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xxMikexx [17]
3 years ago
15

An alpha particle is identical to a:

Physics
1 answer:
Rudik [331]3 years ago
3 0

Answer:

B-hydrogen nucleus

Explanation:

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For the image of the overhead projector to be in focus, the distance from the projector lens to the image, <img src="https://tex
rjkz [21]
Given:
distance from the projector lens to the image, di
projector lens focal length, f
distance from the transparency to the projector lens, do

thin lens equation: 1/f = 1/di + 1/do
do = 4 inches
di = 8 feet

convert feet to inches, for uniformity.
1 foot = 12 inches
8 feet * 12 inches/ft = 96 inches
 
1/f = 1/96 inches + 1/4 inches

Adding fractions, denominator must be the same.

1/f = (1/96 * 1/1) + (1/4 * 24/24)
1/f = 1/96 + 24/96
1/f = 25/96

to find the value of f, do cross multiplication
1*96 = f * 25
96 = 25f
96/25 = f
3.84 = f

The focal length of the project lens is 3.84 inches 

4 0
3 years ago
You watch a house being built in your neighborhood. After a few months, it’s ready to be sold. From start to finish this is an e
jenyasd209 [6]
The answer is A of cours
3 0
3 years ago
Sound waves, water waves, and light waves are all alike in that they all
Paha777 [63]
Sound and water waves are longitudinal waves, they require a medium to travel through and occilate particles 90 degrees to the wave motion
Light is a transverse wave. It doesnt require a medium to travel through.
All three reflect, refract and diffract


Light is difficult to think of because it acts in ways which waves cannot explain in some cirumstances. It acts like a particle (called photons) in some conditions, but acts like a normal sound or water wave does in others. Try not to get too caught up in light being a wave or a particle because even physists dont know how to explain it yet.
7 0
3 years ago
What is the energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b?
xeze [42]

The energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b is 3.06 eV.

<h3>Change in energy level of the electron</h3>

When photons jump from a higher energy level to a lower level, they emit or radiate energy.

The change in energy level of the electrons is calculated as follows;

ΔE = Eb - Ef

ΔE = -2.68 eV - (-5.74 eV)

ΔE = 3.06 eV

Thus, the energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b is 3.06 eV.

Learn more about energy level here: brainly.com/question/14287666

#SPJ1

7 0
2 years ago
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