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pochemuha
3 years ago
5

A laser pulse of duration 25 ms has a total energy of 1.4 J. The wavelength of this radiation is

Physics
1 answer:
SpyIntel [72]3 years ago
8 0

Answer:

n = 4 x 10¹⁸ photons

Explanation:

First, we will calculate the energy of one photon in the radiation:

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

where,

E = Energy of one photon = ?

h = Plank's Constant = 6.625 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of radiation = 567 nm = 5.67 x 10⁻⁷ m

Therefore,

E = \frac{(6.625\ x\ 10^{-34}\ J.s)(3\ x\ 10^8\ m/s)}{5.67\ x\ 10^{-7}\ m}

E = 3.505 x 10⁻¹⁹ J

Now, the number of photons to make up the total energy can be calculated as follows:

Total\ Energy = nE\\1.4\ J = n(3.505\ x\ 10^{-19}\ J)\\n = \frac{1.4\ J}{3.505\ x\ 10^{-19}\ J}\\

<u>n = 4 x 10¹⁸ photons</u>

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"Force" can be defined as a push or a<br> O pull<br> O grab
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Answer:

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4 0
3 years ago
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5.) A 2000 N bear slides down a tree at a constant velocity, what is the
PilotLPTM [1.2K]

Answer:

2000 N

1600 N

Explanation:

Step 1:

It is given that the bear's weight is 2000 N and it slides down the tree with constant velocity. Since it is sliding with constant velocity the overall force  on the bear is zero. The weight of the bear acting downward balances the upward force caused due to friction. Hence the upward force equals the weight 2000 N.  

Step 2:

It is given that the bear slides down with an acceleration 2 m/sec sq  

Weight of the bear = 2000 N

Mass of the bear = 2000/10 (taking g = 10 m/sec sq)=200 kg

Force = Mass * Acceleration

Hence net force acting downward = 200*2=400 N.

Net force = Weight of bear - Force acting upward on the bear

400 = 2000 -  Force acting upward on the bear

Force on the bear acting upward = 2000 - 400 = 1600 N

4 0
3 years ago
A sandbag motionless in outer space is hit by a three-times-as massive sandbag moving at 12 m/s. They stick together. This is an
vekshin1

Answer:

This is an example of inelastic collision and it loses zero(0) % of initial kinetic energy

Explanation:

Let the mass of the motionless sandbag = m₁

Let the mass of the moving sandbag = m₂ = 3m₁

initial velocity of the motionless sandbag, u₁ = 0

initial velocity of the moving sandbag, u₂ = 12 m/s

Let their final velocity, = v

Collision between two particles can either be elastic or inelastic.

Since they stick together after the impact, then the collision is inelastic

Apply the principle of conservation of linear momentum;

Initial kinetic energy = final kinetic energy

¹/₂m₁u₁² + ¹/₂m₂u₂² = ¹/₂v²(m₁ + m₂)

¹/₂m₁(0)² + ¹/₂(3m₁)(12)² = ¹/₂v²(m₁+3m₁)

216m₁ = 2m₁ v²

v² = 108

v = √108

v = 10.392 m/s

Change in kinetic energy = Final kinetic energy - initial kinetic energy

Initial Kinetic energy, KE₁ = ¹/₂m₁u₁² + ¹/₂m₂u₂²

                                   KE₁ = ¹/₂m₁(0)² + ¹/₂(3m₁)(12)²

                                   KE₁ = 216m₁ J

Final kinetic energy, KE₂ = ¹/₂v²(m₁ + m₂)

                                  KE₂ = ¹/₂(108)(m₁ + 3m₁)

                                  KE₂ = 216m₁ J

ΔKE = KE₂ - KE₁ =  216m₁ J -  216m₁ J = 0%

Therefore, this is an example of inelastic collision and it loses zero(0) % of initial kinetic energy

4 0
3 years ago
Which statement accurately describes CI2?
atroni [7]
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8 0
3 years ago
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A particle moves with a velocity    1 v 5j 3j 6k ms      under influence of a contact force  F 10i 10j 20k N.    
katrin2010 [14]

Answer:

P = 200 W

Explanation:

The expression for the power is

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the job

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we substitute

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we substitute in the power equation, remember that the scalar product of the unit vectors is

i.i = j.j = k.k = 1 and the other products are zero

    P = 10 5 + 10 3 + 20 6

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