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padilas [110]
3 years ago
10

What is the purpose of the magnetic field in a cathode ray tube?

Physics
1 answer:
andre [41]3 years ago
4 0

Answer:

In order to determine if the cathode ray consisted of charged particles, Thomson used magnets and charged plates to deflect the cathode ray. He observed that cathode rays were deflected by a magnetic field in the same manner as a wire carrying an electric current, which was known to be negatively charged.

Explanation:

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Who discovered natural magnet for the first time?​
andrew-mc [135]
Answer: Magnes



Explanation: Magnes found that the nails and buckle of his sandals and the tip of his staff were attracted to the rock he was standing on. He dug up the Earth to find lodestones. Lodestones contain magnetite, a natural magnetic material Fe3O4.
8 0
3 years ago
A 4 kg box is on a frictionless 35° slope and is connected via a massless string over a massless, frictionless pulley to a hangi
Anarel [89]

Answer:

(a) 19.62 N

(b) Box moves down the slope

(c) 24.43 N

Explanation:

(a)  

2 Kg box  causes tension

T=mgwhere m is mass, g is gravitational force taken as 9.81T=2*9.81 =19.62 N  (b)  Block mass of 4 Kg  [tex]T'-mg sin \theta=0 hence T'=mg sin \theta where m is mass and g is gravitational force  

T'=4*9.81 sin 35= 22.5071 N  

Since T' is greater than mg sin\theta , then the box moves down the slope  

(c)  

Acceleration a= \frac {forward   force-backward   force}{Total mass}= \frac {mg sin \theta -mg}{m1 + m2}  

a= \frac {22.51-19.62}{2+4}=0.48

When moving, the box will exert force T"= mgsin \theta + ma  

T"= 4*9.81 sin 35 +(4*0.48)= 24.43 N

7 0
3 years ago
Read 2 more answers
AQ5. What is a car’s speed in m/s if it travels: Remember to write the speed number WITH the correct unit.
AVprozaik [17]

Answer:A

Explanation:

7 0
3 years ago
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Two 60 cm parallel disks are separated by 40 cm and are aligned directly on top of each other. Both disks are black surfaces wit
Crazy boy [7]

Answer:

775.48 W

Explanation:

given,

diameter of disk = 0.6 cm

length of the disk = 0.4 m

T₁ = 450 K         T₂ = 450 K      T₃ = 300 K

\dfrac{d}{r_1}=\dfrac{0.4}{0.3} = 1.33

now,

the value of view factor (F₁₂)corresponding to 1.33

F₁₂ = 0.265

F₁₃ = 1 - 0.265 = 0.735

now,

net rate of radiation heat transfer from the disk to the environment:

=\dot{Q_{1-3}+Q_{2-3}} = 2 \dot{Q_{1-3}}

       = 2 F₁₃ A₁ σ (T₁⁴ - T₃⁴)

       = 2 x 0.735 x π x (0.3)² x (5.67 x 10⁻⁸ W/m²) (450⁴ - 300⁴)

       = 775.48 W

Net radiation heat transfer from the disks to the environment = 775.48 W

3 0
4 years ago
An electron with a charge of -1.6 × 10-19 coulombs experiences a field of 1.4 × 105 newtons/coulomb. What is the magnitude of th
RSB [31]
F=Eq. Just sub in the values and you are done.
7 0
4 years ago
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