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GaryK [48]
4 years ago
5

A laser beam of wavelength 600 nm is incident on two slits that are separated by 0.02 mm. What is the separation between adjacen

t bright maxima observed on a screen 5 m away?(A) 1.1 mm (B) 0.625 mm (C) 20 mm (D) 15 cm (E) 0.1 mm
Physics
1 answer:
Liula [17]4 years ago
4 0

Answer:

option D

Explanation:

given,

wavelength = 600 nm

width of separation = 0.02 mm

L = 5 m

for mth order maxima

d \times \dfrac{y_m}{L}=m\lambda

for (m+1)th order maxima

d \times \dfrac{y_{m+1}}{L}=(m+1)\lambda

now,

y_m=\dfrac{mL\lambda}{d}      and

y_{m+1}=\dfrac{(m+1)L\lambda}{d}

hence,

\Delta y = y_{m+1} - y_m

\Delta y =\dfrac{L\lambda}{d}

\Delta y =\dfrac{5 \times 600 \times 10^{-9}}{0.02 \times 10^{-3}}

\Delta y =0.15\ m

\Delta y =15\ cm

hence, the correct answer is option D

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Answer:

It speeds up, and the angle increases.

Explanation:

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4 years ago
Maria is filling a bucket of water from a faucet. After she turns it on, she sees that the cross-sectional area of the water str
GenaCL600 [577]

Answer:

t = 47.62 sec

Explanation:

Given data;

A_1 = 4.62 \times 10^4 m^2

A_2 = 2.52 \times 10^4 m^2

h = 2.50 cm

volume 10 L

from

A_1 v_1 = A_2 v_2

4.62 \times 10^4 v_1 = 2.52 \times 10^4 v_2

4.62 v_1 = 2.52 v_2 ......1

from bernoulli eq

P_1 + \frac{1}{2} \rho v_1^2 + \rho g h = P_2 + \frac{1}{2} \rho v_2^2

P_1 =P_2 = P_{atm}

v_2^2 = v_1^2 +2gh ... 2

from 1 and 2 equation

v_1 = 0.46 m/s

volume flow rate is

Q = A_1 \times v_1 = 4.62 \times 10^[-4} v_1 = 2.1 \times 10^{-4} m^3/s

t  = \frac{v}{Q}

t =\frac{10\times 10^{-3}}{2.1 \times 10^{-4}} = 47.62 s

3 0
3 years ago
Read 2 more answers
A supertanker filled with oil has a total mass of 6.1 x108 kg. If the dimensions of the ship are those of a rectangular box 300
IrinaVladis [17]

Answer:

The bottom of the sea is 25 m below sea level.

Explanation:

Given data

Mass = 6.1 × 10^{8} \ kg

\rho_{sea} = 1020\  \frac{kg}{m^{3} }

We know that Buoyant force on the tank is equal to gravity force of the tank.

F_B = F_g

(\rho_{Fluid}) (g) (V_{disp}) = m g

(\rho_{Fluid})  (V_{disp}) = m

1020 × V_{disp} = 6.1 × 10^{8}

V_{disp} = 598039.21 m^{3}

We know that

V_{disp} = W × L × H

598039.21 = 300 × 80 × H

H = 25 m

Therefore the bottom of the sea is 25 m below sea level.

7 0
4 years ago
(a) You short-circuit a 20 volt battery by connecting a short wire from one end of the battery to the other end. If the current
erica [24]

(a) 1.11 \Omega

When the battery is short-cut, the only resistance in the circuit is the internal resistance of the battery. Therefore, we can apply Ohm's law:

r=\frac{V}{I}

where

V = 20 V is the voltage across the internal resistance of the battery

I = 18 A is the current flowing through it

Solving the equation,

r=\frac{20 V}{18 A}=1.11\Omega

(b) 360 W

The power generated by the battery is given by the equation

P=VI

where

V = 20 V is the voltage of the battery

I = 18 A is the current

Substituting into the formula,

P=(20 V)(18 A)=360 W

(c) 360 J

The energy dissipated by the internal resistance is given by

E=Pt

where

P = 360 W is the power generated

t = 1 s is the time

Solving the equation, we find

E=(360 W)(1 s)=360 J

(d) 1.65 A

The battery is now connected to a R=11 \Omega resistor. This means that the internal resistance of the battery is now connected in series with the other resistor R: so, the total resistance of the circuit is

R_T = r+R=1.11 \Omega +11 \Omega = 12.11 \Omega

And so, the current flowing through the circuit is

I=\frac{V}{R_T}=\frac{20 V}{12.11\Omega}=1.65 A

(e) 29.9 W

The power dissipated in the external resistor is given by

P=I^2 R

where

I = 1.65 A is the current

R=11 \Omega is the resistance

Solving the equation, we find

P=(1.65 A)^2(11 \Omega)=29.9 W

(f) 18.17 V

The terminals of the voltmeter are placed at the two end of the battery. The battery provides an emf of 20 V, however due to the internal resistance, some of this voltage is dropped across the internal resistance. Therefore, the actual potential difference that will be read by the voltmeter will be:

V=\epsilon - Ir =20 V -(1.65 A)(1.11 \Omega)=18.17 V

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3 years ago
The acceleration of a particle is directly proportional to the square of the time t. When t =0, the particle is at x =24 m. Know
AlexFokin [52]

Answer:

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a(t) = C₁t²

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v(t) = ⅓C₁t³ + C₂

position is the integral of velocity

x(t) = (⅟₁₂C₁)t⁴ + C₂t + C₃

x(0) = 24 = (⅟₁₂C₁)0⁴ + C₂0 + C₃

C₃ = 24

x(6) = 96 = (⅟₁₂C₁)6⁴ + C₂6 + 24

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         C₂ = 12 - 18C₁

v(6) = 18 = ⅓C₁6³ + C₂

          18 = 72C₁ + C₂

          18 = 72C₁ + (12 - 18C₁)

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           C₁ = 1/9

           C₂ = 12 - 18(1/9)

           C₂ = 10

4 0
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