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pochemuha
3 years ago
6

If the light strikes the plastic (from the water) at an angle θw, at what angle θa does it emerge from the plastic (into the air

)? Express your answer in terms of nw, np, na, and θw. Remember that the inverse sine of a number x should be entered as asin(x) in the answer box.
Physics
2 answers:
Natasha2012 [34]3 years ago
6 0

Answer:

\theta _a=asin(\frac{sin(\theta _w)\times n_w}{n_a})

Explanation:

According to Snell's law the angle of incidence and angle of refraction are related by:

\frac{sin(\theta _w)}{sin(\theta _a)}=\frac{n_a}{n_w}\\\\\therefore sin(\theta _a)=\frac{sin(\theta _w)\times n_w}{n_a}\\\\\theta _a=sin^{-1}(\frac{sin(\theta _w)\times n_w}{n_a})\\\\\theta _a=asin(\frac{sin(\theta _w)\times n_w}{n_a})

umka2103 [35]3 years ago
6 0

Answer:

θa = asin(nwsin(θw)/na)

Explanation:

We are going to look at the two important angles within the plastic: the angle immediately after the first refraction (the water/plastic interface) and the angle immediately before the second refraction (the plastic/air interface). In order to ascertain their relationship, draw a picture with the path the light follows in the plastic and the normals to both surfaces. Once you have labeled both angles, keep in mind that the surfaces are parallel, and thus their normals are parallel lines. An important theorem from geometry will give you the relationship between the angles.

Using Snell's Law,

θa = asin(nwsin(θw)/na)

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13) The condenser pressure is 417.4 psig for r-410A and the condenser outlet temperature is 108f. how much subcooling is there i
Tanzania [10]

Answer:

12°F

Explanation:

Calculation for how much subcooling is there in the condenser

Since the CONDENSING TEMPERATURE for 417.4 psig discharge pressure is 120 degrees (120°) which means that the amount of subcooling that is there in the condenser will be calculated using this formula

Amount of Condenser subcooling= Condensing Temperature discharge pressure -Condenser outlet temperature

Let plug in the formula

Amount of Condenser subcooling=120°-108f

Amount of Condenser subcooling=12°F

Therefore the amount of subcooling that is there in the condenser will be 12°F

8 0
2 years ago
A car enters a level, unbanked semi-circular hairpin turn of 100 m radius at a speed of 28 m/s. The coefficient of friction betw
meriva

Answer:

As  28m/s = 28m/s

Explanation:

r = the radius of the curve

m =  the mass of the car

μ = the coefficient of kinetic friction

N = normal reaction

When rounding the curve, the centripetal acceleration is

a = \frac{v^{2}}{r}

therefore

\mu mg = m \frac{v^{2}}{r} \\\\ \mu =  \frac{v^{2}}{rg}

v = \sqrt{\mu rg}

\mu = \sqrt{0.8 \times 100\times9.8} \\\\= 28m/s

As  28m/s = 28m/s

8 0
3 years ago
A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the
kow [346]

Answer:

q = - 93.334 nC

Explanation:

GIVEN DATA:

Radius of ring  73 cm

charge on ring 610 nC

ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

E  =  2000 N/C

Electric field due tor ring is guiven as

E = \frac{KQx}{[x^2+ R^2]^{3/2}}

E = \frac{9\time 10^9 \times 610\times 10^[-9} 0.70}{(0.70^2 + 0.73^2)^{3/2}}

E1 = 3714.672 N/C

electric field due to point charge q

E  =\frac[kq}{x^2}

E = \frac{9\times 10^9 \times q}{0.70^2}

E2 = 1.837\times 10^{10}\times q

now the eelctric charge at point P is

E = E1 + E22000 =  3714.672 + 1.837\times 10[10} \times q

solving for q

q = - 93.334 nC

7 0
3 years ago
Where is the potential energy equal to zero?
s2008m [1.1K]

Answer:

im sure your already past this but it's E.

Explanation:

This is because in this case potential energy is linear to height, which means that the higher the more potential energy.

4 0
3 years ago
a swimmer experiences a total (absolute) pressure of 117,500 pa in a pool. how far below the surface are they located?
marta [7]

Answer:

Explanation:

We know that the pressure can be calculated in the following way:

p = d·g·h

with d being the density of the water, g the gravitational acceleration and h the depth.

Also d of the water = 1000 kg/m^3 circa and g = 9.8 m/s^2 circa

117,500 Pa = 1000kg/m³ · 9.8m/s² · h

Therefore h = 11,9 m

4 0
2 years ago
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