How are mixtures and pure substances related?

At the equator, near the surface of the Earth, the magnetic field is approximately 50.0 μT northward, and the electric field is

n200080 [17]

a) $8.93\cdot 10^{-30} N$ , downward

b) $1.76\cdot 10^{-17} N$ , upward

c) $1.20\cdot 10^{-17} N$ , downward

Explanation:

a)

The gravitational force of an object (also known as weight of the object) is the attractive force with which the object is pulled towards the Earth's centre.

For an object near the Earth's surface, the magnitude of the gravitational force is given by the equation

$F=mg$

where

m is the mass of the object

g is the acceleration due to gravity

In this problem, we have:

$m=9.11\cdot 10^{-31} kg$ is the mass of the electron

$g=9.8 m/s^2$ is the acceleration due to gravity

Therefore, the gravitational force on the electron is:

$F=(9.11\cdot 10^{-31})(9.8)=8.93\cdot 10^{-30} N$

And the direction is downward, towards the Earth's centre.

b)

The electric force on a charged particle is the force produced by the presence of an electric field.

In particular, this force is:

- Repulsive (away from the source of the field) if the charge has the same sign of the charge source of the field

- Attractive (towards the source of the field) if the charge has opposite sign to the charge source of the field

The magnitude of the electric force is given by:

$F=qE$

where

q is the charge of the particle

E is the strength of the electric field

In this problem:

$q=1.6\cdot 10^{-19} C$ is the charge of the electron

$E=110 N/C$ is the strength of the electric field

Substituting,

$F=(1.6\cdot 10^{-19})(110)=1.76\cdot 10^{-17} N$

And since the electron has negative charge, the direction of the force is opposite to that of the electric field, so upward.

c)

When a charged particle is moving in a magnetic field, it experiences a force which is perpendicular to both the direction of motion of the charge and to the direction of the magnetic field.

The magnitude of this force is given by (if the charge moves perpendicular to the magnetic field)

$F=qvB$

where

q is the charge

v is the velocity of the particle

B is the strength of the magnetic field

In this problem:

$q=1.6\cdot 10^{-19} C$ is the charge of the electron

$v=1.50\cdot 10^6 m/s$ is the velocity

$B=50.0\mu T = 50.0 \cdot 10^{-6} T$ is the strength of the magnetic field

Substituting,

$F=(1.6\cdot 10^{-19})(1.50\cdot 10^6)(50.0\cdot 10^{-6})=1.20\cdot 10^{-17} N$

And the direction can be determined using the right-hand rule:

- Index finger: direction of velocity (eastward)

- Middle finger: direction of magnetic field (northward)

- Thumb: direction of force (upward) --> however the electron has negative charge, so the direction of the force is reversed --> downward

5 0

2 years ago

Give two example of conditions when two objects (Object A=1 kg and Object B =10 kg) would have the same momentum.

bearhunter [10]

Answer:

they can have the momentum if only they are been divided by another number to see the difference

6 0

2 years ago

According to coulombs law, how is the electric force related to the distance between two charges?

lorasvet [3.4K]

Answer:

D

Explanation:

From the formula of coulombs law F = Kq1q2/square of r, we can say the electric force is indirectly related to square of r

4 0

3 years ago

A light has a wavelength of 5.06x10-7m. what is the frequency of the light?

Artyom0805 [142]

So, the frequency of that light approximately $\sf{\bold{5.93 \times 10^{14} \: Hz}}$

<h3>Introduction</h3>

Hi ! Here I will help you to discuss the relationship between frequency and wavelength, with the velocity constant of electromagnetic waves in a vacuum. We all know that regardless of the type of electromagnetic wave, it will have the same velocity as the speed of light (light is part of electromagnetic wave too), which is 300,000 km/s or $\sf{3 \times 10^8}$ m/s. As a result of this constant property, <u>the shorter the wavelength, the greater the value of the electromagnetic wave frequency</u>. This relationship can also be expressed in this equation:

$\boxed{\sf{\bold{c = \lambda \times f}}}$

With the following condition :

c = the constant of the speed of light in a vacuum ≈ $\sf{3 \times 10^{8} \: m/s}$ m/s
$\sf{\lambda}$ = wavelength (m)
f = electromagnetic wave frequency (Hz)