Answer:
19.3m/s
Explanation:
Use third equation of motion

where v is the velocity at halfway, u is the initial velocity, g is gravity (9.81m/s^2) and h is the height at which you'd want to find the velocity
insert values to get answer
![v^2-0^2=2(9.81m/s^2)(38/2)\\v^2=9.81m/s^2 *38\\v^2=372.78\\v=\sqrt[]{372.78} \\v=19.3m/s](https://tex.z-dn.net/?f=v%5E2-0%5E2%3D2%289.81m%2Fs%5E2%29%2838%2F2%29%5C%5Cv%5E2%3D9.81m%2Fs%5E2%20%2A38%5C%5Cv%5E2%3D372.78%5C%5Cv%3D%5Csqrt%5B%5D%7B372.78%7D%20%5C%5Cv%3D19.3m%2Fs)
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:
1/2 M V^2 = .1 M g H where 10% of PE goes into KE
V^2 = .2 g H = .2 * 9.8 * (2100 - 1600) = 980 m^2 / s^2
V = 31.1 m/s increase in speed during descent
1 km / hr = 1000 m / 3600 sec = .278 m/s
V = 31.1 m/s / (.278 m/s / km /hr)= 112 km/hr
Answer:
The total number of electrons is 
(3) is correct option.
Explanation:
Given that,
The current equation is

We know that,
The formula of charge








We need to calculate the number of electron
Using formula of charge




Hence, The total number of electrons is 