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3 years ago

What are the basics of physics.I want to learn them for an exam?

2 answers:

7 0

If it were possible to give that information over in a few paragraphs or a few mimutes, then you'd see little stores everywhere, where you could drop in for an hour or an afternoon, pay a few bucks, and learn Physics. You don't, because you can't, because it isn't. Here are a few things you definitely need. But I have to tell you: If you don't put time and effort into learning them, they won't do you much good. ==> F=m a. ==> Total mass is conserved. ==> Total energy is conserved. ==> Everything falls at the same rate. ==> Sound needs material to travel through. Light doesn't. ==> Light, radio, heat, X-rays, and ultraviolet radiation are all the same thing. ==> It takes force on an object to make it speed up, slow down, or curve. A moving object with no force acting on it keeps going in a straight line at the same speed forever. ==> Gravity attracts every speck of mass in the universe to every other speck. The forces of attraction are greater when the masses are greater and closer together.

Westkost [7]3 years ago

5 0

One cannot simply give the basics of physics.

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Natali [406]

Heat transfer

Have a good day

6 0

2 years ago

1pt Which is an observation? O A. A person hears the song of a bird. OB. A person suggests that bird calls are a means of commun

Doss [256]

Answer:

Explanation:

he describes as he writes them down

3 0

3 years ago

A 1kg mass is thrown to a height of 2cm. what is the potential energy

prohojiy [21]

Mass=m=1kg
Height=h=2cm=0.02m
Acceleration due to gravity=g=10m/s^2

$\\ \tt\hookrightarrow P.E=mgh$

$\\ \tt\hookrightarrow PE=1(10)(0.02)$

$\\ \tt\hookrightarrow PE=0.2J$

4 0

2 years ago

Read 2 more answers

An average hole drift velocity of 103 cm/sec results when 2 V is applied across a 1 cm long semiconductor bar. What is the hole

Helga [31]

Answer:

ε = 2 V/cm

Explanation:

To calculate the mobility inside this bar, we just need to apply the expression that let us determine the mobility. This expression is the following:

ε = ΔV / L

Where:

ε: Hole mobility inside the bar

ΔV: voltage applied in the bar

L: Length of the bar

We already have the voltage and the length so replacing in the above expression we have:

ε = 2 V / 1 cm

The data of the speed can be used for further calculations, but in this part its not necessary.

Hope this helps

8 0

2 years ago

Three positive charges A, B, and C, and a negative charge D are placed in a line as shown in the diagram. All four charges are o

polet [3.4K]

Answer:

a. charge C experiences the greatest net force, and charge B receives the smallest net force

b. ratio=9

Explanation:

<u>Electrostatic Force</u>

Two point-charges $q_1$ and $q_2$ separated a distance d will exert a force on each other of a magnitude given by the Coulomb's formula

$\displaystyle F=\frac{k\ q_1\ q_2}{r^2}$

Where k is the proportional constant of value

$k=9*10^9\ N.m^2/c^2$

The diagram provided in the question shows four identical charges (let's assume their value is Q) separated by identical distance (of value d). The force between the charges next to others is

$\displaystyle F_1=\frac{k\ Q\ Q}{d^2}$

$\displaystyle F_1=\frac{k\ Q^2}{d^2}$

The force between charges separated 2d is

$\displaystyle F_2=\frac{k\ Q^2}{(2d)^2}$

$\displaystyle F_2=\frac{k\ Q^2}{4d^2}$

And the force between the charges A and D is

$\displaystyle F_3=\frac{k\ Q^2}{(3d)^2}$

$\displaystyle F_3=\frac{k\ Q^2}{9d^2}$

Now, let's analyze each charge and the force applied to them by the others

Let's recall equally signed charges repel each other and differently signed charges attrach each other

Charge A. It receives force to the left from B and C and to the right from D

$\displaystyle F_A=-F_1-F_2+F_3=-\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}$

$\displaystyle F_A=\frac{k\ Q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})$

$\displaystyle F_A=-\frac{41}{36}F_1$

Charge B. It receives force to the right from A and D and to the left from C

$\displaystyle F_B=F_1-F_1+F_2=\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{4d^2}$

$\displaystyle F_B=\frac{1}{4}F_1$

Charge C. It receives forces to the right from all charges.

$\displaystyle F_C=F_2+F_1+F_1=\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{d^2}$

$\displaystyle F_C=\frac{9}{4}F_1$

Charge D. It receives forces to the left from all charges

$\displaystyle F_D=-F_3-F_2-F_1=-\frac{k\ Q^2}{9d^2}-\frac{k\ Q^2}{4d^2}-\frac{k\ Q^2}{d^2}$

$\displaystyle F_D=-\frac{49}{36}F_1$

Comparing the magnitudes of each force is just a matter of computing the fractions

$\displaystyle \frac{41}{36}=1.13,\ \frac{1}{4}=0.25,\ \frac{9}{4}=2.25,\ \frac{49}{36}=1.36$

We can see the charge C experiences the greatest net force, and charge B receives the smallest net force

The ratio of the greatest to the smallest net force is

$\displaystyle \frac{\frac{9}{4}}{\frac{1}{4}}=9$

The greatest force is 9 times the smallest net force

7 0

3 years ago