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Agata [3.3K]
2 years ago
7

What is the photon energy (eV) of indigo light​

Physics
1 answer:
e-lub [12.9K]2 years ago
7 0

Answer:

When dealing with "particles" such as photons or electrons, a commonly used unit of energy is the electron-volt (eV) rather than the joule (J). An electron volt is the energy required to raise an electron through 1 volt, thus a photon with an energy of 1 eV = 1.602 × 10-19 J.

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Momentum is usually not exactly conserved in a real world demonstration of momentum conservation. What is a possible reason for
Maksim231197 [3]

Answer:

For any collision occurring in an isolated system, momentum is conserved. The total amount of momentum of the collection of objects in the system is the same before the collision as after the collision.

Explanation:

Hope this helps

5 0
3 years ago
Standing still, Bruce, the quarterback, gets tackled by Biff, the 90.0-kg tackle, who is traveling at 7.0 m/s. Upon collision, B
tatyana61 [14]

\\ \sf\longmapsto \Delta P=P

\\ \sf\longmapsto m1v1=m2v2

\\ \sf\longmapsto 90(7)=m2(10)

\\ \sf\longmapsto 10m2=630

\\ \sf\longmapsto m2=\dfrac{630}{10}

\\ \sf\longmapsto m2=63kg

Bruces mass is 63kg

3 0
2 years ago
Read 2 more answers
Potassium-40 has a half-life of approximately 1.25 billion years. Approximately how many years will pass before a sample of pota
Maksim231197 [3]

Answer:

3.75 billion years

Explanation:

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

Half-life (t½) = 1.25 billion years

Number of half-lives (n) = 3

Time (t) =?

The time taken for the sample of potassium-40 to contains one-eighth the original amount of parent isotope can be obtained as:

n = t / t½

3 = t / 1.25

Cross multiply

t = 3 × 1.25

t = 3.75 billion years.

Therefore, it will take 3.75 billion years for the sample of potassium-40 to contains one-eighth the original amount of parent isotope

8 0
2 years ago
The 49-g arrow is launched so that it hits and embeds in a 1.45 kg block. The block hangs from strings. After the arrow joins th
Dimas [21]

Answer:

the initial speed of the arrow before joining the block is 89.85 m/s

Explanation:

Given;

mass of the arrow, m₁ = 49 g = 0.049 kg

mass of block, m₂ = 1.45 kg

height reached by the arrow and the block, h = 0.44 m

The gravitational potential energy of the block and arrow system;

P.E = mgh

P.E = (1.45 + 0.049) x 9.8 x 0.44

P.E = 6.464 J

The final velocity of the system after collision is calculated as;

K.E = ¹/₂mv²

6.464 = ¹/₂(1.45 + 0.049)v²

6.464 = 0.7495v²

v² = 6.464 / 0.7495

v² = 8.6244

v = √8.6244

v = 2.937 m/s

Apply principle of conservation of linear momentum to determine the initial speed of the arrow;

P_{initial} = P_{final}\\\\mv_{arrow} + mv_{block} = (m_1 + m_2)V\\\\0.049(v) + 1.45(0) = (0.049 + 1.45)2.937\\\\0.049v = 4.4026\\\\v = \frac{4.4026}{0.049} \\\\v = 89.85 \ m/s

Therefore, the initial speed of the arrow before joining the block is 89.85 m/s

4 0
2 years ago
First, let us consider an object launched vertically upward with an initial speed v. Neglect air resistance. Part B) At the top
harina [27]

Answer:

Option d is the correct option Kinetic energy is minimum while as potential energy is maximum

Explanation:

At the top most point of the flight since it cannot reach any further up in the vertical direction thus the potential energy at this position shall be maximum. Now since the total energy of the projectile is conserved so the remaining kinetic energy shall be minimum at that point so as the sum of the kinetic and potential energies remain constant.

3 0
3 years ago
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