All categories

2 years ago

Describe in brief brief the scope and excitement in Physics

Physics

1 answer:

andriy [413]2 years ago

8 0

Answer:

The scope of physics is quite vast and covers an enormous range of magnitude of physical quantities like mass, length, time, energy etc.

The scope is divided into two sections, where on one side it studies about phenomena at a very small microscopic level involving and analyzing electrons, protons, and neutrons.

You might be interested in

The sun _____.

enyata [817]

The sun appears brighter than any other star.

(It isn't really, but it looks that way because it's much much much much much much closer to us than any other star.)

7 0

3 years ago

What is the total number of pounds of carbon dioxide that 2 trees remove from the air each year?

Vikentia [17]

Answer:

On average, one acre of new forest can sequester about 2.5 tons of carbon annually. Young trees absorb CO2 at a rate of 13 pounds per tree each year. Trees reach their most productive stage of carbon storage at about 10 years at which point they are estimated to absorb 48 pounds of CO2 per year.

5 0

3 years ago

Read 2 more answers

What is electromagnetic waves

Elis [28]

Answer:

waves that are created as a result of vibrations between an electric field and a magnetic field

4 0

2 years ago

Read 2 more answers

If u walk to school at a speed of 1.2 m/s and it takes you 18 min to reach the school, what is the distance from your home to th

Reptile [31]

Answer:

21.6m

Explanation:

Distance=speed×time

=1.2m/s×18min

=21.6m

Hope this helps u

3 0

2 years ago

Read 2 more answers

Question 1 of 4 Attempt 4 The acceleration due to gravity, ???? , is constant at sea level on the Earth's surface. However, the

Evgen [1.6K]

Answer:

g(h) = g ( 1 - 2(h/R) )

*At first order on h/R*

Explanation:

Hi!

We can derive this expression for distances h small compared to the earth's radius R.

In order to do this, we must expand the newton's law of universal gravitation around r=R

Remember that this law is:

$F = G \frac{m_1m_2}{r^2}$

In the present case m1 will be the mass of the earth.

Additionally, if we remember Newton's second law for the mass m2 (with m2 constant):

$F = m_2a$

Therefore, we can see that

$a(r) = G \frac{m_1}{r^2}$

With a the acceleration due to the earth's mass.

Now, the taylor series is going to be (at first order in h/R):

$a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R}$

a(R) is actually the constant acceleration at sea level

and

$a(R) =G \frac{m_1}{R^2} \\ \frac{da(r)}{dr}_{r=R} = -2 G\frac{m_1}{R^3}$

Therefore:

$a(R+h) \approx G\frac{m_1}{R^2} -2G\frac{m_1}{R^2} \frac{h}{R} = g(1-2\frac{h}{R})$

Consider that the error in this expresion is quadratic in (h/R), and to consider quadratic correctiosn you must expand the taylor series to the next power:

$a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R} + \frac{h^2}{2!} \frac{d^2a(r)}{dr^2}_{r=R}$

6 0

3 years ago

Read 2 more answers