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lapo4ka [179]
3 years ago
10

Are electric files lines due to an isolated positive point charge emanate radially away from the charge?

Physics
1 answer:
lara31 [8.8K]3 years ago
4 0

Answer:

6

Explanation:

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In Milgram's experiment, compliance, or doing what the experimenter asked,
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In Milgram's experiment, compliance, or doing what the experimenter asked,

the teacher and the learner were in the same room. -C.

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How can a gas become a good conductor? Simpler answers would be helpful.
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A gas has to become ionisied in order to become a conductor. It must have a chain reaction in which atoms in it became unstable, in which they loose stabile electronic configuration. In order for a gas to become a conductor, it must have free particles, and it can happen only in ionisied gas.
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A plant cell is no longer capable of capturing energy from sunlight and converting it into chemical energy. Which organelle is m
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8 0
1 year ago
Describe a technology used in space exploration.
Alex777 [14]

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High speed optical communication technology

To be able to communicate from the space to the earth and from earth to space is one of the most essential features required during space exploration.

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The key to analyse the studies and observations is being able to communicate the data collected, photos taken etc to the launch centers or space centers on earth. The space centers on earth should also be able to communicate with the persons or the satellites in space.

This is made possible using the optical communication technology which involves the use of optical fibers, lasers etc, since high speeds are more efficient during communication

3 0
3 years ago
A transverse wave on a string is described by the following wave function.y = (0.090 m) sin (px/11 + 4pt)(a) Determine the trans
alukav5142 [94]

Explanation:

(a) It is known that equation for transverse wave is given as follows.

                 y = (0.09 m)sin(\pi \frac{x}{11} + 4 \pi t)

Now, we will compare above equation with the standard form of transeverse wave equation,

                 y = A sin(kx + \omega t)

where,    A is the amplitude = 0.09 m

              k is the wave vector = \frac{\pi}{11}

              \omega is the angular frequency = 4\pi

              x is displacement = 1.40 m

              t is the time = 0.16 s

Now, we will differentiate the equation with respect to t as follows.

The speed of the wave  will be:

                   v(t) = \frac{dy}{dt}

                v(t) = A \omega cos(kx + \omega t)

        v(t) = (0.09 m)(4\pi) cos(\frac{\pi \times 1.4}{11} + 4 \pi \times 0.16)

          v(t) = -0.84 m/s

The acceleration of the particle in the location is

            a(t) = \frac{dv}{dt}

           a(t) = -A \omega 2sin(kx + \omega t)

           a(t) = -(0.09 m)(4 \pi)2 sin(\frac{\pi \times 1.4}{11} + 4\pi \times 0.16)

           a(t) = -9.49 m/s^{2}

Hence, the value of transverse wave is 0.84 m/s and the value of acceleration is 9.49 m/s^{2} .

(b)  Wavelength of the wave is given as follows.

               \lambda = \frac{2\pi}{k}

              \lambda = (frac{2\pi}{\frac{\pi}{11})


              \lambda = 22 m

The period of the wave is

             T = \frac{2 \pi}{\omega}

             T = \frac{2 \pi}{4 \pi}

                = 0.5 sec

Now, we will calculate the speed of propagation of wave as follows.

                    v = \frac{\lambda}{T}

                       = \frac{22 m}{0.5 s}

                       = 44 m/s

therefore, we can conclude that wavelength is 22 m, period is 0.5 sec, and speed of propagation of wave is 44 m/s.

7 0
3 years ago
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