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marusya05 [52]
3 years ago
11

When light is incident normally on the interface between two transparent optical media, the intensity of the reflected light is

given by the expression(equation 1)In this equation S1 represents the average magnitude of the Poynting vector in the incident light (the incident intensity), S’1 is the reflected intensity, and n1 and n2 are the refractive indices of the two media. The fraction of incident intensity is 0.0246 or 2.46 precent for 589-nm light.For light normally incident on an interface between vacuum and a transparent medium of index n is given by S2/S1 = 4n/(n + 1)2 (equation 2). (a) Light travels perpendicularly through a diamond slab, surrounded by air, with parallel surfaces of entry and exit. Apply the tramsmission fraction in equation 2 to find the approximate overall transmission through the slab of diamond, as a percentage. Ignore light relected back and forth within the slab.(b) Light travels perpendicularly through a diamond slab, surrounded by air, with parallel surfaces of entry and exit. The intensity of the transmitted light is what fraction of the incident intensity? Include the effects of light reflected back and forth inside the slab.If someone could answer part (a) and (a) I would gladly appreciate it.
Physics
1 answer:
Alex777 [14]3 years ago
8 0

i need to ask sum so oycixgixigdtdidit

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Answer:

b

Explanation:

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A weightlifter liftsa 1,250-N barbell 2 m in 3 s. how much power was used to lift the barbell?
STatiana [176]
Power = Force * Distance/ time
P = 1,250 * 2/3
P = 2,500/3
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So, your final answer is 833.33 Watts
5 0
3 years ago
A copper rod of cross-sectional area 11.6 cm2 has one end immersed in boiling water and the other in an ice-water mixture, which
julia-pushkina [17]

Answer:

0.686 g of ice melts each second.

Solution:

As per the question:

Cross-sectional Area of the Copper Rod, A = 11.6\ cm^{2} = 11.6\times 10^{- 4}\ m^{2}

Length of the rod, L = 19.6 cm = 0.196 m

Thermal conductivity of Copper, K = 390\ W/m.^{\circ}C

Conduction of heat from the rod per second is given by:

q = \frac{KA\Delta T}{L}

where

\Delta T = 100^{\circ} - 0^{\circ} = 100^{\circ}C = temperature difference between the two ends of the rod.

Thus

q = \frac{390\times 11.6\times 10^{- 4}\times 100}{0.196} = 228.48\ J/s

Now,

To calculate the mass, M of the ice melted per sec:

M = \frac{q}{L_{w}}

where

L_{w} = Latent heat of fusion of water = 333 kJ/kg

M = \frac{228.48}{333\times 10^{3}} = 6.86\times 10^{- 4}\ kg = 0.686\ g

5 0
3 years ago
A 2000-kg car moving with a speed of 20 m/s collides with and sticks to a 1500-kg car at rest at a stop sign. Show that because
amid [387]

Answer:

13.33m/s

Explanation:

Given data

m1= 2000kg

u1= 20m/s

m2= 1500kg

u2= 0m/s

v1= 10m/s

Required

The speed of the sticks

We know that  from the expression for the conservation of momentum

m1u1+m2u2= m1v1+m2v2

2000*20+1500*0=2000*10+1500*v2

40000=20000+1500v2

collect like terms

40000-20000= 1500v2

20000= 1500v2

v2= 20000/1500

v2= 13.33 m/s

Hence the velocity of the sticks is 13.33m/s

8 0
3 years ago
William WangHomework #33 Regents Review (3)0989Assignment Mode : Open (Time On Task) 9 of 25 ListenDuring a collision, an 84-kil
Lemur [1.5K]

Answer:

1.7\cdot 10^3 N

Explanation:

The impulse theorem states that the product between the force and the time interval of the collision is equal to the change in momentum:

F \Delta t = m \Delta v

where

F is the force

\Delta t is the time interval

m is the mass

\Delta v is the change in velocity

Here we have

m = 84 kg

\Delta t = 1.2 s

\Delta v = 24 m/s

So we can solve the equation to find the force:

F= \frac{m \Delta v}{\Delta t }=\frac{(84 kg)(24 m/s)}{1.2 s}=1680 N \sim 1.7\cdot 10^3 N

4 0
2 years ago
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