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sveticcg [70]
3 years ago
5

1.A Radio station broadcasts modern song on medium wave 350 Hz every day at ten o’clock in the morning. The velocity of radio wa

ve is 3X108 ms-1. The wavelength of another wave created in water is one percent of the radio wave and the velocity of sound in water is 1450 ms-1.
How many times of the frequency of the radio wave that of the wave created in the water? Analyze mathematically.
Physics
1 answer:
love history [14]3 years ago
8 0

Answer:

ans \:  = \boxed{{4.8 \times 10}^{ - 4}  Hz}

Explanation:

given \to \\  f_{r} = 350 \:  \\ v_{r} =  {3 \times 10}^{8}  \\ but \to \\ v = f \gamma   \to \:  \gamma  =  \frac{v}{f}  : hence \to \\  \gamma _{r} =  \frac{v_{r}}{f_{r}}   =  \frac{3 \times 10^{8} }{350}   =  \boxed{857,142.85714 \: m}\\ therefore \to \\ given \to \\  f_{w} = water \: frequency = \:  \boxed{  ?}\:  \\ v_{w} =  14 50 \\ but \to \\ v = f \gamma   \to \:  \gamma  =  \frac{v}{f}  : hence \to \\  \gamma _{w} =  \frac{v_{w}}{f_{w}}   =  \frac{1}{100}  \times \gamma _{r}  =  \frac{1}{100}  \times 857,142.85714  \\\gamma _{w}  =  \boxed{8,571.4285714 \: m} : hence \to \:  \\ f_{w} =  \frac{v_{w}}{ \gamma _{w}}  =  \frac{1450}{8,571.4285714}  =  \boxed{0.1691666667} \\ if \: the \: number \: of \: times = \boxed{ x} \\ f_{r} (x)=f_{w} \\ (x) =  \frac{f_{w}}{f_{r}}  =  \frac{0.1691666667}{350}  = 0.0004833333 \\ hence \to \\ the  \: frequency  \: of \:  the \:  radio  \: wave  \: is \to \:   \boxed{{4.8 \times 10}^{ - 4}  }\:  \\ that  \: of  \: the \:  wave  \: created  \: in  \: the  \: water.

♨Rage♨

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Answer:

The wavelength of the light is 555 nm.

Explanation:

according to Bragg's law..

n×λ = d×sin(θ)

n is the fringe number

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A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per un
Zarrin [17]

Answer:

steady state temperature =88.7deg C

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Explanation:

A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per unit length of 0.01 Ω/m. If a current of 100 A flows through the wire and the convection coefficient is 500W/m2K, what is the steady state temperature of the wire? From the time the current is applied, how long does it take for the wire to reach a temperature within 1-degC of the steady state value? The density of the wire is 8,000kg/m3, its heat capacity is 500 J/kgK and its thermal condu

The diameter of the wire is known to be=1mm

properties=

The density of the wire is 8,000 kg/m3,

heat capacity is 500 J/kgK

themal conductivity is 20W/m.K

electrical resistance per unit length of 0.01 Ω/m

from lump capavity method

B_{i} =\frac{hr/2}{k}

500*(2.5*10^-4)/20

0.006<0.1

we know also, to find steady state temperature

\piDh(T-Tinf)=I^{2} R_{e}

make T the subject of the equation , we have

T=25+\frac{100^2*0.01}{\pi*0.001*500 }

T=88.7 degC

rate of chnage in temperature

dT/dt=\frac{I^2*Re}{rho*c*\pi*D^2/4 } -\frac{4h}{rho*c*D} (T-Tinf)

at t=o and integrating both sides\frac{T-Tinf-(I^2*Re/\pi*Dh) }{Ti-Tinf-(I^2*Re/\pi*Dh } =exp\frac{-4ht}{rho*c*D}

we have

\frac{87.7-25-63.7}{25-25-63.7} =exp\frac{4*500t}{8000*500*0.001}

t=8.31s

steady state temperature =88.7deg C

t=time within  1 degC of it steady stae is 8.31s

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Explanation:

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2 years ago
How much work would have to be done to bring a 1150kg automobile traveling at 86km/h to a stop?​
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Explanation:

We have,

Mass of an automobile is 1150 kg

The automobile traveling at 86 km/h and then it comes to stop.

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Concept used : Work energy theorem

Th change in kinetic energy of an object is equal to the work done by it. The work done is then given by :

W=\dfrac{1}{2}m(v^2-u^2)

Here, v = 0

W=-\dfrac{1}{2}mu^2\\\\W=-\dfrac{1}{2}\times 1150\times (23.88)^2\\\\W=-327896.28\ J

or

W=3.27\times 10^5\ J

Therefore, the work done by the automobile is -3.27\times 10^5\ J.

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