The moon goes in front of the sun
or the earth blocks the sun from the mon
<span>lunar eclipses</span>
Answer:
C = 2.9 10⁻⁵ F = 29 μF
Explanation:
In this exercise we must use that the voltage is
V = i X
i = V/X
where X is the impedance of the system
in this case they ask us to treat the system as an RLC circuit in this case therefore the impedance is
X =
tells us to take inductance L = 0.
The angular velocity is
w = 2π f
the current is required to be half the current at high frequency.
Let's analyze the situation at high frequency (high angular velocity) the capacitive impedance is very small
→0 when w → ∞
therefore in this frequency regime
X₀ = 
the very small fraction for which we can despise it
X₀ = R
to halve the current at f = 200 H, from equation 1 we obtain
X = 2X₀
let's write the two equations of inductance
X₀ = R w → ∞
X= 2X₀ =
w = 2π 200
we solve the system
2R = \sqrt{R^2 +( \frac{1}{wC} )^2 }
4 R² = R² + 1 / (wC) ²
1 / (wC) ² = 3 R²
w C =
C =
let's calculate
C =
C = 2.9 10⁻⁵ F
C = 29 μF
a) uniform velocity
b) zero or no acceleration
c) (see picture)
EXPLANATION:
(see picture)
Answer:
Explanation:
The equilibrium equation is:

The initial concentration of A can be calculated from the ideal gas equation:

Determine the conversion of substance A to substance C using an ICE table and the Kc constant:
A ⇄ 3C
I 0.304 0
C - x + 3x
E 0.304 - x 3x

Solve for x:
You need to use a graphing calculator:
Then:

The equilbrium conversion is:
![\% = [0.304mol/liter-0.1845mol/liter]/(0.304mol/liter)\times 100](https://tex.z-dn.net/?f=%5C%25%20%3D%20%5B0.304mol%2Fliter-0.1845mol%2Fliter%5D%2F%280.304mol%2Fliter%29%5Ctimes%20100)

Answer:
Total work done is 2606.08 J.
Explanation:
Given :
Mass of box , m = 23 kg .
Force applied , F = 100 N .
Angle from horizon ,
.
Coefficient of kinetic friction ,
.
Distance travelled by box , d = 34 m .
Now ,
Total work done = work done by boy + work done by friction.
Hence , this is the required solution.