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ehidna [41]
3 years ago
5

what is the relationship between high frequency and short wavelength? Low frequency and long wavelength?​

Physics
1 answer:
larisa [96]3 years ago
5 0

Answer:

The frequency of a wave is inversely proportional to its wavelength. That means that waves with a high frequency have a short wavelength, while waves with a low frequency have a longer wavelength.

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A parallel-plate capacitor is held at a potential difference of 250 V. A proton is fired toward a small hole in the negative pla
MissTica

Answer:

The speed of proton when it emerges through the hole in the positive plate is 2.05\times 10^5\ m/s.

Explanation:

Given that,

A parallel-plate capacitor is held at a potential difference of 250 V.

A A proton is fired toward a small hole in the negative plate with a speed of, u=3\times 10^5\ m/s

We need to find the speed when it emerges through the hole in the positive plate. It can be calculated using the conservation of energy as :

qV=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2\\\\1.6\times10^{-19}\times250=\dfrac{1}{2}mv^2-\frac{1}{2}\cdot1.67\times10^{-27}\cdot(3\times10^{5})^{2}\\\\\dfrac{1}{2}mv^2=3.515\cdot10^{-17}\\\\v=\sqrt{\dfrac{3.515\cdot10^{-17}\cdot2}{1.67\times10^{-27}}}\\\\v=2.05\times 10^5\ m/s

So, the speed of proton when it emerges through the hole in the positive plate is 2.05\times 10^5\ m/s.

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3 years ago
When a front wheel drops off the roadway you should?
nikdorinn [45]
<span>braking and returning suddenly to the roadway</span>
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Where is the energy in a glucose molecule stored?
Semmy [17]

Answer:

Energy is stored in the bonds between atoms

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Behavior that is harmful to one's self is? 10 pts
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Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 1.00 s, it rotates 21.0 rad. Du
ELEN [110]

With constant angular acceleration \alpha, the disk achieves an angular velocity \omega at time t according to

\omega=\alpha t

and angular displacement \theta according to

\theta=\dfrac12\alpha t^2

a. So after 1.00 s, having rotated 21.0 rad, it must have undergone an acceleration of

21.0\,\mathrm{rad}=\dfrac12\alpha(1.00\,\mathrm s)^2\implies\alpha=42.0\dfrac{\rm rad}{\mathrm s^2}

b. Under constant acceleration, the average angular velocity is equivalent to

\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2

where \omega_f and \omega_i are the final and initial angular velocities, respectively. Then

\omega_{\rm avg}=\dfrac{\left(42.0\frac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)}2=42.0\dfrac{\rm rad}{\rm s}

c. After 1.00 s, the disk has instantaneous angular velocity

\omega=\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)=42.0\dfrac{\rm rad}{\rm s}

d. During the next 1.00 s, the disk will start moving with the angular velocity \omega_0 equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle \theta according to

\theta=\omega_0t+\dfrac12\alpha t^2

which would be equal to

\theta=\left(42.0\dfrac{\rm rad}{\rm s}\right)(1.00\,\mathrm s)+\dfrac12\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)^2=63.0\,\mathrm{rad}

5 0
4 years ago
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