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maxonik [38]
2 years ago
6

You are standing 10 meters from a light source. Then, you back away from the light source until you are 20 meters away from it.

From your perspective, what has happened to the intensity of the wave?
A. It has decreased by a factor of two.
B. It has increased by a factor of two.
C. It has decreased by a factor of four.
D. It has increased by a factor of four.
Physics
1 answer:
bogdanovich [222]2 years ago
5 0
Inverse Square Law of Light states that light intensity falls off rapidly with distance from its source. T<span>he intensity varies with the square of the flash-to-subject distance.
 intensity at distance 1/intensity at distance 2=distance2^2/distance1^2
In our case, distance1=10m and distance2=20m, 20^2/10^2= 400/100=4, which means that t</span>he intensity of the wave<span> has increased by a factor of four.
Answer:D</span>
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Explain how the forces need to change so the aeroplane can land
Fofino [41]
When an airplane is flying straight and level at a constant speed, the lift it produces balances its weight, and the thrust it produces balances its drag. However, this balance of forces changes as the airplane rises and descends, as it speeds up and slows down, and as it turns.
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a lady bug walks 10 cm forward then 5 cm backwards in 20 seconds. what is the average speed of the ladybug ?
igor_vitrenko [27]

A lady bug moves 10 cm forward and 5 cm backwards

so total distance moved by lady bug = 10 + 5 = 15 cm

total time taken by the lady bug

t =  20 s

so the average speed is given as

v = \frac{d}{t}

v = \frac{15}{20}

v = 0.75 cm/s

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2 years ago
A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate
diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

\Delta S = \frac{Q}{T}

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

W = Q_{source}-Q_{sink}

According to the data given we have to,

Q_{source} = 200000Btu

T_{source} = 1500R

Q_{sink} = 100000Btu

T_{sink} = 600R

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}

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On the other hand,

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The total change of entropy would be,

S = \Delta S_{source}+\Delta S_{sink}

S = -133.33+166.67

S = 33.34Btu/R

Since S\neq   0 the heat engine is not reversible.

PART B)

Work done by heat engine is given by

W=Q_{source}-Q_{sink}

W = 200000-100000

W = 100000 Btu

Therefore the work in the system is 100000Btu

4 0
2 years ago
Steam at 0.6 MPa, 200 oC, enters an insulated nozzle with a velocity of 50 m/s. It leaves at a pressure of 0.15 MPa and a veloci
Rudiy27

Answer:

x2 = 0.99

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at pressure p1 = 0.6MPa and temperature 200 degree celcius

h1 = 2850.6 kJ/kg

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\dot m( h1+\frac{v1^2}{2}+ gz1 )+ Q = \dot m( h2+\frac{v2^2}{2}+ gz1) + W

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at pressure p2 = 0.15MPa

specific enthalpy of fluid hf = 467.13 kJ/kg

enthalpy change hfg = 2226.0 kJ/kg

specific enthalpy of the saturated gas hg = 2693.1 kJ/kg

as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have

quality of steam x2

h2 = hf + x2(hfg)

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x2 = 0.99

6 0
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snow_tiger [21]

As we know that range of the projectile motion is given by

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here we know that range will be same for two different angles

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so it is given that one of the projection angle is 75 degree

so other angle for same range must be 90 - 75 = 15 degree

so other projection angle must be 15 degree

5 0
2 years ago
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