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We are able to see the color whose visible ray falls into our eyes. The rays from the sun get scattered by small dust particles or water droplets or other molecules in the air . Blue light gets scattered the most because of its shorter wavelength. So, even though rays ofndifferent colors fall into our eyes.. blue rays are lot more in no. So the optical nerve of our eye recieves the signal for blue the most.

Given, the actual path length travelled is 23 km, displacement is 10 km and the time taken is 28 min.

(a)

Average speed of the taxi is given as,

v avg = Total Path length Time

Here, s is the total path length

Substitute the values in above equation.

v avg = Total Path length Time = 23 km ( 28 min )( 1 h 60 min ) =49.3  km/h

Hence, the average speed of the taxi is 49.3  km/h .

(b)

Magnitude of the average velocity is given as,

v avg = Displacement Time

Substitute the values in the above equation.

v avg = Displacement Time = 10 km ( 28 min )( 1 h 60 min ) =21.4  km/h

(c)

From the calculation in (a) and (b), it is obtained that the average speed is not equal to the magnitude of average velocity.

Here, Volume of water =30 m3
t=15 min=15×60 s=900 s;Height,h=40 m
Efficiency, η=30%
Density of water =103kg m−3
∴ Mass of water pumped =Volume×Density
=(30 m3)(103 kg m−3)=3×104kg
Poutput=Wt=mght=(3×104kg)(10 ms−2)(40 m)900 s
=43×104W
Efficiency, η=PoutputPinput
Pinput=Poutputη=4×1043×30100=49×105
=44.4×103W=44.4 kW.

Linear momentum of particle, →p=px^i+py^j+pz^k

Mass of the big structure, M = 50,000 kg
Inner radius of the column, r = 30 cm = 0.3 m
Outer radius of the column, R = 60 cm = 0.6 m
Young’s modulus of steel, Y=2×1011Pa

Total force exerted, F=Mg=50000×9.8N

Stress = Force exerted on a single column =50000×9.84=122500N
Young’s modulus, Y=StressStrain

Strain =(F/A)Y
Where,
Area, A=π(R2−r2)=π((0.6)2−(0.3)2)

Strain =122500/[π((0.6)2−(0.3)2)×2×1011]=7.22×10−7

Hence, the compressional strain of each column is 7.22×10−7.

Water is flowing at a rate of 3.0 litre/min.
The geyser heats the water, raising the temperature from 27oC to 77oC.
Initial temperature, T1=27oC
Final temperature, T2=77oC
Rise in temperature,T=T1−T2=77−27=50oC
Heat of combustion =4×104J/g
Specific heat of water, C=4.2J/goC
Mass of flowing water, m= 3.0 litre/min = 3000 g/min
Total heat used, Q=mcT
=3000×4.2×50
=6.3×105J/min
Rate of consumption =6.3×105/(4×104) = 15.75 g/min.