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andrey2020 [161]
3 years ago
9

A single slit is illuminated by light of wavelengths λa and λb, chosen so that the first diffraction minimum of the λa component

coincides with the second minimum of the λb component. (a) If λb = 370 nm, what is λa? For what order number mb (if any) does a minimum of the λb component coincide with the minimum of the λa component in the order number (b) ma = 4 and (c) ma = 5?
Physics
1 answer:
ElenaW [278]3 years ago
6 0

Answer:

Explanation:

First minimum of λa will be formed at 1 x λa D/d distance from the centre where D is distance of screen and d is width of slit.

As per given statement

λa D/d = 2 x λb D/d

λa  = 2 x λb

= 2 x 370 nm

= 740 nm

Let ma  th order of λa component coincides with mb the order of λb component

ma λa D/d = mb λb D/d

ma / mb = λb / λa

ma / mb = 1/ 2

mb = 2 ma

if ma = 4

mb = 8

if ma = 5

mb = 10

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How much work is done (by a battery, generator, or some other source of potential difference) in moving Avogadro's number of ele
konstantin123 [22]

Answer:

-1486 KJ

Explanation:

The work done by an electric field on a charged body is:

W = ΔV * q

where ΔV = change in voltage

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The total charge of Avogadro's number of electrons is:

6.0221409 * 10^(23) * -1.6023 * 10^(-19) = -9.65 * 10^(4)

The change in voltage, ΔV, is:

9.20 - (6.90) = 15.4

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A horizontal spring-mass system has low friction, spring stiffness 165 N/m, and mass 0.6 kg. The system is released with an init
AURORKA [14]

a) 19.4 cm

b) 3.2 m/s

Explanation:

a)

A horizontal spring-mass system has a motion called simple harmonic motion, in which the mass oscillates following a periodic function (sine or cosine) around an equilibrium position.

As the system oscillates back and forth, its total mechanical energy (sum of elastic potential energy and kinetic energy) will remain conserved (since we consider friction negligible). The elastic potential energy at any point is given by:

U=\frac{1}{2}kx^2

where

k is the spring constant

x is the displacement of the system

While the kinetic energy at any point is

K=\frac{1}{2}mv^2

where

m is the mass

v is the speed

So the total mechanical energy of the system is

E=K+U=\frac{1}{2}mv^2+\frac{1}{2}kx^2

For this system, when it is initially released,

m = 0.6 kg

k = 165 N/m

x = 7 cm = 0.07 m

v = 3 m/s

So the total energy is

E=\frac{1}{2}(0.6)(3)^2+\frac{1}{2}(165)(0.07)^2=3.1 J

Since friction is negligible, this total energy remains constant. Therefore, when the system reaches its maximum stretch during the motion, the kinetic energy will be zero and all the mechanical energy will be elastic potential energy; so we will have:

E=U=\frac{1}{2}kx_{max}^2

where x_{max} is the maximum stretch. Solving for x_{max},

x_{max}=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(3.1)}{165}}=0.194 m

So, 19.4 cm.

b)

The maximum speed in a spring-mass oscillating system is reached when the kinetic energy is maximum, and therefore, since the total energy is conserved, when the elastic potential energy is zero:

U=0

which means when the displacement is zero:

x = 0

So, when the system is transiting through the equilibrium position.

Therefore, the total mechanical energy is equal to the maximum kinetic energy:

E=K=\frac{1}{2}mv_{max}^2

where

m is the mass

v_{max} is the maximum speed

Here we have:

E = 3.1 J

m = 0.6 kg

Therefore, solving for the maximum speed,

v_{max}=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(3.1)}{0.6}}=3.2 m/s

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