The work done is equal to the change in potential energy which is:
P.E = mgh
P.E = 500 x 9.81 x 15
P.E = 73,575 J
Power = work / time
Power = 73,575 / 20
Power = 3,700 Watts
784 Newtons or 176.37 lbs
Formula for terminal
velocity is:
Vt = √(2mg/ρACd)
<span>Vt = terminal velocity = ?
<span>m = mass of the falling object = 72 kg
<span>g = gravitational acceleration = 9.81 m/s^2
<span>Cd = drag coefficient = 0.80
<span>ρ = density of the fluid/gas = 1.2 kg/m^3</span>
<span>A = projected area of the object (feet first) = 0.21 m * 0.41
m = 0.0861 m^2
Therefore:</span></span></span></span></span>
Vt = √(2 * 72
* 9.81 / 1.2 * 0.0861 * 0.80)
<span>Vt = 130.73 m/s</span>
Answer:
The following options are true based on the properties of electric field;
a) Electric field lines near positive point charges radiate outward.
b) The electric force acting on a point charge is proportional to the magnitude of the point charge.
d) In a uniform electric field, the field lines are straight, parallel, and uniformly spaced.
Explanation:
From option b) From coulomb's law F = Kq1q2r/r2
Answer:
It compares the the difference between a radioactive element remaining in specimen to the amount of the radioactive element that would have been originally trapped in the specimen. This is done by comparing the ratio of the relative abundance of this radioactive element to its non radioactive isotope in nature to their ratio remaining in the specimen and comparing it to the half-life of the radioactive isotope.
