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2 years ago

13

Which statement about nuclear fusion is correct? A. Two hydrogen electrons become protons during fusion. B. Helium nuclei can fu

se to form elements such as nitrogen.
C. The fusion reaction lasts for several hundred years. D. Large nuclei break apart to form smaller nuclei.

2 answers:

velikii [3]2 years ago

8 0

Answer:

A is the answer

Explanation:

Yes, Two hydrogen electrons become protons during fusion.

By far, the most common fusion reaction in nature combines two hydrogen atoms to make a helium atom.

prohojiy [21]2 years ago

3 0

Answer:

B

Explanation:

A no

B Yes ....can fuse all the way up to making Iron

C no

D no .... the opposite

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How far does light travel in 13 s?

Zolol [24]

Answer:

2,421,666 miles

Explanation:

light travels 186,282 miles per second in a vaccum.

So 186,282 × 13 = 2,421,666

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4 years ago

When a surface is illuminated with electromagnetic radiation of wavelength 480 nm, the maximum kinetic energy of the emitted ele

algol [13]

Answer:

Max kinetic energy for 340 nm wavelength will be $2.238\times 10^{-19}j$

Explanation:

In first case wavelength of electromagnetic radiation $\lambda =480nm=480\times 10^{-9}m$

Plank's constant $h=6.6\times 10^{-34}J-s$

Maximum kinetic energy = 0.54 eV

Energy is given by $E=\frac{hc}{\lambda }=\frac{6.6\times 10^{-34}\times 3\times 10^8}{480\times 10^{-9}}=4.125\times 10^{-19}J$

We know that energy is given

$E=K_{MAX}+\Phi$ , here $\Phi$ is work function

So $4.125\times 10^{-19}=0.54\times 10^{-19}+\Phi$

$\Phi =3.585\times 10^{-19}J$

Now wavelength of second radiation = 340 nm

So energy $E=\frac{hc}{\lambda }=\frac{6.6\times 10^{-34}\times 3\times 10^8}{340\times 10^{-9}}=5.823\times 10^{-19}J$

So $K_{MAX}=5.823\times 10^{-19}-3.585\times 10^{-19}=2.238\times 10^{-19}j$

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4 years ago

Isla made ice by putting water in the freezer. Freezing is an example of (2 points)

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Answer:

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3 years ago

Read 2 more answers

Why are fossils more commonly found in sedimentary rocks and not igneous rocks?

stiv31 [10]

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3 years ago

A) A real object 4.0 cm high stands 30.0 cm in front of a converging lens of focal length 23 cm. Find the image distance, the im

vfiekz [6]

Explanation:

Given that,

Height of object = 4.0 cm

Distance of the object u= -30.0 cm

Focal length = 23 cm

We need to calculate the image distance

Using lens's formula

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

Where, u = object distance

v = image distance

f = focal length

Put the value into the formula

$\dfrac{1}{v}-\dfrac{1}{-30}=\dfrac{1}{23}$

$\dfrac{1}{v}=\dfrac{1}{23}-\dfrac{1}{30}$

$\dfrac{1}{v}=\dfrac{7}{690}$

$v=98.57\ cm$

We need to calculate the height of the image

Using formula of height

$\dfrac{h'}{h}=\dfrac{-v}{u}$

Where, h' = height of image

h = height of object

$\dfrac{h'}{4}=\dfrac{-98.57}{-30}$

$h'=\dfrac{98.57\times4}{30}$

$h'=13.14\ cm$

The image is real and inverted.

(b). Now, object distance u = 13.0 cm

We need to calculate the image distance

Using lens's formula

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

Put the value into the formula

$\dfrac{1}{v}=\dfrac{1}{23}-\dfrac{1}{13.0}$

$\dfrac{1}{v}=-\dfrac{10}{299}$

$v=-29.9\ cm$

We need to calculate the height of the image

$\dfrac{h'}{4}=\dfrac{-(-29.9)}{-13.0}$

$h=-\dfrac{29.9\times4}{13.0}$

$h=-9.2\ cm$

The image is virtual and erect.

Hence, This is required solution.

6 0

3 years ago