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Tema [17]
3 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]3 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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An artist working on a piece of metal in his forging studio plunges the hot metal into oil in order to harden it. The metal piec
lutik1710 [3]

Answer:

The temperature of the metal is  T_m  =  376.8 ^o C

Explanation:

From the question we are told that

     The mass of the metal is  M =  60 \ kg

     The specific heat of the metal is  c_p  =  0.1027 kcal/(kg \cdot ^oC)

       The mass of the oil is M_o  =  810 \ kg

       The temperature of the oil is  T_o  =  35^oC

       The specific heat of oil is  c_o  =  0.7167 kcal/(kg \cdot ^oC )

       The equilibrium temperature is T_e  =  39 ^oC

According to the law of energy conservation

     Heat lost by metal  =  heat gained by the oil

So  

   The quantity  of heat lost by the metal is mathematically represented as

               Q =  - Mc_p \Delta T

=>            Q =  -Mc_p (T_m  -  T_c)

Where T_ m  the temperature of metal before immersion

The negative sign show heat lost

The quantity  of gained t by the metal is mathematically represented as      

           Q =  M_o c_o \Delta T

=>        Q =  M_o c_o (T_c - T_o)

So  

         Mc_p (T_m  -  T_c)   =   M_o c_o (T_c - T_o)

substituting values

          - 60 * 0.1027 (T_m  - 39)   =   810 * 0.7167 *  (39 - 35)

=>       T_m  =  376.8 ^o C

         

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A trombone can produce pitches ranging from 85 Hz to 660 Hz approximately. When the trombone is producing a 562 Hz tone, what is
tester [92]

To solve this problem we will apply the concept of wavelength, which warns that this is equivalent to the relationship between the speed of the air (in this case in through the air) and the frequency of that wave. The air is in standard conditions so we have the relation,

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