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Tema [17]
2 years ago
8

Cathode ray tubes in old television sets worked by accelerating electrons and then deflecting them with magnetic fields onto a p

hosphor screen. The magnetic fields were created by coils of wire on either side of the tube carrying large currents. In one such TV set, the phosphor screen is 51.2 cm wide, and is 11.1 cm away from the center of the magnetic deflection coils (that is, the center of the region of magnetic field). The electron beam is first accelerated through a 22,000 V potential difference before it enters the magnetic field region, which is 1.00 cm wide. The field is approximately uniform and perpendicular to the velocity of the electrons. If the field were turned off, the electrons would hit the center of the screen. What magnitude of magnetic field (in mT) is needed to deflect the electrons so that they hit the far edge of the screen

Physics
1 answer:
Roman55 [17]2 years ago
3 0

Answer:

B = 0.046T

Explanation:

given

size of the screen = 51.2cm

distance from center = 11.1cm

region of magnetic field = 1.00cm

V= 22000V= 22kV

 

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Lexy used the formula shown to calculate the force of gravity on a space shuttle. Fg = G What does 3 × 105 kg represent? the dif
STALIN [3.7K]
<h2>Answer:</h2>

<u>This term shows the </u><u>mass of the space shuttle</u>

<h2>Explanation:</h2>

We know that the mass of the Earth is 5.972 × 10^24 kg. Similarly the sum of  mass of earth and the mass of shuttle must be a greater number as compared to the number given. It simply means that the mass of earth is itself 5.972 × 10^24 kg and the value given is 3 × 105 kg so it is obvious that if was the sum then it must be greater than the mass of earth. Therefore we can say that this not the mass of earth, neither the sum of mass of earth and shuttle, but this is only the mass of space shuttle which is the last multiple choice.

5 0
2 years ago
A kangaroo jumps to a vertical height of 2.8 m. How long was it in the air before returning to earth
BaLLatris [955]
The answer would be 2.8m height on earth takes 
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8 0
2 years ago
Two identical horizontal sheets of glass have a thin film of air of thickness t between them. The glass has refractive index 1.4
Gre4nikov [31]

Answer:

the wavelength λ of the light when it is traveling in air = 560 nm

the smallest thickness t of the air film = 140 nm

Explanation:

From the question; the path difference is Δx = 2t  (since the condition of the phase difference in the maxima and minima gets interchanged)

Now for constructive interference;

Δx= (m+ \frac{1}{2} \lambda)

replacing ;

Δx = 2t   ; we have:

2t = (m+ \frac{1}{2} \lambda)

Given that thickness t = 700 nm

Then

2× 700 = (m+ \frac{1}{2} \lambda)     --- equation (1)

For thickness t = 980 nm that is next to constructive interference

2× 980 = (m+ \frac{1}{2} \lambda)     ----- equation (2)

Equating the difference of equation (2) and equation (1); we have:'

λ = (2 × 980) - ( 2× 700 )

λ = 1960 - 1400

λ = 560 nm

Thus;  the wavelength λ of the light when it is traveling in air = 560 nm

b)  

For the smallest thickness t_{min} ; \ \ \ m =0

∴ 2t_{min} =\frac{\lambda}{2}

t_{min} =\frac{\lambda}{4}

t_{min} =\frac{560}{4}

t_{min} =140 \ \  nm

Thus, the smallest thickness t of the air film = 140 nm

7 0
2 years ago
Read 2 more answers
Consider a variety of colors of visible light (say 400 nm to 700 nm) falling onto a pair of slits.
babymother [125]

Answer:

Explanation:

The relationship between angle and wavelength for maxima and minima in Young's double slit experiment is given by

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d\sin \theta =m\lambda

For Destructive interference

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where \lambda =wavelength

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for second order maxima i.e. m=2

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6 0
2 years ago
The study of alternating electric current requires the solutions of equations of the form i equals Upper I Subscript max Baselin
KiRa [710]

Answer:

Explanation:

i = Imax sin2πft

given i = 180 , Imax = 200 , f = 50  , t = ?

Put the give values in the equation above

180 = 200 sin 2πft

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sin2π x 50t = .9

sin 360 x 50 t = sin ( 360n + 64 )

360 x 50 t = 360n + 64

360 x 50 t =  64 ,  ( putting n = 0 for least value of t )

18000 t = 64

t = 3.55 ms  .

8 0
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