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nata0808 [166]
2 years ago
15

A light ray passing from medium 1 to medium 2 is bent away from the perpendicular normal to the boundary surface. What happens t

o the velocity of the refracted wave as it enters medium 2?
The velocity increases.


The velocity decreases.


The velocity becomes zero.


The velocity does not change.
Physics
2 answers:
Gre4nikov [31]2 years ago
8 0

Answer: Velocity Increases

Explanation: Plato

SVETLANKA909090 [29]2 years ago
6 0

Answer:

The answer is a for Plato users.

Explanation:

Since the angle of the refracted ray moves away from the normal, it must be traveling in a faster medium.

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Hi:) anyone able to help with part - aii the one on cooling fins ? Thankss!
taurus [48]

The cooling find are there to keep the temperature low. They are mostly black in color, and hence, radiate heat, which is how the temperature is controlled.

5 0
3 years ago
What is the difference between a bound orbit and an unbound orbit around the sun?
Anit [1.1K]

Answer:

Explanation:

The difference between a bound orbit and an unbound orbit around the sun is that:

An object on a bound orbit pursues the same way around the Sun again and again, while an object on an unbound orbit moves toward the Sun only a single time and afterward stays away forever &  never returns.

8 0
3 years ago
Refrigerant-134a enters the expansion valve of a refrigeration system at 160 psia as a saturated liquid and leaves at 30 psia. D
KatRina [158]

Answer:

Temperature : 92.9 F

Internal Energy change: -2.53 Btu/lbm

Explanation:

As

mh1=mh2

h1=h2

In table A-11 through 13E

p2=120Psi, h1= 41.79 Btu/lbm,

u1=41.49

So T1=90.49 F

P2=20Psi

h2=h1= 41.79 Btu/lbm

T2= -2.43F

u2= 38.96 Btu/lbm

T2-T1 = 92.9 F

u2-u1 = -2.53 Btu/lbm

3 0
2 years ago
A horizontal wire of length 0.53 m. carrying a current of 7.5 A. is placed in a uniform external magnetic field. When the wire i
Lady_Fox [76]

Answer:

3.4 mT

Explanation:

L = 0.53 m

i = 7.5 A

Theta = 19 degree

F = 4.4 × 10^-3 N

Let B be the strength of magnetic field.

Force on a current carrying conductor placed in a magnetic field.

F = i × L × B × Sin theta

4.4 × 10^-3 = 7.5 × 0.53 × B × Sin 19

B = 3.4 × 10^-3 Tesla

B = 3.4 mT

6 0
2 years ago
To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following general equation y(
DerKrebs [107]

Answer:

0.0549 m

Explanation:

Given that

equation y(x,t)=Acos(kx−ωt)

speed  v = 8.5 m/s

amplitude A = 5.5*10^−2 m

wavelength λ   = 0.5 m

transverse displacement = ?

v = angular frequency / wave number

and

wave number = 2π/ λ

wave number =  2 * 3.142 / 0.5

wave number = 12.568

angular frequency = v k

angular frequency = 8.5 * 12.568

angular frequency = 106.828 rad/sec ~= 107 rad/sec

so

equation y(x,t)=Acos(kx−ωt)

y(x,t)= 5.5*10^−2 cos(12.568 x−107t)

when x =0 and and t = 0

maximum y(x,t)= 5.5*10^−2 cos(12.568 (0) − 107 (0))

maximum y(x,t)= 5.5*10^−2  m

and when x =  x = 1.52 m and t = 0.150 s

y(x,t)= 5.5*10^−2 cos(12.568 (1.52) −107(0.150) )

y(x,t)= 5.5*10^−2 × (0.9986)

y(x,t) = 0.0549 m

so the transverse displacement is  0.0549 m

5 0
3 years ago
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