<h2>
Answer:</h2>
1.8 x 10⁻⁵J
<h2>
Explanation:</h2>
The energy (E) stored in a capacitor of capacitance, C, when a voltage, V, is supplied is given by;
E = 
x C x V² -------------------(i)
Now, from the question;
C = 2.00μF = 2.00 x 10⁻⁶F
V = 18.0V
Substitute these values into equation (i) as follows;
E = x 2.00 x 10⁻⁶ x 18.0
E = 1.8 x 10⁻⁵J
Therefore, the quantity of energy stored in the capacitor is 1.8 x 10⁻⁵J
Answer:
a)0.024
b)0.148
Explanation:
Let 's represent the set of deer ticks Carrying Lyme disease with L and the set of deer ticks carrying Human Granulocytic Ehrlichiosis with H
Given:
P(L) = 0.16
P(H) = 0.10
P(L n H) = 0.1 ·P( L u H )
Hence, P( L u H) = 10 ·P( L nH)
(a)
Hence. using the equation. P(L U H) = P(L) + P(H) - P(L n H)
Hence, 10 · P(L n H ) = 0.16 + 0.1 - P(L n H )
Hence, 11 · P(L n H) = 0.16 + 0.1 = 0.26
Hence, P(L n H) =
0.26/11=0.024
(b)
We know that condition probability P(H ║ L) = p(L n H)/P(L)
hence, P(H ║ L) =(0.26/11)/0.16 =0.148
Answer: Frequency= 0.5 Hz
Period= 2s
Explanation:
2 sec= 0.5 Hz
0.5Hz= 2 sec
Answer:
See explanations
Explanation:
Given that,
Electric field E=135V/m
Energy stored in 1m³of air=?
The energy stored in an electric field is given as
u = ½ εo E²
Where
U is the energy stored
εo is permissivity and it value is 8.85×10^-12C²/N..m²
And E is the electric field
Then,
U=½×8.85×10^-12×135²
U=8.06×10^-8J/m³
Then, the energy stored in 1m³ of air is 8.06×10^-8 J/m³
Acceleration = 75/3 = 25 m/s^2
