Answer:
x = 11.23 m
Explanation:
For this interesting exercise, we must use angular kinematics, linear kinematics and the relationship between angular and linear quantities.
Let's reduce to SI system units
θ = 155 rev (2pi rad / rev) = 310π rad
α = 2.00rev / s2 (2pi rad / 1 rev) = 4π rad / s²
Let's look for the angular velocity at the time the piece is released, with starting from rest the initial angular velocity is zero (wo = 0)
w² = w₀² + 2 α θ
w =√ 2 α θ
w = √(2 4pi 310pi)
w = 156.45 rad / s
The relationship between angular and linear velocity
v = w r
v = 156.45 0.175
v = 27.38 m / s
In this part we have the linear speed and the height that it travels to reach the floor, so with the projectile launch equations we can find the time it takes to arrive
y =
t - ½ g t²
As it leaves the highest point its speed is horizontal
y = 0 - ½ g t²
t = √ (-2y / g)
t = √ (-2 (-0.820) /9.8)
t = 0.41 s
With this time we calculate the horizontal distance, because the constant horizontal speed
x = vox t
x = 27.38 0.41
x = 11.23 m
Think it would be c, 5.1 g
Answer:
<h3>The answer is 30.43 L</h3>
Explanation:
The new volume can be found by using the formula for Boyle's law which is

Since we are finding the new volume

From the question we have

We have the final answer as
<h3>30.43 L</h3>
Hope this helps you
Answer:
4.5 s, 324 ft
Explanation:
The object is projected upward with an initial velocity of

The equation that describes its height at time t is
(1)
where t, the time, is measured in seconds.
In order to find the time it takes for the object to reach the maximum height, we must find an expression for its velocity at time t, which can be found by calculating the derivative of the position, s(t):
(2)
At the maximum heigth, the vertical velocity is zero:
v(t) = 0
Substituting into the equation above, we find the corresponding time at which the object reaches the maximum height:

And by substituting this value into eq.(1), we also find the maximum height:

Answer: C. 1.64 x 10-3 m/s2