Answer:
0.074 V
Explanation:
Parameters given:
Number of turns, N = 121
Radius of coil, r = 2.85 cm = 0.0285 m
Time interval, dt = 0.179 s
Initial magnetic field strength, Bin = 55.1 mT = 0.0551 T
Final magnetic field strength, Bfin = 97.9 mT = 0.0979 T
Change in magnetic field strength,
dB = Bfin - Bin
= 0.0979 - 0.0551
dB = 0.0428 T
The magnitude of the average induced EMF in the coil is given as:
|Eavg| = |-N * A * dB/dt|
Where A is the area of the coil = pi * r² = 3.142 * 0.0285² = 0.00255 m²
Therefore:
|Eavg| = |-121 * 0.00255 * (0.0428/0.179)|
|Eavg| = |-0.074| V
|Eavg| = 0.074 V
Answer:
(a) 
(b) 
Solution:
According to the question:
Capacitance, C = 
Voltage of the battery, 
(a)The Energy stored in the Capacitor is given by:



(b) When the voltage of the battery is 6.00 V, the the energy stored in the capacitor is given by:



plasmas are a lot like gases
hope this helps.
The angular momentum is defined as,

Acording to this text we know for conservation of angular momentum that

Where
is initial momentum
is the final momentum
How there is a difference between the stick mass and the bug mass, we define that
Mass of the bug= m
Mass of the stick=10m
At the point 0 we have that,

Where l is the lenght of the stick which is also the perpendicular distance of the bug's velocity
vector from the point of reference (O), and ve is the velocity
At the end with the collition we have

Substituting




Applying conservative energy equation we have


Replacing the values and solving

Substituting
l=\frac{13}{0.54(9.8)}

A.)
False
<u>
</u><u>T</u>here reason why it's false is because Nitrogen is not responsible for clouds and precipitation. The real answer is that Water (Ocean / Other) is responsible for clouds and precipitation.
Good Day / Night :D