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:
60mph=26.8224meters per second
Explanation:
- - <span><span>D. It includes payroll records.
This is because it does not apple to a business plan, when all other choices do.
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The electrical power supplied to the lamp is 100W when the p.d. across it is 24V and the current is 4.2 A.
<h3>What is electrical power?</h3>
Electrical power is the rate at which electrical work is done.
The unit of electrical power is Watts.
- Power = Current × Voltage
The experiment will require a 24V Battery, a Lamp marked 24V, 100 W, a key, wires, a rheostat, a voltmeter and an ammeter.
The circuit is completed and the rheostat is adjusted to provide until the voltmeter reads 24V.
The ammeter reading is taken as well.
When the voltmeter reading is 24V, it can be seen that' the ammeter reading for current is approximately 4.2 A.
Calculating, the power of the lamp:
Power = 24 × 4.2 = 100 W
Therefore, the electrical power supplied to the lamp is 100W when the p.d. across it is 24V and the current is 4.2 A.
Learn more about power and p.d. at: https://brainly.in/question/13629965
Answer:
a. 192 m/s
b. -17,760 kPa
Explanation:
First let's write the flow rate of the liquid, using the following equation:
Q = A*v
Where Q is the flow rate, A is the cross section area of the pipe (A = pi * radius^2) and v is the speed of the liquid. The flow rate in both parts of the pipe (larger radius and smaller radius) needs to be the same, so we have:
a.
A1*v1 = A2*v2
pi * 0.02^2 * 12 = pi * 0.005^2 * v2
v2 = 0.02^2 * 12 / 0.005^2
v2 = 192 m/s
b.
To find the pressure of the other side, we need to use the Bernoulli equation: (600 kPa = 600000 N/m2)
P1 + d1*v1^2/2 = P2 + d1*v2^2/2
Where d1 is the density of the liquid (for water, we have d1 = 1000 kg/m3)
600000 + 1000*12^2/2 = P2 + 1000*192^2/2
P2 = 600000 + 72000 - 1000*192^2/2
P2 = -17760000 N/m2 = -17,760 kPa
The speed in the smaller part of the pipe is too high, the negative pressure in the second part means that the inicial pressure is not enough to maintain this output speed.
