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AlladinOne [14]
3 years ago
15

Monochromatic light from a point source is directed toward the sharp edge of a solid object. How does diffraction change the ima

ge that appears on a screen beyond the object?
Why are diffraction patterns not commonly observed in everyday situations?
Physics
2 answers:
Anna71 [15]3 years ago
6 0

Did you get the answers the quiz??? If so please write out the answers in the comment section :)

Otrada [13]3 years ago
5 0

Monochromatic light from a point source is directed toward the sharp edge of a solid object.

  • A spot will be seen on the screen after diffraction.
  • The size of the obstructing object is greater than the wavelength of the incident light in everyday situations, so we are not able to view diffraction patterns in everyday situations.

<u>Explanation:</u>

When a monochromatic light from a point source is directed toward the sharp edge of a solid object, the light beam will get diffracted from the edge as the body of the solid object will not be passing the light. So the incident light will occur as a spots on the screen beyond the object after diffraction.

Diffraction of light occurs when the incident light falls on a grater and the width of the grater or the obstructing object should be comparable or lesser to the size of the wavelength. So, in everyday situations we do not observe diffraction patterns as the obstructing objects in real world.

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A unit mass of an ideal gas at temperature T undergoes a reversible isothermal process from pressure P1 to pressure P2 while los
mafiozo [28]

Answer:

Option B

Change in entropy of the process is \Delta S= Rln(\frac{P_{1}}{P_{2}})

Explanation:

The entropy of a system is a measure of the degree of disorderliness of the system.

The entropy of a system moving from process 1 to 2 is given as

\Delta S = \int\limits^2_1 {\frac{\delta q}{T}}

recall from first law, \delta q =du +Pdv

hence we have, \Delta S = \int\limits^2_1 {\frac{du +Pdv}{T}}

since the process is isothermal, du= 0

this gives us \Delta S = \int\limits^2_1 {\frac{Pdv}{T}}

integrating within the limits of 1 and 2, will give us

\Delta S = R ln (\frac{V_{2}}{V_{1}})

also from ideal gas laws,

\frac{V_{2}}{V_{1}}=\frac{P_{1}}{P_{2}}

hence we have    \Delta S = R ln (\frac{P_{1}}{P_{2}})

This makes the correct option B

4 0
3 years ago
Read 2 more answers
A water discharge of 9 m3/s is to flow through this horizontal pipe, which is 0.98 m in diameter. If the head loss is given as 1
Mashutka [201]

Answer: The power required by the pump to produce a discharge of 9m³/s is 5990joules/secs.

Explanation: The given parameters from the questions are:

Flow rate Q = 9m³/s, Diameter D = 0.98m, acceleration a = 1.0, head loss(Pressure P) is given by the function 10v²/2g.

STEP 1. Find the velocity of water in the pipe from the equation:

Diameter D = (√4.Q/π.v), where v is the velocity, and Q is flow rate

Making v subject of the formula gives:

v = 4Q/π.√D =[ 4 × 9m³/s / 3.142 × (√0.98m)] = 11.69m/s.

STEP 2. Find the pressure from the relationship, P = 10v²/2g, NB. g = a

P = 10 × (11.69m/s)² / 2× 1.0m/s²

P = 683.25N/m² or Pascal.

STEP 3. Find force exerted by the pump;

Recall that Pressure P = Force/Area

But Area A = π.r², where r = D/2

Therefore, A = π.(D/2)²

A = 3.142 × [0.98m/2]² = 0.75m²

Therefore, Force = Pressure × Area

Force F = 683.25N/m² × 0.75m²

F = 512.44N.

STEP 4. Find work done

Work done W by the pump is = Force × distance d moved by the water

W = F . d

Also recall that flow rate Q = Velocity/time.

Q = v/t, we can write t = v/Q.

Time t = 11.69m/s / 9m³/s = 1.298s

Also recall that velocity v = distance d/time t, v = d/t, making d subject of formula gives v × t

Distance d = v × t = 11.69m/s × 1.298s = 15.17m.

Hence,

Work Done W = Force × distance

W = 512.44N × 15.17m = 7775.56Nm or joules.

Lastly, Power P = Work done/ time

P = 7775.56joules/1.298s

P = 5990.4joules/s.

8 0
2 years ago
Question on Resistance. WORTH 20 POINT
kirill [66]

From Ohm's law:  R = V / I

Resistance = (voltage) / (current)

The first paragraph TELLS you that the current is always 0.5 A, and the table tells you the voltage across each piece of wire.

Again . . .  <em>R = V / I</em>

6 0
2 years ago
Read 2 more answers
to what circuit element is an ideal inductor equivalent for circuits with constant currents and voltages?
Marat540 [252]

Answer:

Short circuit

Explanation:

In an ideal inductor circuit with constant current and voltage, it implies that the voltage drop in the circuit is zero (0).

Also, In circuit analysis, a short circuit is defined as a connection between two nodes that forces them to be at the same voltage.

In an ideal short circuit, this means there is no resistance and thus no voltage drop across the connection. That is voltage drop is zero (0).

Therefore, the circuit element is short circuit.

8 0
2 years ago
Ice fishermen sit on top of frozen lakes in the winter and catch fish in the liquid water below through holes cut in the ice she
melomori [17]

Answer:

This is because below 4°c, water unlike other materials becomes less dense when it's temperature is further lowered.

Explanation:

Due to the unusual nature of water; at about 4°c, the behavior of the density of water in relation to its temperature reverses. This means that water becomes less dense as it becomes colder below 4°c. The colder parts therefore floats to the top of the water body while the warmer part sinks allowing the top to freeze and the remaining body below to remain in its liquid state.

The freezing of the top of the lake alone protects the remaining depth of water from freezing by acting as an insulator and preventing further heat loss from the water to the ambient space. If this had not been the case, and water froze all through, marine lives will freeze to death and it will be more difficult to melt the ice come the next summer.

This behavior is due to the hydrogen bonding of the water molecules.

8 0
2 years ago
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