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zvonat [6]
4 years ago
13

Reflections from a thin layer of air between two glass plates cause constructive interference for a particular wavelength of lig

ht λ. By how much must the thickness of this layer be increased for the interference to be destructive?
A. λ/8
B. λ/4
C. λ/2
D. λ
Physics
1 answer:
gavmur [86]4 years ago
4 0

Answer:

C

Explanation:

Thin film interference is most constructive or most destructive when the path length difference for the two rays is an integral or half-integral wavelength, respectively. That is, for rays incident perpendicularly, 2t = λn, 2λn, 3λn, . . . or  2t = λn/2, 3λn/2, 5λn/2 , .....

We see a constant phase shift off λn/2. Hence option C

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A rubber rod rubbed with fur acquires a charge of -4,8×10(-9)C. What is the charge on the fur? How much mass is transferred to r
mote1985 [20]
1) The charge left on the fur is equal and opposite to the charge transferred to the rod:
Q=+4.8 \cdot 10^{9} C
In fact, when the rod is rubbed with the fur, a net charge of Q=-4.8 \cdot 10^{-9} C has been transferred to the rod, leaving it negatively charged. If we assume the fur was initially neutral, this means that we have now an excess of positive charges on the fur, and the amount of this charge must be equal (in magnitude, but with opposite sign) to the charge transferred to the rod.

2) The mass transferred to the rod is equal to the total mass of the electrons transferred to the rod.
The charge transferred to the rod is
Q=-4.8 \cdot 10^{-9} C
The charge of 1 electron is
e=-1.6 \cdot 10^{-19} C
So the number of electrons transferred is
N= \frac{Q}{e}= \frac{-4.8 \cdot 10^{-9} C}{-1.6 \cdot 10^{-19} C}=3.0 \cdot 10^{10}

The mass of 1 electron is m=9.1 \cdot 10^{-31} kg, therefore the total mass transferred to the rod is
M=Nm=(3 \cdot 10^{10})(9.1 \cdot 10^{-31} kg)=2.73 \cdot 10^{-20} kg

7 0
3 years ago
Air breaks down and conducts charge as a spark if the electric field magnitude exceeds V/m. Determine the maximum charge that ca
Ymorist [56]

Answer:

Explanation:

Suppose

Magnitude of Electric Field is E V/m

Area of the cross-section is A

capacitor C=\frac{\epsilon A}{d}

Distance between Area of capacitor is d

Maximum Charge stored is

Q_{max}=capacitor\times Potential\ Difference

Potential\ Difference=electric\ Field\times distance=E\times d

Q_{max}=C\times Ed=CEd

Q_{max}=\frac{\epsilon _0A}{d}\times Ed

Q_{max}=\epsilon _0AE

 

3 0
4 years ago
A flat coil of wire consisting of 20 turns, each with an area of 50 cm2, is positioned perpendicularly to a uniform magnetic fie
Scilla [17]

Answer:

Induced current, I = 0.5 A

Explanation:

It is given that,

number of turns, N = 20

Area of wire, A=50\ cm^2=0.005\ m^2

Initial magnetic field, B_i=2\ T

Final magnetic field, B_f=6\ T

Time taken, t = 2 s

Resistance of the coil, R = 0.4 ohms

We know that due to change in magnetic field and emf will be induced in the coil. Its formula is given by :

\epsilon=\dfrac{-d\phi}{dt}

Where

\phi=BA

\epsilon=\dfrac{-d(NBA)}{dt}

\epsilon=NA\dfrac{B_f-B_i}{t}

\epsilon=20\times 0.005\times \dfrac{6-2}{2}

\epsilon=0.2\ V

Let I is the induced current in the wire. It can be calculated using Ohm's law as :

\epsilon=I\times R

I=\dfrac{\epsilon}{R}

I=\dfrac{0.2}{0.4}

I = 0.5 A

So, the magnitude of the induced current in the coil is 0.5 A. Hence, this is the required solution.

5 0
3 years ago
How to find acceleration?
Marat540 [252]

Answer:

a=(v-u)/t is how you find acceleration.

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elena55 [62]
Liquid = B
plasma = C
solid = A
gas = D
8 0
3 years ago
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