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pickupchik [31]

2 years ago

10

Two beams of coherent light are shining on the same piece of white paper. with respect to the crests and troughs of such waves,

darkness will occur on the paper where

1 answer:

defon2 years ago

7 0

Answer:

"where crests and troughs have their maxima at the same time"

Crests and troughs are 180 deg out of phase and when they have their maxima at the same time and place, their net contribution will be zero"

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A 42 Watt light bulb is connected to a 12 volt source of potential

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Answer:

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Explanation:

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Describing a Physical Change

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Answer:

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3 years ago

Why does sound travel slower at a cold temperature?

tatyana61 [14]

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7 0

3 years ago

Read 2 more answers

Block A is accelerating with Block B at a rate of 0.800 m/s2 along a frictionless surface. It suddenly encounters a surface that

Natalija [7]

Answer:

$a=0.6\ m/s^2$

Explanation:

The attached figure shows the whole description. Considering the applied force is 100 N.

The acceleration of both blocks A and B, $a=0.8\ m/s^2$

Firstly calculating the mass m using the second law of motion as :

F = ma

m is the mass

$m=\dfrac{F}{a}$

$m=\dfrac{100\ N}{0.8\ m/s^2}$

m = 125 kg

It suddenly encounters a surface that supplies 25.0 N a friction, F' = 25 N

$(F-F')=ma$

$(100-25)=125\times a$

$a=\dfrac{75}{125}=0.6\ m/s^2$

So, the new acceleration of the block is $0.6\ m/s^2$ . Hence, this is the required solution.

7 0

3 years ago

A parallel-plate capacitor has square plates that are 8.00cm on each side and 4.20mm apart. The space between the plates is comp

Sergeu [11.5K]

Answer:

$U=1.29\times 10^{-7}\ J$

Explanation:

Given that

a= 8 cm (square)

A= a ² = 64 cm²

d= 4.2 mm

d₁= 2.1 mm ,K₁= 4.7

d₂=2.1 mm , K₂=2.6

We know that capacitance given as

$C_1=\dfrac{K_1\varepsilon _oA}{d_1}$

$C_1=\dfrac{4.7\times 8.85\times 10^{-12}\times 64\times 10^{-4}}{2.1\times 10^{-3}}$

$C_1=1.26\times 10^{-10}\ F$

$C_2=\dfrac{K_2\varepsilon _oA}{d_2}$

$C_2=\dfrac{2.6\times 8.85\times 10^{-12}\times 64\times 10^{-4}}{2.1\times 10^{-3}}$

$C_2=0.701\times 10^{-10}\ F$

Net capacitance

$C=\dfrac{C_1C_2}{C_1+C_2}$

$C=\dfrac{1.26\times 10^{-10}\times 0.701\times 10^{-10}}{1.26\times 10^{-10}+0.701\times 10^{-10}}\ F$

$C=4.5\times 10^{-11}\ F$

We know that stored energy given as

$U=\dfrac{CV^2}{2}$

V= 76 V

$U=\dfrac{4.5\times 10^{-11}\times 76^2}{2}\ J$

$U=1.29\times 10^{-7}\ J$

3 0

3 years ago