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xxMikexx [17]
3 years ago
8

Give one real-life example of the Doppler Effect and explain what causes this phenomenon. Please write at least 3 complete sente

nces, using correct punctuation and spelling. (12pts)
Physics
1 answer:
Elanso [62]3 years ago
7 0
Dipper effect of an oncoming train get louder as it approaches and sound diminishes as it goes away sound traveling
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Star A and Star B have different apparent brightnesses but identical luminosities. Star A is 10 light years away from earth and
STALIN [3.7K]

intensity of a star is inversely depends on the square of the distance from the star

we can say it is given as

\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}

here we know that

\frac{I_1}{I_2} = 36 times

also we know that

r_1 = 10 Ly

now we will have

\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}

36 = \frac{r_2^2}{10^2}

r_2 = 60 Ly

so other star is at distance 60 Light years

8 0
3 years ago
As a diligent physics student, you carry out physics experiments at every opportunity. At this opportunity, you carry a 1.33 m l
deff fn [24]

Answer: 62 μT

Explanation:

Given

Length of rod, l = 1.33 m

Velocity of rod, v = 3.19 m/s

Induced emf, e = 0.263*10^-3 V

Using Faraday's law, the induced emf of a rod can be gotten by the formula

e = blv where,

e = induced emf of the rod

b = magnetic field of the rod

l = length of the rod

v = velocity of the rod. On substituting, we have

0.263*10^-3 = b * 1.33 * 3.19

0.263*10^-3 = b * 4.2427

b = 0.263*10^-3 / 4.2427

b = 0.0000620 T

b = 62 μT

Thus, the strength of the magnetic field is 62 μT

8 0
3 years ago
Read 2 more answers
Two identical conducting spheres, A and B, carry equal charge. They are separated by a distance much larger than their diameters
Vesnalui [34]

Answer:

C. \frac{3F}{8}

Explanation:

Let initial charges on both spheres be,q

F=\frac{Kq^2}{d^2}   \ \ \  \ \ \  \ \ \  \ \_i

When the sphere C is touched by A, the final charges on both will be,\frac{q}{2}

#Now, when C is touched by B, the final charges on both of them will be:

q_c=q_d=\frac{q/2+q}{2}\\\\=\frac{3q}{4}\\

Now the force between A and B is calculated as:

F\prime=\frac{k\times\frac{q}{2}\times \frac{3q}{4}}{d^2}\\F\prime=\frac{3F}{8}

Hence the electrostatic force becomes 3F/8

4 0
2 years ago
A battery-operated car utilizes a 12.0 V system. Initially, the car is at rest at the base of a 195 m high hill. Some time later
ale4655 [162]

Answer:

140265.8 C = 1.403 × 10⁵ C

Explanation:

The battery's electric potential energy is used to account for the kinetic and potential work done in moving the car up this hill.

Potential work required to move the 757 kg car up a vertical height of 195 m = mgh

P.E = 757 × 9.8 × 195 = 1446627 J

Kinetic work done = (1/2)(m)(v²)

K.E = (1/2)(757)(25²) = 236562.5 J

Total work done in moving the car up that height = 1446627 + 236562.5 = 1683189.5 J

And this would be equal to the potential of the battery.

For the battery, potential difference = (electric potential energy)/(charges moved)

ΔV = ΔU/q

q = ΔU/ΔV

ΔU = 1683189.5 J

ΔV = 12.0 V

q = 1683189.5/12 = 140265.8 C

7 0
3 years ago
Read 2 more answers
A wave with a speed of 9 m/s and a frequency of 0.5 Hz has a λ of what?
patriot [66]

Wave speed = (wavelength) x (frequency)

Wavelength = (wave speed) / (frequency)

Wavelength = (9 m/s) / (0.5 Hz)

<em>Wavelength = 18 m</em>

6 0
2 years ago
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