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) i = -9.63 cm
, h ’= .0.24075 cm erect
b) i = 259.74 cm
,
Explanation:
For this exercise let's start by finding the focal length of the lens
1 / f = (n-1) (1 / R₁ - 1 / R₂)
1 / f = (1.70 -1)) 1 / ∞ - 1/13)
1 / f = 0.0538
f = - 18.57 cm
Now we can use the constructor equation
1 / f = 1 / o + 1 / i
1 / i = 1 / f - 1 / o
1 / i = -1 / 18.57 -1/20
1 / i = -0.1038 cm
I = -9.63 cm
For the height of the
image let's use magnification
m = h '/ h = - i / o
h ’= -h i / o
h ’= - 0.5 (-9.63) / 20
h ’= .0.24075 cm
b) we invert the lens
The focal length is
1 / f = (1.70 -1) (1/13 - 1 / int)
1 / f = 0.0538
f = 18.57 cm
1 / i = 1 / f -1 / o
1 / I = 1 / 18.57 - 1/20
1 / I = 3.85 10-3
i = 259.74 cm
h ’= - 0.5 259.74 / 20
h ’= 6.4935 cm
You will have to fly around the whole earth to get to your landing station
Answer is A because the speed and velocity would change. Think of it as GTA, your going 30+ miles per hour and you take a left turn, the speed and velocity would change in an instant..
Hope this helped.
