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Sedbober [7]
2 years ago
13

A 75,000-watt radio station transmits at 88 MHz. Determine the number of joules transmitted per second. 88 J/s 8800 J/s 2,000,00

0 J/s 75,000 J/s
Physics
1 answer:
Free_Kalibri [48]2 years ago
6 0

Answer:

the first one is the correct answer

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TRUE OR FALSE: The following drops were most likely dropped from a 90 degree angle.
horsena [70]

Answer:

True

Explanation:

If it weren't from a 90 degree angle then the circle would be a bit more oval shaped

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3 years ago
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Why is the H-R diagram useful in plotting the life cycles of stars?
Roman55 [17]
The sun provides a handy benchmark for describing other stars. The mass of this solar system's sun gives us a unit for measuring other stars' masses.
4 0
4 years ago
A square coil with a side length of 16.0 cm and 29 turns is positioned in a region with a horizontally directed, spatially unifo
tensa zangetsu [6.8K]

Answer:

A

  \epsilon_{max}  = 0.774 \  V

B

wt  =  0^o

Explanation:

From the question we are told that

   The  length of the side is  l  =  16.0 \ cm  = 0.16 \ m

    The  number of turns is  N = 29  \ turns

    The  magnetic field is  B =  83.0 mT  =  83 *10^{-3} \ T

    The  angular speed is  w  =  1.20 * 10^2  rev/min =  \frac{1.20 *10^2 *  2\pi}{60 } = 12.6 \  rad/s

Generally  the area is  A =  l^2

Generally the induced emf is mathematically represented as

         \epsilon  =  N * w  *  B  *  A  * cos(wt)

At maximum  cos(wt) =  1

    So

       \epsilon_{max}  =  N * w  *  B  *  A

       \epsilon_{max}  = 29  *  12.6 *  83*10^{-3}* (l^2)

=>   \epsilon_{max}  = 29  *  12.6 *  83*10^{-3}* ((0.16)^2)

=>  \epsilon_{max}  = 0.774 \  V

At maximum emf

      cos (wt)  =  1

=>   (wt)  =  cos^{-1} (1)

=>   wt  =  0^o

6 0
3 years ago
In a double slit experiment, the intensity of light at the center of the central bright fringe is measured to be 6.2 µW/m2. What
-BARSIC- [3]

Answer:

   I_FWHW = 3.2 μW / m²

Explanation:

In the analysis of optics and electricity a very useful magnitude is the width at half height (FWHW) and the intensity at this height, which is given by

               I_FWHW = I₀ / 2

corresponds to the width of the line for this intensity.

In this case they give the maximum intensity for which

               I_FWHW = 6.2 / 2

               I_FWHW = 3.2 μW / m²

You do not give more data in your exercise, but the most interesting calculation is to find the angle values ​​for which you have this intensity since it is this range is 50% of the energy of the system, have I write the equation for this calculation

             I = Io cos² x₁   (sin x / x)²

             x₁ = π d sin θ /λ

             x = π b sin θ /λ

where d is the separation of the slits and b the width of each slit

5 0
4 years ago
1. Which statement best describes the motion of the scooter? *
salantis [7]

Answer:

HI

Explanation:

its E

4 0
3 years ago
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