Answer:
Approximately 
. (Assuming that the drag on this ball is negligible, and that 
.)
Explanation:
Assume that the drag (air friction) on this ball is negligible. Motion of this ball during the descent:
- Horizontal: no acceleration, velocity is constant (at
is constant throughout the descent.) - Vertical: constant downward acceleration at , starting at
.
The horizontal velocity of this ball is constant during the descent. The horizontal distance that the ball has travelled during the descent is also given: 
. Combine these two quantities to find the duration of this descent:
.
In other words, the ball in this question start at a vertical velocity of 
, accelerated downwards at , and reached the ground after 
.
Apply the SUVAT equation 
to find the vertical displacement of this ball.
![\begin{aligned}& x(\text{vertical}) \\[0.5em] &= -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t\\[0.5em] &= - \frac{1}{2} \times 10\; \rm m \cdot s^{-2} \times (7.5\; \rm s)^{2} \\ & \quad \quad + 0\; \rm m \cdot s^{-1} \times 7.5\; s \\[0.5em] &= -281.25\; \rm m\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%26%20x%28%5Ctext%7Bvertical%7D%29%20%5C%5C%5B0.5em%5D%20%26%3D%20-%5Cfrac%7B1%7D%7B2%7D%5C%2C%20g%20%5Ccdot%20t%5E%7B2%7D%20%2B%20v_0%5Ccdot%20t%5C%5C%5B0.5em%5D%20%26%3D%20-%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%2010%5C%3B%20%5Crm%20m%20%5Ccdot%20s%5E%7B-2%7D%20%5Ctimes%20%287.5%5C%3B%20%5Crm%20s%29%5E%7B2%7D%20%5C%5C%20%26%20%5Cquad%20%5Cquad%20%2B%200%5C%3B%20%5Crm%20m%20%5Ccdot%20s%5E%7B-1%7D%20%5Ctimes%207.5%5C%3B%20s%20%5C%5C%5B0.5em%5D%20%26%3D%20-281.25%5C%3B%20%5Crm%20m%5Cend%7Baligned%7D)
.
In other words, the ball is below where it was before the descent (hence the negative sign in front of the number.) The height of this cliff would be 
.
A wave can be described as the disturbance of particles in an area. Think about it this way: particles (matter) carry energy. For all the laws of physics to work, this energy must be "traded" somehow. This happens by miniscule vibrations in the particles, which are apparent disturbances. This creates a wave, and therefore a wave is, indeed, a disturbance.<span />
Answer:
Explanation:
From the exercise we have:
To find the velocity after 2.4s we need to use the following formula:
The negative sign means that the kit is going down.
Answer:
11.3 m/s
Explanation:
First, find the time it takes for the first stone to fall 3.2 m.
Given:
Δy = 3.2 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
(3.2 m) = (0 m/s) t + ½ (9.8 m/s²) t²
t = 0.81 s
Next, find the time for the first stone to land.
Given:
Δy = 15 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
(15 m) = (0 m/s) t + ½ (9.8 m/s²) t²
t = 1.75 s
The difference in time is 1.75 s − 0.81 s = 0.94 s. Find the initial velocity needed for the second stone to land after that amount of time.
Given:
Δy = 15 m
a = 9.8 m/s²
t = 0.94 s
Find: v₀
Δy = v₀ t + ½ at²
(15 m) = v₀ (0.94 s) + ½ (9.8 m/s²) (0.94 s)²
v₀ = 11.3 m/s
Answer:
Kinetic Generating Plants
Hydro-electric plants and wind-mills also convert energy into electricity. Instead of heat energy, they use kinetic energy, or the energy of motion. Moving wind or water (sometimes referred to as "white coal") spins a turbine, which in turn spins the rotor of a generator?
