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blsea [12.9K]
2 years ago
8

What type of modulation is typically used by broadcasting stations to transmit pictures on television screens?

Physics
2 answers:
umka21 [38]2 years ago
4 0
Analog

Television transmitters use one of two different technologies: analog, in which the picture and sound are transmitted by analog signals modulated onto the radio carrier wave, and digital in which the picture and sound are transmitted by digital signals.
Naddika [18.5K]2 years ago
4 0

Answer:

A. Amplitude

Explanation:

The type of modulation that is typically used by broadcasting stations to transmit pictures on television screens is amplitude. Therefore, the answer to your question is A.

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If you push a 4-kg mass...
Lena [83]

Answer:

B

Explanation:

F = ma , a = F/m

a1 = F/10 and a2 = F/4

Since Force is constant, a2 will we greater than a1

4 0
2 years ago
A standing wave pattern is created on a string with mass density μ = 3.4 × 10-4 kg/m. A wave generator with frequency f = 61 Hz
uranmaximum [27]

Answer:

1) λ = 0.413 m , 2)v = 25,213 m / s , 3)  T = 0.216 N , 4) m = 22.04 10-3 kg

Explanation:

1) The resonance occurs when the traveling wave bounces at the ends and the two waves are added, the ends as they are fixed have a node, the wavelength and the length of the string are related

         λ = 2L / n               n = 1, 2, 3 ...

In this case L = 0.62 m and n = 3

Let's calculate

        λ = 2 0.62 / 3

        λ = 0.413 m

2) the velocity related to wavelength and frequency

      v =  λ f

      v = 0.413 61

      v = 25,213 m / s

3) let's use the equation

     v = √T /μ

     T = v² μ

     T = 25,213² 3.4 10⁻⁴

     T = 0.216 N

4) the rope tension is proportional to the hanging weight

      T-W = 0

     T = W

    W = m g

    m = W / g

    m = 0.216 / 9.8

    m = 22.04 10-3 kg

5) n = 2

     λ = 2 0.62 / 2

     λ = 0.62 m

6) v =  λ f

     v = 0.62 61

     v = 37.82 m / s

7) T = v² μ

   T = 37.82² 3.4 10⁻⁴

   T = 0.486 N

8) m = W / g

   m = 0.486 / 9.8

   m = 49.62 10⁻³ kg

9) n = 1

    λ = 2 0.62

    λ = 1.24 m

    v = 1.24 61

    v = 75.64 m / s

    T = v² miu

    T = 75.64² 3.4 10⁻⁴

    T = 2.572 10⁻² N

    m = 2.572 10⁻² / 9.8

    m = 262.4 10⁻³ kg

5 0
3 years ago
An archer pulls back the string of a bow to release an arrow at a target. Which kind of potential energy is transformed to cause
Ugo [173]
Elastic Potential Energy because the elasticity in the string stores up the energy.
3 0
3 years ago
Read 2 more answers
When does gravitational lensing occur? High concentrations of dark matter cause length contraction of nearby objects. The gravit
Stella [2.4K]

Answer:

A massive object (like a galaxy cluster) bends the light from an object (like a quasar) that lies behind it.

Explanation:

A massive object, like a galaxy cluster, is able to deform the space-time shape as a consequence of its own gravity, so the light that it is coming from a source that is behind it in the line of sight will be bend or distorts in a way that will be magnified, making small arcs around the cluster with the image of the background object.

This technique is useful for astronomers since they make research of faraway objects (at hight redshift) that otherwise will difficult to detect with a telescope.

4 0
3 years ago
A fish takes the bait and pulls on the line with a force of 2.3 N. The fishing reel, which rotates without friction, is a unifor
natulia [17]

Explanation:

The given data is as follows.

     Pulling force on the reel is F = T = 2.3 N

     Mass of cylinder (m) = 0.82 kg

    Radius of cylinder (r) = 0.045 m

Formula for torque pulling force on the cylinder is as follows.

           \tau = Fr

                     = I \times \alpha

Moment of inertia of the cylinder (I) is as follows.

            I = \frac{mr^{2}}{2}

  \alpha = angular acceleration of the cylinder

Hence,

                    Fr = (\frac{mr^{2}}{2}) \alpha

or,   \alpha = \frac{2F}{mr}

                   = \frac{2 \times 2.3 N}{0.82 kg \times 0.045 m}

                   = 124.66 rad/s^{2}

Amount of line pulled is h and it is in a times of 0.2 sec.

Now, linear acceleration is calculated as follows.

             a = r \alpha

                = 0.045 m \times 124.66 rad/s^{2}

                 = 5.61 m/s^{2}

Now, relation between h and acceleration as follows.

             h = v_{o}t + \frac{1}{2}at^{2}

here,   v_{o} = 0

Hence, calculate the value of h as follows.

             h = v_{o}t + \frac{1}{2}at^{2}

                = 0 + \frac{1}{2} \times 5.61 m/s^{2} \times (0.2s)^{2}

                = 0.1121 m

Thus, we can conclude that acceleration of the fishing reel is 5.61 m/s^{2} and it will pull 5.61 m/s^{2} line from the reel in 0.20 s.

7 0
3 years ago
Read 2 more answers
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