Answer:
a = 52s²
Explanation:
<u>How to find acceleration</u>
Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2). Acceleration is also a vector quantity, so it includes both magnitude and direction.
<u>Solve</u>
We know initial velocity (u = 16), velocity (v = 120) and acceleration (a = ?)
We first need to solve the velocity equation for time (t):
v = u + at
v - u = at
(v - u)/a = t
Plugging in the known values we get,
t = (v - u)/a
t = (16 m/s - 120 m/s) -2/s2
t = -104 m/s / -2 m/s2
t = 52 s
Answer:
W = 19.845 J
Explanation:
Work is defined as W = Fdcos
, where F is the force exerted and d is the distance. Because the direction the ball is falling is the same direction as the force itself,
= 0 deg, and since cos(0) = 1, this equation is equivalent to W = Fd. In this case, the force exerted is the weight force, which is equivalent to m * g. Substituting you get:
W = mgd = 0.810 kg * 9.8 m/s^2 * 2.5m
W = 19.845 J
B <span> of Earth’s surface is covered by water. Very little or no light penetrates beyond a few hundred feet in water</span>
Density = (mass) / (volume)
4,000 kg/m³ = (mass) / (0.09 m³)
(4,000 kg/m³) x (0.09 m³) = mass
mass = 360 kg
force of gravity = (mass) x (acceleration of gravity) = (360 kg) x (9.8 m/s²) = (360 x 9.8) kg-m<span>/s² </span><span>= </span>3,528 newtons .
The normal stress follows the formula written below:
σ = F/A
There are two types of stress, axial and tangential. Since we are only given with the dimension of the radius (and not the length), the possible stress is axial. So, the area is,
A = πr² = π(0.75 in)² = 1.767 in²
So,
σ = F/A = 500 lb/1.767 in² = <em>282.94 psi</em>