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Sloan [31]
4 years ago
8

The state highway patrol radar guns use a frequency of 8.66 GHz. If you're approaching a speed trap driving 35.0 m/s, what frequ

ency shift will your FuzzFoiler 2000 radar detector see
Physics
1 answer:
otez555 [7]4 years ago
3 0

Answer:

The frequency shift detected is \Delta  f  =1010.3 Hz

Explanation:

From the question we are told that

    The normal frequency of the state highway patrol radar guns f = 8.66 GHz =   8.66 *10^{9} \ Hz

     The speed of approach is v  =  35 .0 \ m/s

       

Now the frequency measure by the the state highway patrol radar guns as your car approaches is mathematically represented as

        f_n  =  f (1 + \frac{v}{c} )

Where c is the speed of light which a has constant value of

     c =  3.0 *10^{8 } \ m/s

  Now

       f_n - f  =  \frac{fv}{c} )

=>    \Delta  f  =   \frac{fv}{c}

substituting values

        \Delta  f  = \frac{ 8.66 *10^{9} *  35}{3.0 *10^8 }

        \Delta  f  =1010.3 Hz

 

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sin(\theta_{min}) = \frac{m\lambda}{D}

Here,

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D = Distance between slits

\lambda = Wavelength

Replacing with our values we have,

sin(\theta_{min}) = \frac{(1)(737*10^{-9})}{71.7*10^{-6}m}

sin(\theta_{min}) = 0.01028

Through the relationship between distances then we have that the basic amplification distance is given by the relationship between the distance of the slit L and the angle, then

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6 0
4 years ago
An electric generator converts kinetic energy into which kind of energy?
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Which of the following was NOT a department in Washington's first cabinet?
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Explanation:

6 0
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A spherical balloon has a radius of 7.15 m and is filled with helium. The density of helium is 0.179 kg/m^3, and the density of
tekilochka [14]

The largest mass of cargo the balloon can lift is 791.06 kg

First, we need to calculate the mass of helium.

Since the radius of the spherical balloon is r = 7.15 m, its volume is V = 4πr³/3.

The volume of the balloon also equals the volume of helium present.

Now, the mass of helium m = density of helium, ρ × volume of helium, V

m = ρV

Since ρ = 0.179 kg/m³

m = ρV

m = ρ4πr³/3.

m = 0.179 kg/m³ × 4π(7.15 m)³/3

m = 0.179 kg/m³ × 4π(365.525875 m³)/3

m = 0.179 kg/m³ × 1462.1035π m³/3

m = 261.7165265π/3 kg

m = 822.207/3 kg

m = 274.07 kg

Since the mass of the skin and structure of the balloon is 910 kg, the total mass, M of the balloon = mass of skin and structure + mass of helium gas is 910 kg + 274.07 kg = 1184.07 kg.

The weight of this mass W = Mg where g = acceleration due to gravity.

The buoyant force on the balloon due to the air is the weight of air displaced, W' = mass of air, m' × acceleration due to gravity, g.

W' = m'g

Now, the mass of air m' = density of air, ρ' × volume of air displaced, V'

We know that the volume of air displaced, V' = volume of balloon, V

So, V' = V = 4πr³/3.

Since the density of air, ρ' = 1.29 kg/m³,

m' = ρ'V

m = 1.29 kg/m³ × 4π(7.15 m)³/3

m = 1.29 kg/m³ × 4π(365.525875 m³)/3

m = 1.29 kg/m³ × 1462.1035π m³/3

m = 1886.113515π/3 kg

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So, the net weight W" that the balloon can lift is W" = W' - W = m'g - Mg = (m' - M )g = (1975.13 kg - 1184.07 kg)g = 791.06g.

So, the net mass m" = W"/g = 791.06g/g = 791.06 kg

This net mass is the largest mass of cargo that the balloon can lift.

Thus, the largest mass of cargo the balloon can lift is 791.06 kg

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