / -9.8 m/s^2 = 4.08 s
The height reached is the average velocity times this time 4.08 s, with v(avg) = [v(initial) + v(final)] / 2 with v(final) = 0. v(avg) = v(initial) / 2 = 40 m/s / 2 = 20 m/s.
So the distance d of maximum height is
d = v(avg)•t
d = 20 m/s • 4.08 s = 81.6 m.
Answer:
the correct statement is 2. The solid plate will have the greater angular acceleration.
the correct phrase is 4. The plate with the hole has its mass distributed further out from the axis of rotation, which will increase its moment of inertia.
Explanation:
Newton's second law expression for rotational motion is
τ = I α (1)
where the torque is
τ = F r
in this case, as the discs have the same radius and the applied force is the same, the torque is the same on the two discs.
The moment of inertia is given by the expression
I =∫ r² dm
for bodies with high symmetry are tabulated
the moment of inertia for in disk solid is I₁ = ½ m R₂²
the moment for a disk with a hole I₂ = ½ m (R₁² + R₂²)
We can see that the moment of inertia of the disk with the hole is greater than the moment of inertia of the solid disk.
Let's use equation 1
α = τ/I
therefore the angular acceleration is lower for the body with the higher moment of inertia, consequently the solid disk has higher angular acceleration
the correct statement is 2
The reason is because the moment of inertia is higher for the hollow disk.
the correct phrase is 4
a) 12.5 m
Explanation:
During the acceleration phase, the distance travelled by the elevator is given by:
where
v = 5 m/s is the final speed of the elevator
u = 0 is the initial speed
a = 1.0 m/s^2 is the acceleration
d is the distance travelled
Using the equation, we can find d:
b) 45 s
Explanation:
The distance covered during the acceleration phase is 12.5 m. The distance covered during the de-celeration phase, at the end of the trip, is exactly the same (because the values of u, v and a are the same). The total distance is 200 m, so the distance covered at a constant speed of 5 m/s is
This distance is covered at a speed of v=5 m/s, so the time taken is
The time the elevator takes to cover the 12.5 m in the acceleration phase is
and the same for the deceleration phase; therefore, the total time for the trip is
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