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3 years ago

How many photons with 10 ev are required to produce 20 joules of energy?

1 answer:

6 0

One electron Volt (eV) is equal to 1.6 x 10^-19 Joules. Therefore, 10 eV is equal to 1.6 x 10^-18 Joules. In order to produce 20 Joules of energy from 10 eV photons, we would require 20 x 1/(1.6 x 10^-18) = 1.25 x 10^19 particles. This demonstrates that in the world of particle physics, the Joule is a massive energy unit relative to the commonly used electron Volt.

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A 25.0-g sample of copper at 363 K is placed in I 00.0 g of water at 293 K. The copper and water quickly come to the sa me tempe

Simora [160]

Answer:

Final temperature is 295K

Explanation:

Where the sample of copper is placed in the water, the heat transferred from the copper is equal that the heat absorbed by the water.

The heat transferred from the copper is:

C× $\frac{1mol}{63,546g}$ ×mass×ΔT

Where C is molar heat capacity of copper (24,5J/molK)

Mass is 25,0g

And ΔT is final temperature - initial temperature (X-363K)

Also, the heat absorbed by the water is:

-C× $\frac{1mol}{18,02g}$ ×mass×ΔT

Where C is molar heat capacity of water (75,2J/molK)

Mass is 100,0g

And ΔT is final temperature - initial temperature (X-293K)

As heat transferred is equal to heat absorbed:

24,5J/molK× $\frac{1mol}{63,546g}$ ×25,0g×(X-363K) = -75,2J/molK× $\frac{1mol}{18,02g}$ ×100,0g× (X-293K)

9,64X J/K - 3499J = - 417X J/K + 122273J

426,64X J/K = 125772 J

Final temperature is 295K

I hope it helps!

6 0

3 years ago

a piece of metal with a mass of 15.3 grams has a temperature of 50.0°C. When the metal is placed in 80.2 grams of water at 21.0°

AleksAgata [21]

Answer:

1.21

Explanation:

Heat rise in the body happens due to heat supplied by water to the body.

Heat rise in body = m₁ c₁ ΔT₁

Where m₁ is mass of body and c₁ is its specific heat of body

Heat lost from water to the body = m₂ c₂ ΔT₂

Where m₂ is mass of water and c₂ is its specific heat of water ( c₂ =1 (since water))

Equating both:

15.3 x c₁ x 4.3 = 80.2 x 1 x 4.3

⇒ c₁ = 80.2 / (15.3 x 4.3) = 1.21

6 0

3 years ago

Read 2 more answers

A proton that has a mass m and is moving at 270 m/s in the i hat direction undergoes a head-on elastic collision with a stationa

Nataly_w [17]

Answer:

$V_p = 267.258 m/s$

$V_n = 38.375 m/s$

Explanation:

using the law of the conservation of the linear momentum:

$P_i = P_f$

where $P_i$ is the inicial momemtum and $P_f$ is the final momentum

the linear momentum is calculated by the next equation

P = MV

where M is the mass and V is the velocity.

so:

$P_i = m(270 m/s)$

$P_f = mV_P + M_nV_n$

where m is the mass of the proton and $V_p$ is the velocity of the proton after the collision, $M_n$ is the mass of the nucleus and $V_n$ is the velocity of the nucleus after the collision.

therefore, we can formulate the following equation:

m(270 m/s) = m $V_p$ + 14m $V_n$

then, m is cancelated and we have:

270 = $V_p$ + $14V_n$

This is a elastic collision, so the kinetic energy K is conservated. Then:

$K_i = \frac{1}{2}MV^2 = \frac{1}{2}m(270)^2$

and

Kf = $\frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2$

then,

$\frac{1}{2}m(270)^2$ = $\frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2$

here we can cancel the m and get:

$\frac{1}{2}(270)^2$ = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$

now, we have two equations and two incognites:

270 = $V_p$ + $14V_n$ (eq. 1)

$\frac{1}{2}(270)^2$ = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$

in the second equation, we have:

36450 = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$ (eq. 2)

from this last equation we solve for $V_n$ as:

$V_n = \sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }$

and replace in the other equation as:

270 = $V_p +$ 14 $\sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }$

so,

$V_p = -267.258 m/s$

Vp is negative because the proton go in the -i hat direction.

Finally, replacing this value on eq. 1 we get:

$V_n = \frac{270+267.258}{14}$

$V_n = 38.375 m/s$

3 0

3 years ago

4. All electromagnetic radiation moves at the speed of ________________, which is _________________ m/s.

VARVARA [1.3K]

Answer:

All electromagnetic radiation, regardless of its wavelength, travels at what is called the "speed of light". In a vacuum, it is measured as 299,792,458 meters per second. 3.0 x 108 meters per second.

7 0

3 years ago

I NEED some help please!

andrew11 [14]

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3 0

3 years ago