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Tema [17]
3 years ago
14

A pair of narrow slits, separated by 1.8 mm, is illuminated by a monochromatic light source. Light waves arrive at the two slits

in phase. A fringe pattern is observed on a screen 4.8 m from the slits.
In Situation 26.2, there are 5.0 bright fringes/cm on the screen. The wavelength of the monochromatic light is closest to:_______.
A) 550 nm
B) 600 nm
C) 650 nm
D) 700 nm
E) 750 nm
Physics
1 answer:
Katyanochek1 [597]3 years ago
5 0

Answer:

Answer:

750 nm

Explanation:

= separation of the slits = 1.8 mm = 0.0018 m

λ = wavelength of monochromatic light

= screen distance = 4.8 m

= position of first bright fringe =

= order = 1

Position of first bright fringe is given as

λ = 7.5 x 10⁻⁷ m

λ = 750 nm

Explanation:

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8. A 40.0kg block of metal is suspended from a scale and is immersed in water. The dimensions of the block are 12.0cm x 10.0cm x
Alja [10]

Answer:

a.1017.9N

b.1029.7N

c.T=380.6N

d.11.8 N

Explanation:

(a) The absolute pressure at the level of the top of the block is

Po=atmospheric pressure

g=gravity

h1=height

rh0=density of water

P1=Po+rho*g*h1

1.013*10^5+1000(9.81)(0.05)

1.0179*10^5Pa

at the level of the bottom of the block we have

P2=Po+rho*g*h2

1.013*10^5+1000(9.81)(0.17)

1.0297* 10^5 Pa

the downward force exerted on the top by the water is

Ftop=P*A== × = 1.0179* 10^5* 0.100

= 1017.9 N

and the upward force the water exerts on the bottom of the block , which is the buoyant force

Fbot== × = 1.0297 10^5 * 0.100 m

= 1029.7 N

(b) The scale reading is the tension, T, in the cord supporting the block.

if the block is at equilibrium, then sum of vertical forces

EFy=T+Fbot-Ftop-mg=0

T=mg+Ftop-Fbot

T=40*9.81-(1029.7-1017.9)

T=380.6N

(c)  Archimedes principle state that, the buoyant force on the block equals the weight of the displaced water. Thus,

Buoyant force=rho *g*h

= = 1000* 0.100^2 * 0.120 m *9.80 m s=11.8 N

from the answer a Ftop-Fbot

1029.7-1017.9=11.8N, the same as the buoyant force

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3 years ago
5.54 Two kilograms of air within a piston–cylinder assembly WP execute a Carnot power cycle with maximum and minimum temper atur
Step2247 [10]

Answer:

A) 60%

B) p2 = 1237.2 kPa

   v2 = 0.348 m^3

C) w1-2 = w3-4 = 1615.5 kJ

   Q2-3 = 60 kJ

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  Л = 1 - \frac{Tl}{Th}  

where Tl = 300 k

            Th = 750 k

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= P3v3 = mRT3  ----- (1)

v3 = 0.4m , mR = 2* 0.287, T3 = 750

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60 = 1076.25 * 0.4 ln(p2/p3)

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calculate volume ( V2 )

p2v2 = p3v3

v2 = p3v3 / p2

   = (1076.25 * 0.4 ) /  1237.2  

  = 0.348 m^3

C) calculate the work and heat transfer for each four processes

work :

W1-2 = mCv( T2 - T1 )

        = 2*0.718 ( 750 - 300 ) = 1615.5 kJ

W3-4 = 1615.5 kJ

heat transfer

Q2-3 = W2-3 = 60KJ

Q3-4 = 0

D ) sketch of the cycle on p-V coordinates

attached below  

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