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3 years ago

I want ti know how to study

Physics

2 answers:

loris [4]3 years ago

5 0

Answer:

There are tons of different ways to study.

Explanation:

If unclear on a subject, or you just want to try to get a good grade on a test that is coming up, it would be helpful to study. Find good ways that YOU like to study.

1rst you want to find your style. EX: If you are a logical learner, study maps are helpful.

One good way is to review your notes. Sometimes i do not like reveiwing my notes if they have messy handwriting, or just look bad, so you might want to make your notes look fun and pretty. After you have studied your notes, ask yourself some questions about the subject you would imagine your teacher asking.

Another way is flashcards. You can get some cool flashcard and info from this cool website called quizlet.

Also, you NEED breaks. If you study for more than 2 hours, you might get a lack of motivation and burnout.

So if you want to study for 30 mins take a 5-10 min break and then keep going

If you are reading a book and are trying to go fast, but still understand, read the first sentence of every paragraph, because those are like topic sentences.

When studying, pretend you are going to teach the material afterwords "If you can teach it, you know you have done right."

arlik [135]3 years ago

3 0

Answer:

Make sure everything is organized have a planner it can help

Get rid of all distractions

Listen to music if it helps you concentrate

Have your notes

Being willing to stay focus on what you are doing

Understand what you are doing

And most off all Be Happy and Remain Calm : )

You might be interested in

A mover loads a 100 kg box into the back of a moving truck by

NeX [460]

Answer:

2.7

Explanation:

The following data were obtained from the question:

Mass (m) of box = 100 Kg

Length (L) of ramp = 4 m

Height (H) of ramp = 1.5 m

Mechanical advantage (MA) of ramp =?

Mechanical advantage of a ramp is simply defined as the ratio of the length of the ramp to the height of the ramp. Mathematically, it is given by:

Mechanical Advantage = Lenght / height

MA= L/H

With the above formula, we can obtain the mechanical advantage of the ramp as follow:

Length (L) of ramp = 4 m

Height (H) of ramp = 1.5 m

Mechanical advantage (MA) of ramp =?

MA = 4/1.5

MA = 2.7

Therefore, the mechanical advantage of the ramp is 2.7

3 0

2 years ago

A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur

muminat

Answer:

a) E = 0

b) $E = \dfrac{k_e \cdot q}{ r^2 }$

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) $\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}$

From which we have;

$E \cdot A = \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}} = \dfrac{0}{\varepsilon _{0}} = 0$

E = 0/A = 0

E = 0

b) $\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}$

$E \cdot A = \dfrac{+q }{\varepsilon _{0}}$

$E = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}$

By Gauss theorem, we have;

$E\oint dS = \dfrac{q}{\varepsilon _{0}}$

Therefore, we get;

$E \cdot (4 \cdot \pi \cdot r^2) = \dfrac{q}{\varepsilon _{0}}$

The electrical field outside the spherical shell

$E = \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }= \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }$