Answer:
D) D =
, E) (C, D) = (
Explanation:
Part D) two expressions are indicated
3C + 4D = 5
2C +5 D = 2
let's simplify each expression
3C + 4D = 5
4D = 5 - 3C
we divide by 4
D =
The other expression
2C +5 D = 2
2C = 2 - 5D
C =
we can see that the correct result is 1
Part E.
It is asked to solve the problem by the substitution method, we already have
D =
we substitute in the other equation
2C +5 D = 2
2C +5 (5/4 - ¾ C) = 2
we solve
C (2 - 15/4) + 25/4 = 2
-7 / 4 C = 2 - 25/4
-7 / 4 C = -17/4
7C = 17
C =
now we calculate D
D =
D = 5/4 - 51/28
D =
D = - 16/28
D =
the result is (C, D) = (
)
Answer:
A) V_rms = 29 V
B) Vav = 0 V
Explanation:
A) We are told that;
V = V_o cos ωt
voltage amplitude; V = V_o = 41.0V
Now, the formula for the root-mean-square potential difference Vrms is given as;
V_rms = V/√2
Thus plugging in relevant values, we have;
V_rms = 41/√2
V_rms = 29 V
B) Due to the fact that the voltage is sinusoidal from the given V = V_o cos ωt, we can say that the average potential difference Vav between the two terminals of the power supply would be zero.
Thus; Vav = 0 V
They got back in the Lunar Explorer Module
The application of force over distance is known as [WORK]
Answer:
If this is electrical currents , make the wire longer, smaller diameter wires, heat it up