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3 years ago

Explanation's of E=MC²

Physics

2 answers:

Semmy [17]3 years ago

7 0

Answer:

E = How much energy which we can get from mass M of atoms

M - Mass of the atom

C - Speed of light

Explanation:

Energy equals mass times the speed of light squared." On the most basic level, the equation says that energy and mass (matter) are interchangeable; they are different forms of the same thing. Under the right conditions, energy can become mass, and vice versa. We humans don't see them that way how can a beam of light and a walnut, say, be different forms of the same thing but Nature does

natita [175]3 years ago

4 0

Answer:

E means energy

M= Mass

C=speed of light squared (the exponent means squared)

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Problem 21-40a:

Greeley [361]

Answer:

Part a)

$\frac{F_e}{F_g} = 2.74 \times 10^{12}$

Part b)

$q = 3.37 \times 10^{-4} C$

Explanation:

As we know that electric force on electric charge is given as

$F = qE$

here we have

$q = 1.6 \times 10^{-19}C$

E = 153 N/C

now force is given as

$F = (1.6 \times 10^{-19})(153) = 2.45 \times 10^{-17} N$

Gravitational force on electric charge near surface of earth is given as

$F_g = mg$

$F_g = (9.1 \times 10^{-31})(9.81) = 8.93 \times 10^[-30} N$

now the ratio of two forces is given as

$\frac{F_e}{F_g} = \frac{2.45 \times 10^{-17}}{8.93 \times 10^{-30}}$

$\frac{F_e}{F_g} = 2.74 \times 10^{12}$

Part b)

Now the ball is balanced by the electric force and the force of gravity on it

so here we have

$F_g = qE$

$mg = qE$

$(5.25 \times 10^{-3})(9.81) = q(153)$

here we have

$q = 3.37 \times 10^{-4} C$

8 0

4 years ago

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photoshop1234 [79]

Answer:

Explanation:

4 0

3 years ago

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A force of 13 N is used to slide a 0.8 kg book across a table. What is the magnitude of the book’s acceleration?

Mkey [24]

Explanation:

acceleration=F/m

a=13/0.8

a=16 25m/s²

8 0

4 years ago

A real (non-Carnot) heat engine, operating between heat reservoirs at temperatures of and performs 4.1 kJ of net work, and rejec

Sati [7]

There are some missing data in the problem. The full text is the following:
"A real (non-Carnot) heat engine, operating between heat reservoirs at temperatures of 710 K and 270 K performs 4.1 kJ of net work, and rejects 9.7 kJ of heat, in a single cycle. The thermal efficiency of a Carnot heat engine, operating between the same heat reservoirs, in percent, is closest to.."

Solution:
The efficiency of a Carnot cycle working between cold temperature $T_C$ and hot temperature $T_H$ is given by
$\eta = 1 - \frac{T_C}{T_H}$
and it represents the maximum efficiency that can be reached by a machine operating between these temperatures. If we use the temperatures of the problem, $T_C=270 K$ and $T_H=710 K$ , the efficiency is
$\eta = 1 - \frac{270 K}{710 K}=0.62 = 62%$

Therefore, the correct answer is D) 62 %.

6 0

3 years ago

Find the minimum value of n in the Balmer series for which the predicted wavelength is in the ultraviolet region of the spectrum

d1i1m1o1n [39]

Answer:

λ =365.4 nm

Explanation:

Boh's atomic model of the Hydrogen atom the energy of each level is

Eₙ = - 13.606 / n²

where the synergy is in electonvotes and the value of E₀ = 13.606 eV is the energy of the base state of hydrogen.

An atomic transition occurs when an electron goes from an excited state and joins everything of lower energy.

ED = 13.606 (1 / n₀² - 1 / $n_{f}^{2}$ )

we are going to apply this relationship to answer slash.

At the beginning of the studies of atomic transitions, each group did not consider having a different name

name Initial state

Lymman 1

Balmer 2

the final state is any other state sta the continuum that corresponds to n = inf

Let's look for the highest energy of the Balmer series

ΔE = 13.606 (1/2² - 1 /∞)

ΔE = 3.4015 eV

Let's use the Planck relation for the energy

E = h f = h c /λ

λ = h c / E

Let's reduce the energy to J

E = 3.4015 eV (1.6 10⁻¹⁹ J / 1 eV) = 5.4424 10⁻¹⁹

λ = 6.63 10⁻³⁴ 3 10⁸ / 5.4424 10⁻¹⁹

λ = 3.654 10⁻⁷ m

λ = 3,654 10⁻⁷ m (10⁹ nm / 1m)

λ =365.4 nm

this eta radiation in the ultraviolet range

5 0

3 years ago