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Ivenika [448]
2 years ago
8

What happens to light when it travels from air into water

Physics
1 answer:
spin [16.1K]2 years ago
8 0

Answer:

Water is more dense than air. When water goes through a denser thing, the light is "bent" more towards the "normal" which is a straight, vertical line.

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A spring of spring constant k is attached to a support at the bottom of a ramp that makes an angle θ with the horizontal. A bloc
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Answer:

x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}

Explanation:

From the law of conservation of energy

Energy lost  by the spring, W=Kinetic energy gained, KE+Potential energy gained, PE+Work done by friction, Fr

0.5kd^{2}=0.5mv^{2}+mgLsin\theta+\mu_{k}mgcos\theta x

x(mgsin\theta+\mu_{k}mgcos\theta)=0.5kd^{2}

x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}

The required distance from A to B is x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}

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3 years ago
Which rf characteristic best determines the range of a 2.4 ghz ism signal?
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3 0
3 years ago
What subatomic particle acts like a mini-magnet?
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 The subatomic particles that acts like a mini-magnet is electron. Electrons are negatively charged sub atomic particles in an atom. The electron spin is a property of an electron that makes it behave like it's spinning; a spinning electron produces a magnetic field that makes it behave like a tiny magnet in an atom. 
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3 years ago
The parallel plates in a capacitor, with a plate area of 8.00 cm2 and an air-filled separation of 2.70 mm, are charged by a 8.70
MrRa [10]

Answer:

a)  ΔV₁ = 21.9 V, b) U₀ = 99.2 10⁻¹² J, c) U_f = 249.9 10⁻¹² J,  d)  W = 150 10⁻¹² J

Explanation:

Let's find the capacitance of the capacitor

         C = \epsilon_o \frac{A}{d}

         C = 8.85 10⁻¹² (8.00 10⁻⁴) /2.70 10⁻³

         C = 2.62 10⁻¹² F

for the initial data let's look for the accumulated charge on the plates

          C = \frac{Q}{\Delta V}

          Q₀ = C ΔV

           Q₀ = 2.62 10⁻¹² 8.70

           Q₀ = 22.8 10⁻¹² C

a) we look for the capacity for the new distance

          C₁ = 8.85 10⁻¹² (8.00 10⁻⁴) /6⁴.80 10⁻³

          C₁ = 1.04 10⁻¹² F

       

          C₁ = Q₀ / ΔV₁

          ΔV₁ = Q₀ / C₁

          ΔV₁ = 22.8 10⁻¹² /1.04 10⁻¹²

          ΔV₁ = 21.9 V

b) initial stored energy

          U₀ = \frac{Q_o}{ 2C}

          U₀ = (22.8 10⁻¹²)²/(2  2.62 10⁻¹²)

          U₀ = 99.2 10⁻¹² J

c) final stored energy

          U_f = (22.8 10⁻¹²) ² /(2  1.04 10⁻⁻¹²)

          U_f = 249.9 10⁻¹² J

d) the work of separating the plates

as energy is conserved work must be equal to energy change

          W = U_f - U₀

          W = (249.2 - 99.2) 10⁻¹²

          W = 150 10⁻¹² J

note that as the energy increases the work must be supplied to the system

6 0
3 years ago
What is the electric potential at a point between the 2.5-ω and 5.5-ω resistors, if the electric potential at the positive termi
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