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Sonja [21]
3 years ago
14

The spreading of waves behind an aperture ismore for long wavelengths and less for short wavelengths.less for long wavelengths a

nd more for short wavelengths.the same for long and short wavelengths.not discussed in this chapter.
Physics
1 answer:
kompoz [17]3 years ago
6 0
<h2>Answer: The spreading of waves behind an aperture ismore for long wavelengths and less for short wavelengths</h2>

Here we are talking about Diffraction and, in fact, waves diffract the most when their wavelength is about the same size of the gap or aperture.

Diffraction happens when a wave (mechanical or electromagnetic wave) meets an obstacle or a slit .When this occurs, <u>the wave bends around the corners of the obstacle or passes through the opening of the slit that acts as an obstacle, forming multiple patterns with the shape of the aperture of the slit. </u>

<u />

Note that the principal condition for the occurrence of this phenomena is that the obstacle must be comparable in size (similar size) to the size of the wavelength.

In other words, when the gap (or slit) size is larger than the wavelength, the wave passes through the gap and does not spread out much on the other side, but when the gap size is equal to the wavelength, maximum diffraction occurs and the waves spread out greatly.

This means the smaller the slit or obstacle (diffracting object), the wider the resulting diffraction pattern, and the greater the obstacle, the narrower de resulting patter.

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The potential at location A is 382 V. A positively charged particle is released there from rest and arrives at location B with a
jarptica [38.1K]

Answer: 247.67 V

Explanation:

Given

Potential At A V_a=382\ V

Potential at V_c=785\ V

when particle starts from A it reaches with velocity v_b at Point while when it starts from C it reaches at point B with velocity 2v_b

Suppose m is the mass of Particle

Change in Kinetic Energy of particle moving under the Potential From A to B

q\cdot \left ( V_a-V_b\right )=0.5m\cdot (v_b)^2----1

Change in Kinetic Energy of particle moving under the Potential From C to B

q\cdot \left ( V_c-V_b\right )=0.5m\cdot (2v_b)^2-----2

Divide 1 and 2 we get

\frac{V_a-V_b}{V_c-V_b}=\frac{v_b^2}{4v_b^2}

on solving we get

V_b=\frac{4}{3}\cdot V_a-\frac{1}{3}\cdot V_c

V_b=\frac{743}{3}=247.67\ V

                     

4 0
3 years ago
What is nuclear fission? (Points : 1)
mixas84 [53]
The splitting of the atomic nucleus into parts
4 0
3 years ago
Read 2 more answers
How is the percent efficiency of a machine determined?
NemiM [27]

Since work is the change in kinetic energy, the efficiency of a machine can be stated as the percentage of the output work divided by the input work minus the work lost from to friction and heat. Multiply Eff by 100% to get the efficiency percentage.

3 0
3 years ago
A capacitor is charged until its stored energy is 7.54 J. A second capacitor is then connected to it in parallel. If the charge
Ivan

Answer:

2 J

Explanation:

A charged capacitor of capacitance C_1 with energy of 7.54 J, is connected in parallel with another capacitor C_2 , so the charge is equally distributed between them.

(a) The energy stored in the capacitor before it being connected to the other capacitor is:

U_O=q_0^2/2C_1=7.54 J\\

The energy stored in the electric field is the sum of the energies of the two capacitors:

U=U_1+U_2\\U=q_1^2/2C_1+q_2^2/2C_2

since the charge equally distributed,  q_1 = q_2 = q_o/2. and since they are connected in parallel the potential difference on both of them is the same :

V_1=V_2\\q_1/C_1=q_2/C_2\\q_0/C_1=q_0/C_2\\C_1=C_2=C_3\\

hence,

U=q_0^2/8C+q_0^2/8C\\U=q_0^2/4C\\U=2J

8 0
2 years ago
A student drops a ball from the top of a 10-meter tall building. The ball leaves the thrower's hand with a zero speed. What is t
Sergio [31]

Answer:

14 m/s

Explanation:

u = 0, h = 10 m, g = 9.8 m/s^2

Use third equation of motion

v^2 = u^2 + 2 g h

Here, v be the velocity of ball as it just strikes with the ground

v^2 = 0 + 2 x 9.8 x 10

v^2 = 196

v = 14 m/s

7 0
2 years ago
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