Why does Marx’s workers’ paradise resolve the problems of capitalism?

In Thomson’s experiment, why was the glowing beam repelled by a negatively charged plate?

Svetllana [295]

The glowing beam was repelled by a negatively charged plate because they were negatively charged

<h3>What are the nature of charges?</h3>

The nature of charges refers to the properties of charges.

There are two types of charges:

negative charges
positive charges

The law of electricity states that opposite charges attract whereas like charges repel.

Therefor, in Thomson’s experiment, the glowing beam was repelled by a negatively charged plate because they were negatively charged

In conclusion, like charges repel while opposite charges attract.

Learn more about charges at: brainly.com/question/12781208

#SPJ1

5 0

2 years ago

Car A, moving in a straight line at a constant speed of 20. meters per second, is initially 200 meters behind car B, moving in t

MA_775_DIABLO [31]

First, let's express the movement of Car A and B in terms of their position over time (relative to car B)
For car A: y=20x-200 Car A moves 20 meters every second x, and starts 200 meters behind car B
For Car B: y= 15x Car B moves 15 meters every second and starts at our basis point

Set the two equations equal to one another to find the time x at which they meet:
20x - 200 = 15x
200 = 5x
x= 40
At time x=40 seconds, the cars meet. How far will Car A have traveled at this time?
Car A moves 20 meters every second:
20 x 40 = 800 meters

8 0

3 years ago

The small currents in axons corresponding to nerve impulses produce measurable magnetic fields. a typical axon carries a peak cu

Gemiola [76]

Answer:

$6.66\cdot 10^{-12}T$

Explanation:

The magnetic field produced by a current-carrying wire is given by

$B=\frac{\mu_0 I}{2\pi r}$

where

$\mu_0$ is the vacuum permeability

I is the current

r is the distance from the wire

In this problem we have

$I=0.040 \mu A=4\cdot 10^{-8}A$

r = 1.2 mm = 0.0012 m

So the magnetic field strength is

$B=\frac{(4\pi \cdot 10^{-7} H/m)(4\cdot 10^{-8}A)}{2\pi (0.0012 m)}=6.66\cdot 10^{-12}T$

5 0

3 years ago

What is the force on an electron in a CRT when it’s moving at 2.5 × 105 meters/second perpendicular to a magnetic field of 1.5 t

Firdavs [7]

F = Magnetic Force
B = Magnetic Field
V = Velocity

*The vectors from the photo you get doing the left-hand rule.

The magnetic force is always perpendicular to the magnetic field.

And as told in the statement, the electron is moving perpendicular to a magnetic field, that is, the Velocity forms an 90 degree angle / Right angle with the magnetic field.

The formula to find the Magnetic Force is:

$f = |q| \times v \times b \times sin \: \theta$

Where "q" is the Charge and the sin theta is the angle formed by the Velocity and Magnetic Field, in this case it's 90°. Sin 90° = 1.

$f = |- 1.6 \times {10}^{ - 19} | \times 2.5 \times {10}^{5} \times 1.5 \times 1 \\ f = 6 \times {10}^{ - 19 + 5} \\ f = 6 \times {10}^{ - 14} \: newtons$
Newton (N) = C x m/s x T = (C x m x T)/s

4 0

4 years ago

(4) 10 m/s

siniylev [52]

Answer:

13

Explanation:

6+4=10s

50+35=85

av speed=85/10=8.5m/s

4 0

3 years ago