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VladimirAG [237]
3 years ago
7

In a perfect vacuum, light waves travel at 300,000 kilometers every second. When light waves travel through materials, such as a

ir, they
A.speed up

B.slow down

C.increase in amplitude

D.Heat up
Physics
2 answers:
ruslelena [56]3 years ago
3 0
They slow down so the answer is B
yanalaym [24]3 years ago
3 0

c = speed of light in vacuum = 300,000 kilometers/second

n = index of refraction of material

v = speed of light in the material

Speed of light in a given material is given as

v = C/n

v = 300,000/n

for air, the value of index of refraction "n" is slightly greater, hence the speed of light in air is smaller than 300,000 kilometers/second. hence the light waves slow down.

so correct choice is

B.slow down

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ANSWER AND I WILL BUY A CAR
Natali [406]
You are exerting 100N. Since there’s no NET force, then there must be exactly 100N pushing exactly back on your 100N to cancel it to exactly zero. Newton's first law states that whether a body is at rest or travelling in a straight line at a constant speed, it will remain at rest or continue to move in a straight line at a constant speed unless acted upon by a force.
8 0
2 years ago
Consider two parallel plate capacitors. The plates on Capacitor B have half the area as the plates on Capacitor A, and the plate
vichka [17]

Answer:

CB = 4.45 x 10⁻⁹ F = 4.45 nF

Explanation:

The capacitance of a parallel plate capacitor is given by the following formula:

C = ε₀A/d

where,

C = Capacitance

ε₀ = Permeability of free space

A = Area of plates

d = Distance between plates

FOR CAPACITOR A:

C = CA = 17.8 nF = 17.8 x 10⁻⁹ F

A = A₁

d = d₁

Therefore,

CA = ε₀A₁/d₁ = 17.8 x 10⁻⁹ F   ----------------- equation 1

FOR CAPACITOR B:

C = CB = ?

A = A₁/2

d = 2 d₁

Therefore,

CB = ε₀(A₁/2)/2d₁

CB = (1/4)(ε₀A₁/d₁)

using equation 1:

CB = (1/4)(17.8 X 10⁻⁹ F)

<u>CB = 4.45 x 10⁻⁹ F = 4.45 nF</u>

5 0
2 years ago
Physics 1
Degger [83]

The emf is induced in the wire will be 1.56 ×10 ⁻³ V. The induced emf is the product of the magnetic field,velocity and length of the wire.

<h3>What is induced emf?</h3>

Emf is the production of a potential difference in a coil as a result of changes in the magnetic flux passing through it.

When the flux coupling with a conductor or coil changes, electromotive Force, or EMF, is said to be induced.

The given data in the problem is;

B is the magnitude of the magnetic field,= 5.0 ×10⁻⁵ T

V(velocity)=125 M/SEC

L(length)=25 cm=0.25 m

The maximum emf is found as;

E=VBLsin90°

E=125 × 5.0 × 10⁻⁵ ×0.25

E=1.56 ×10 ⁻³ V

Hence, the emf is induced in the wire will be 1.56 ×10 ⁻³ V

To learn more about the induced emf, refer to the link;

brainly.com/question/16764848

#SPJ1

7 0
2 years ago
An empty graduated cylinder weighs 55.26 g. When filled with 48.1 mL of an unknown liquid, it weighs 92.39 g. The density of the
katrin [286]
<h2>Density of the unknown liquid is 771.93 kg/m³</h2>

Explanation:

An empty graduated cylinder weighs 55.26 g

Weight of empty cylinder = 55.26 g = 0.05526 kg

Volume of liquid filled = 48.1 mL = 48.1 x 10⁻⁶ m³

Weight of cylinder plus liquid = 92.39 g = 0.09239 kg

Weight of liquid = 0.09239 - 0.05526

Weight of liquid = 0.03713 kg

We have

        Mass = Volume x Density

        0.03713 = 48.1 x 10⁻⁶ x Density

        Density = 771.93 kg/m³

Density of the unknown liquid is 771.93 kg/m³

5 0
3 years ago
Two long, parallel wires are attracted to each other by a force per unit length of 350 µN/m. One wire carries a current of 22.5
pishuonlain [190]

Answer

given,

force per unit length = 350 µN/m

current, I = 22.5 A

y = y = 0.420 m

\dfrac{F}{L}= \dfrac{KI_1I_2}{d}

I_2 = \dfrac{F}{L}\dfrac{d}{KI_1}

I_2 = 350\times 10^{-6}\times \dfrac{0.42}{2 \times 10^{-7}\times 22.5}

    I₂ = 32.67 A

distance where the magnetic field is zero

\dfrac{4\pi \times 10^{-7}\times 32.67}{2\pi y_1}=\dfrac{4\pi \times 10^{-7}\times 22.5}{2\pi (0.42-y_1)}

y_1 = 0.248\ m

there the distance at which the magnetic field is zero in the two wire is at 0.248 m.

3 0
3 years ago
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