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3 years ago

This version of Einstein’s equation is often used directly to find what value?

2 answers:

8 0

This version of Einstein’s equation is often used directly to find what value? E = ∆mc2

Answer: This version of Einstein’s equation is often used directly to find the mass that is lost in a fusion reaction. Therefore the correct answer to this question is answer choice C).

I hope it helps, Regards.

Sindrei [870]3 years ago

5 0

<u>Answer</u>

A. the energy that is released in a nuclear reaction

<u>Explanation</u>

Robert Einstein devised the equation in 1905. He believed that some radioactive elements such as uranium constantly converted some of their mass to energy.

From the reactions of radioactive elements there came the production of nuclear energy. This energy is given by the Einstein’s equation, E = mc².

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Will a substance that contains an o-h group dissolve in water?

Stella [2.4K]

Usually, it increases the solubility in water.

4 0

3 years ago

Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.400mm wide. The diffraction pattern is observed

kogti [31]

Answer:

a) $y_{first}=5.3mm$

b) $y_{second}=10.6-5.3 =5.3 mm$

Explanation:

The width of the central bright in this diffraction pattern is given by:

$y=\frac{m\lambda D}{a}$ when m is a natural number.

here:

m is 1 (to find the central bright fringe)
D is the distance from the slit to the screen
a is the slit wide
λ is the wavelength

So we have:

$y_{first}=\frac{633*10^{9}*3.35}{0.0004}$

$y_{first}=5.3mm$

Now, if we do m=2 we can find the distance to the second minima.

$y_{2}=\frac{2*633*10^{9}*3.35}{0.0004}$

$y_{2}=10.6 mm$

Now we need to subtract these distance, to get the width of the first bright fringe :

$y_{second}=10.6-5.3 =5.3 mm$

I hope it heps you!

4 0

2 years ago

Which of the following statements are true concerning the reflection of light?

natka813 [3]

Answer:

b. The reflection of light from a smooth surface is called specular reflection.

c. The reflection of light from a rough surface is called diffuse reflection.

Explanation:

a. The angle of incidence is equal to the angle of reflection only when a ray of light strikes a plane mirror.

This is wrong: Based on law of reflection "The angle of incidence is equal to the angle of reflection when light strikes any plane surface" examples plane mirrors, still waters, plane tables, etc

b. The reflection of light from a smooth surface is called specular reflection.

This is correct

c. The reflection of light from a rough surface is called diffuse reflection.

This is correct

d. For diffuse reflection, the angle of incidence is greater than the angle of reflection.

This is wrong: the angle of incident is equal to angle of reflection. The only difference between this type of reflection and specular reflection, is that the normal for diffuse reflection is not parallel to each due to the rough surface in which the light incidents.

For specular reflection, the angle of incidence is less than the angle of reflection.

This is wrong: the angle of incident is equal to angle of reflection

4 0

3 years ago

The restoring force on a pendulum bob is proportional to sin(θ), where θ is the angle which the pendulum makes with the vertical

anyanavicka [17]

Answer:

The angle is 4.44º

Explanation:

If:

$\frac{sin\theta -\theta}{\theta } \leq 0.1o/o\\\frac{sin\theta -\theta}{\theta }\leq 10^{-3}$

According the Taylor`s series:

$sin\theta =\theta -\frac{\theta ^{3} }{3!} ....$

$\frac{\theta ^{2} }{6} \leq 10^{-3} \\\theta ^{2} \leq 6x10^{-3} \\\theta \leq \sqrt{6x10^{-3} } \\\theta \leq 0.0775\\\theta = 0.0775rad=4.44$

8 0

3 years ago

DUE BY MIDNIGHT

olga2289 [7]

Answer:

Option D. 1000 J.

Explanation:

From the question given above, the following data were obtained:

Force (F) applied = 200 N

Distance (s) = 5 m

Time (t) = 10 s

Workdone (Wd) =?

Workdone (Wd) is simply defined as the product of force (F) and distance (s) moved in the direction of the force. Mathematically, it is expressed as:

Wd = F × s

With the above formula, we can calculate the Workdone as illustrated below:

Force (F) applied = 200 N

Distance (s) = 5 m

Workdone (Wd) =?

Wd = F × s

Wd = 200 × 5

Wd = 1000 J

Thus, the Workdone is 1000 J

8 0

2 years ago