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Debora [2.8K]
3 years ago
9

In which way are the cathode rays deflected in the x-plates of the Cathode Ray Oscilloscope​

Physics
2 answers:
DerKrebs [107]3 years ago
5 0

Answer:

vertically

Explanation:

  • Cathode ray Oscilloscope or Cathode ray discharge tube is made of glass is taken with two electrodes.
  • It was used by JJ Thompson to discover electrons
  • The experiment is popularily called cathode ray discharge tube experiment
MrRissso [65]3 years ago
4 0

Answer:

The cathode ray is deflected vertically to the fluorescent screen

Explanation:

.

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When Trinity pulls on the rope with her weight, Newton's Third Law of Motion tells us that the rope will _____
Bingel [31]

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7 0
3 years ago
Two identical speakers are emitting a constant tone that has a wavelength of 0.50 m. Speaker A is located to the left of speaker
Marianna [84]

Answer:

a) and c).

Explanation:

For a complete destructive interference occur, it must be met the following condition relating the wavelength, and the difference in the paths taken by the sound emitted by the sources until arriving to the listening point:

d = |dA- dB| = (2n-1)*(λ/2)

For n= 1,  d = λ/2 = 0.25 m, it doesn't meet any of the cases.

For n=2, d= 3*(λ/2) = 0.75 m

In the case a) we have dA = 2.15 m and dB = 3.00 m, so dB-dA = 0.75 m, which means that in the location stated by case a) a complete destructive interference would occur.

For n=3, d= 5*(λ/2) = 5*0.25 m = 1.25 m.

This is just the case c) because we have dA = 3.75 m and dB = 2.50 m, so dA-dB = 1.25 m, which means that in the location stated by case c) a complete destructive interference would occur also.

The remaining cases don't meet the condition stated above, so the statements found to be true are a) and c),

5 0
3 years ago
A very thin circular disk of radius R = 20.00 cm has charge Q = 30.00 mC uniformly distributed on its surface. The disk rotates
lutik1710 [3]

Answer:

B= 7.5*10^{-15}T

Explanation:

The magnetic field strenght on the z-axis at a distance d from the center is,

B= \frac{\mu_0 Q\omega R^2}{8\pi d^3}

Our values are:

R=20cm\\Q=30mc\\w=5rad/s\\d=2*10^3cm=20m

Replacing,

B= \frac{(4\pi*10^{}-7)(30*10^{-3})(5)(0.2)^2}{8\pi(20)}

B= 7.5*10^{-15}T

3 0
3 years ago
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