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2 years ago

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

Physics

1 answer:

andre [41]2 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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Atoms with a low ionization energy hold tight to their outer valence electrons.

Anestetic [448]

True would be the Correct answer

5 0

3 years ago

Read 2 more answers

A rock of mass m is thrown straight up into the air with initial speed |v0 | and initial position y = 0 and it rises up to a max

REY [17]

Answer:

Explanation:

Case 1:

mass = m

initial velocity = vo

final velocity = 0

height = y

Use third equation of motion

v² = u² - 2as

0 = vo² - 2 g y

y = vo² / 2g ... (1)

Case 2:

mass = 2m

initial velocity = 2vo

final velocity = 0

height = y '

Use third equation of motion

v² = u² - 2as

0 = 4vo² - 2 g y'

y ' = 4vo² / 2g

y' = 4 y

Thus, the second rock reaches the 4 times the distance traveled by the first rock.

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3 years ago

An ultrasonic tape measure uses frequencies above 20 MHz todetermine dimensions of structures such as buildings. It does so byem

ohaa [14]

Answer:

a) 1m

b) 2μs

c) 3mm

Explanation:

3 0

3 years ago

CNA Secondary and Preparatory School

ale4655 [162]

Answer:

2:13 2) cm' c6462.

Explanation:

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3 years ago

A hiker determines the length of a lake by listening for the echo of her shout reflected by a cliff at the far end of the lake.

ArbitrLikvidat [17]

Answer:

L = 499 m

Explanation:

If we assume that the speed of sound is constant, that travels along a straight line, and that the echo is instantaneous, we can find the total distance travelled by the sound, as follows, just applying the definition of average velocity:

$\Delta x = v_{s} * t = 343 m/s* 2.91 s = 998 m$

If we assume that the time needed to reach to the cliff, is the same used for the return travel, the length of the lake will be exactly half of the total distance calculated:

$l_{lake} = \frac{\Delta x}{2} = \frac{998m}{2} = 499 m$

The length of the lake is 499 m.

8 0

3 years ago