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Natasha2012 [34]
3 years ago
13

What type of interference is happening when light waves add together to

Physics
1 answer:
Eva8 [605]3 years ago
3 0

Answer:

They become brighter in constructive interference because the wave paths donot cancel out, but just strengthen to give a brighter light.

No diagram

You might be interested in
The energy from light is used by plants to oxidize which molecule?
Natasha2012 [34]
Photoautotrophs use light energy to convert carbon dioxide into organic compounds. This process is called photosynthesis. Chemoautotrophs extract energy from inorganic compounds by oxidizing them and use this chemical energy, rather than light energy, to convert carbon dioxide into organic compounds
5 0
4 years ago
A gas expands from an initial volume of 0.040 m^3 and an initial pressure of 210 kPa to a final volume of 0.065 m^3 while its te
USPshnik [31]

Answer:

286

Explanation:

p1v1/T1=p2v2/T2

then subtitude your values

T2=760*0.65*273/210*0.040

T2=135/8.4=16+273=289

7 0
4 years ago
In trial 1 of an experiment, a cart moves with a speed of vo on a frictionless, horizontal track and collides with another cart
marta [7]

Answer:

1) elastic shock, the velocity of the center of mass does not change

2) inelastic shock, he velocity of the mass center   change

Explanation:

The position of the center of mass of your system is defined by

          x_{cm} = \frac{1}{M} \sum x_i m_i

in this case we have two bodies

          x_{cm} = \frac{1}{M} (x₁m₁ + x₂ m₂)

the velocity of the center of mass is

          x_{cm} = dx_{cm} / dt = \frac{1}{M} ( m_1 \frac{dx_1}{dt} \ + m_2 \frac{dx_2}{dt} )

          x_{cm} = \frac{1}{M} ( m_1 v_1 + m_2 v_2 )

where M is the total mass of the system.

Therefore to answer this question we have to find the velocity of the body after the collision.

Let's use momentum conservation, where the system is formed by the two bodies, so that the forces have been internal during the collision.

Let's solve each case separately.

2) inelastic shock

initial instant. Before the crash

         p₀ = m₁ v₀ + 0

final instant. After the collision with the cars together

        p_f = (m₁ + m₂) v

         p₀ = p_f

         m₁ v₀ = (m₁ + m₂) v

         v = \frac{m_1}{m_1+m_2}  v₀

let's find the velocity of the center of mass

         M = m₁ + m₂

initial.

         v_{cm o} = \frac{1}{m_1 +m_2} (m₁ vo)

final

         v_{cm f} = \frac{1}{M} ( \frac{m_1}{m_1 + m_2} v_o ) ( v) = v

         v_{cm f} =  \frac{m_1}{M^2} v_o

Let's find the ratio of the velocities of the center of mass

          vcmf / vcmo = \frac{1}{M} = \frac{1}{m_1 +m_2}

           

           

therefore the velocity of the mass center   change

1) elastic shock

initial instant.

           p₀ = m₁ v₀

final moment

           p_f = m₁ v_{1f} + m₂ v_{2f}

           p₀ = p_f

           m₁ v₀ = m₁ v_{1f} + m₂ v_{2f}

           m₁ (v₀ - v_{2f}) = m₂ v_{2f}

in this case the kinetic energy is conserved

           K₀ = K_f

          ½ m₁ v₀² = ½ m₁ v_{1f}² + ½ m₂ v_{2f}²

           m₁ (v₀² - v_{1f}²) = m₂ v_{2f}²

           m₁ (v₀ + v_{1f}) (v₀ - v_{1f}) = m₂ v_{2f}

we write our system of equations

           m₁ (v₀ - v_{1f}) = m₂ v_{2f}             (1)

           m₁ (v₀ - v_{1f}) (v₀ + v_{1f}) = m₂ v_{2f}²

we solve the system

             v₀ + v_{1f} = v_{2f}

we substitute and look for the final speeds

             v_{1f} = \frac{m_1 -m_2}{m1 +m2 } v_o

             v_{2f} = \frac{2 m_1}{m-1+m_2} vo

now let's find the velocity of the center of mass

initial

          v_{cm o} = \frac{1}{M} m₁ v₀

final

          v_{cm f} = \frac{1}{M}  (m₁ v_{1f} + m₂ v_{2f} )

          v_{cm f} = \frac{1}{M} [  m_1  \frac{m_2}{M} + m_2  \frac{2 m_1}{M} ] v₀

          v_{cm f} = \frac{1}{M^2} ( m₁² - m₁m₂ +2 m₁m₂) v₂

          v_{cm f} = \frac{1}{M^2} (m₁² + m₁ m₂) v₀

let's look for the relationship

         v_{cm f} / v_{cm o} = \frac{1}{M} M

         v_{cm f} / v_{cm o} = 1

therefore the velocity of the center of mass does not change

we see in either case the velocity of the center of mass does not change.

4 0
3 years ago
Jane on her way to work walks 3 meters east, 8 meters north, and, another 3 meters east. As shown in the diagram. What is the ma
kifflom [539]
<h2><u><em> 30 meters wide and its current flows </em></u></h2><h2><u><em>Explanation:</em></u></h2><h2><u><em>Why?, Because the area of those numbers are clear to be 30. Now i'm not sure of this answer on being correct so. You should also look into other answers as well.</em></u></h2>
3 0
3 years ago
Using the method of dimension, derive an expression for the velocity of sound waves. assume that the velocitu depends on:
Soloha48 [4]

Answer:

v = K √(E / ρ)

Explanation:

Modulus of elasticity has units of N/m², or kg/m/s².

Density has units of kg/m³.

Velocity has units of m/s.

If we divide modulus of elasticity by density, we can eliminate kg:

E / ρ = [kg/m/s²] / [kg/m³]

E / ρ = [m²/s²]

Taking the square root gets us units of velocity:

√(E / ρ) = [m/s]

Multiply by the constant K:

v = K √(E / ρ)

7 0
3 years ago
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