1. The algebraic letter 'dy' stands for

Nearly monochromatic light of wavelength 420 nm is split in two beams. The first beam traveled 105 nm to the screen more than th

leva [86]

Answer:

phase difference = π / 2

constructive interference

Explanation:

Given data

wavelength = 420 nm

1st beam = 105 nm

path difference = 105 nm

to find out

phase difference and interference pattern of the two beams

solution

we use here equation of phase difference that is

phase difference = 2π / wavelength × Δx

put here value

phase difference = 2π / 420 × 105

phase difference = π / 2

and

we know that here path difference Δx is the integral multiple of the wavelength so it will be constructive interference

Δx is wavelength / 4

8 0

3 years ago

Key concepts (Example | scientific method

nekit [7.7K]

Vocabulary should be, I think:
I. Hypothesis
II. Evidence, data
III. Experiment

What is your question exactly?

3 0

3 years ago

Read 2 more answers

g A 1.5-kg mass attached to spring with a force constant of 20.0 N/m oscillates on a horizontal, frictionless track. At t = 0, t

jok3333 [9.3K]

Answer:

(a) f = 0.58Hz

(b) vmax = 0.364m/s

(c) amax = 1.32m/s^2

(d) E = 0.1J

(e) x(t)=0.1m*cos(2π(0.58s^{-1})t)

Explanation:

(a) The frequency of the oscillation, in a spring-mass system, is calculated by using the following formula:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$ (1)

k: spring constant = 20.0N/m

m: mass = 1.5kg

you replace the values of m and k for getting f:

$f=\frac{1}{2\pi}\sqrt{\frac{20.0N/m}{1.5kg}}=0.58s^{-1}=0.58Hz$

The frequency of the oscillation is 0.58Hz

(b) The maximum speed is given by the following relation:

$v_{max}=\omega A=2\pi f A$ (2)

A: amplitude of the oscillations = 10.0cm = 0.10m

$v_{max}=2\pi (0.58s^{-1})(0.10m)=0.364\frac{m}{s}$

The maximum speed of the mass is 0.364 m/s.

The maximum speed occurs when the mass passes trough the equilibrium point of the oscillation.

(c) The maximum acceleration is given by the following formula:

$a_{max}=\omega^2A=(2\pi f)^2 A$

$a_{max}=(2\pi (0.58s^{-1}))(0.10m)=1.32\frac{m}{s^2}$

The maximum acceleration is 1.32 m/s^2

The maximum acceleration occurs where the elastic force is a maximum, that is, where the mass is at the maximum distance from the equilibrium point, that is, the acceleration.

(d) The total energy of the system is calculated with the maximum potential elastic energy:

$E=\frac{1}{2}kA^2=\frac{1}{2}(20.0N/m)(0.10m)^2=0.1J$

The total energy is 0.1J

(e) The displacement as a function of time is:

$x(t)=Acos(\omega t)=Acos(2\pi ft)\\\\x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)$

6 0

2 years ago

These questions !plz !! i need help!!!

Nataly_w [17]

(6) Wagon B is at rest so it has no momentum at the start. If v is the velocity of the wagons locked together, then

(140 kg) (15 m/s) = (140 kg + 200 kg) v

==> v ≈ 6.2 m/s

(7) False. If you double the time it takes to perform the same amount of work, then you halve the power output:

E / (2t ) = 1/2 × E/t = 1/2 P

3 0

3 years ago

The partial negative charge in a molecule of water occurs because

just olya [345]

the electrons shared between the oxygen and hydrogen atoms spend more time around the oxygen atom nucleus than around the hydrogen atom nucleus

4 0

3 years ago