A waterbed has a force of 1300N on the floor. It exerts 347 Pa of pressure. What is the area of the waterbed?
2 answers:
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Bohr's equation for the change in energy is
where
h = Planck's constant
c == the velocity of light
λ = wavelength.
The velocity is related to wavelength and frequency, f, by
c = fλ
Let us examine the given answers on the basis of the given equations.
a. As λ increases, f decreases and ΔE decreases.
TRUE
b. As λ increases, f increases and ΔE increases.
FALSE
c. As λ increases, f increases and ΔE decreases.
FALSE
Answer:
As the wavelength increases, the frequency decreases and energy decreases.
where
h = Planck's constant
c == the velocity of light
λ = wavelength.
The velocity is related to wavelength and frequency, f, by
c = fλ
Let us examine the given answers on the basis of the given equations.
a. As λ increases, f decreases and ΔE decreases.
TRUE
b. As λ increases, f increases and ΔE increases.
FALSE
c. As λ increases, f increases and ΔE decreases.
FALSE
Answer:
As the wavelength increases, the frequency decreases and energy decreases.
Answer:
<h2>a) 50°</h2><h2>b) 40°</h2>Explanation:
Check the complete diagram n the attachment below
a) The angle of incidence on a plane surface is the angle between the incidence ray and the normal ray acting on a plane surface. The normal ray is the ray perpendicular to the surface while the incidence ray is the ray striking a plane surface.
According to the diagram, the angle of reflection r₂ on M₂ is 90°-g where g is the angle of glance.
Given angle of glance on M₂ to be 40°, r₂ = 90-40 = 50°
According the second law of reflection, the angle of incidence = angle of reflection, therefore i₂ = r₂ = 50° (on M₂)
Also ∠OO₂O₁ = ∠OO₁O₂ = 40° (angle of glance on M₁){alternate angle}
The angle of incidence on M₁ = 90° - 40° = 50°
b) The angle of incidence to the surface of M₁(∠PO₁A)will be the angle of glance on M₁ which is equivalent to 40°
