The task is to show that the right side of the equation has units of [Time], just like the left side has.
The right side of the equation is . . . 2 π √(L/G) .
We can completely ignore the 2π since it has no units at all, so it has no effect on the units of the right side of the equation. Now the task is simply to find the units of √(L/G) .
L . . . meters
G . . . meters/sec²
(L/G) = (meters) / (meters/sec²)
(L/G) = (meters) · (sec²/meters)
(L/G) = (meters · sec²) / (meters)
(L/G) = sec²
So √(L/G) = seconds = [Time]
THAT's what we were hoping to prove, and we did it !
Answer:
61 degrees, I just did the test.
Explanation:
We know that: 1 L = 100 cL. Or 1 cL = 0.01 L. Then we will make the conversion: 34.9 cL = 34.9 / 100 L = 0.349 L. Also: 1 hL = 100 L. 0.349 L = 0.349 / 100 hL = 0.00349 hL. This can be also written as: 3.49 * 10^(-3) hL ( in the scientific notation ). Answer: 3.49 cL = 0.00349 <span>hL </span>
Answer:
a) 
, b) 
, c) 
.
Explanation:
a) Let assume that car travel on a horizontal surface. The equations of equilibrium of the car are:
After some algebraic handling, the following expression for the propulsion force is constructed:
b) The power require to move the car at a speed of 5 meters per second is:
c) The efficiency of the car is:
Given Information:
Voltage of circuit A = Va = 208 Volts
Current of circuit A = Ia = 40 Amps
Voltage of circuit B = Vb = 120 Volts
Current of circuit B = Ib = 20 Amps
Required Information:
Ratio of power = Pa/Pb = ?
Answer:
Ratio of power = Pa/Pb = 52/15
Explanation:
Power can be calculated using Ohm's law
P = VI
Where V is the voltage and I is the current flowing in the circuit.
The power delivered by circuit A is
Pa = Va*Ia
Pa = 208*40
Pa = 8320 Watts
The power delivered by circuit B is
Pb = Vb*Ib
Pb = 120*20
Pb = 2400 Watts
Therefore, the ratio of the maximum power delivered by circuit A to that delivered by circuit B is
Pa/Pb = 8320/2400
Pa/Pb = 52/15
