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Evgen
2 months ago
5

What can happen to an electron when sunlight hits it? select all that apply. select all that apply. it can drop down to a lower

electron shell. it can move out to a higher electron shell. it can collide with the nucleus. it can stay in its original shell?
Physics
2 answers:
ValentinkaMS [3.4K]2 months ago
8 0
When sunlight strikes an electron, it absorbs energy, making moving down to a lower shell impossible. Instead, it can jump to a shell with higher energy. It won't collide with the nucleus, and it could also remain where it is if it doesn't gain sufficient energy to move upwards. 
ValentinkaMS [3.4K]2 months ago
4 0
Two outcomes are possible:
<span>- the electron might jump to a higher energy level (electron shell)
- or it may remain in its current shell
</span><span>
Sunlight is composed of photons; when these photons strike an electron, the electron can absorb one, gaining energy. If the energy absorbed equals or exceeds the gap between shells, the electron elevates to a higher shell. If the photon's energy falls short, the electron stays put within its original shell.</span>
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A string is stretched by two equal but opposite forces f newton each what is tension in string
Maru [3345]

The string does not experience any force of tension, as it balances two forces acting in the same direction. Hence, the tension is zero.

Explanation:

If tension existed in the string, it would mean that two equal but opposite forces are exerting pull in contrary directions.

When a force of f newtons is applied from the right and another force of f newtons from the left, the resulting action occurs through one force. Because there is action on the same string in opposing directions, the tension in the string can only be equal to the magnitude of the string itself.

Therefore, the string indeed has no tension since it is dealing with two forces acting in the same direction. Thus, the tension is zero.

8 0
1 month ago
A turntable of radius R1 is turned by a circular rubberroller of radius R2 in contact with it at their outeredges. What is the r
serg [3582]
The ratio is defined as \frac{w_1}{w_2} = \frac{R_2}{R_1}

. Explanation: The problem indicates that

is the first radius while R_1 is the second radius. The angular velocity of the turntable can be mathematically denoted as , and for the rubber roller it is represented as . In both cases, v_k represents the speed of both the turntable and rubber roller.
5 0
19 days ago
A 100-watt light bulb radiates energy at a rate of 100 J/s. (The watt, a unit of power or energy over time, is defined as 1 J/s.
Keith_Richards [3271]

Answer:

2.64\times 10^{20} The number of photons emitted each second is

Explanation:

Let 'n' stand for the quantity of photons released by the bulb.

Provided Information:

The bulb radiates energy at a rate of 100 J per second (E).

Wavelength of emitted light is (λ) = 525 nm = 525\times 10^{-9}\ m

The energy of a photon is calculated by:

Where,

E_0=\frac{hc}{\lambda}

Now, if we have 'n' photons, the total energy is equivalent to the energy of a single photon multiplied by the count of photons. Thus,

h\to Planck's\ constant=6.626\times 10^{-34}\ Js\\\\c\to Speed\ of \ light=3\times 10^{8}\ m/s

To express in terms of 'n', we find:

E=nE_0\\\\E=\frac{nhc}{\lambda}

Insert the provided values and solve for 'n'. The resulting calculation yields

n=\frac{E\lambda}{hc}

Consequently,

photons are discharged every second.n=\frac{100\times 525\times 10^{-9}}{6.626\times 10^{-34}\times 3\times 10^{8}}\\\\n=2.64\times 10^{20}

2.64\times 10^{20}

8 0
1 month ago
Dane is holding an 8 kilogram box 2 metres above the ground. How much energy is in the box's gravitational potential energy stor
Sav [3153]

Answer:160j

Explanation:

PE=mgh

where m=8kg

g=10N/kg

h=2m

Thus, PE=8*10*2

This results in PE=160j

7 0
1 month ago
Two identical loudspeakers are driven in phase by the same amplifier. The speakers are positioned a distance of 3.2 m apart. A p
Keith_Richards [3271]
The frequency is calculated to be 735 Hz. Given: Person's distances to speakers are r₁ = 4.1 m and r₂ = 4.8 m. The path difference calculates to d = r₂ - r₁ = 4.8 - 4.1 = 0.7 m. In terms of destructive interference, we derive: λ = v/f, where v is the sound speed of 343 m/s. Using n = 1 gives f = 245 Hz, and for n = 3, f = 735 Hz. Thus, the second lowest frequency for destructive interference is 735 Hz.
7 0
17 days ago
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