1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Arada [10]
3 years ago
13

What is the distance, in meters, between adjacent fringes produced by a diffraction grating having 125 lines per centimeter

Physics
1 answer:
inysia [295]3 years ago
8 0

Answer:

The correct answer will be "9×10⁻³ m".

Explanation:

The given values are:

Number of lines

N = 125

As we know

The distance between the adjacent fringes is:

⇒ \frac{\lambda x}{d}

Where, λx = Screen distance

              d = slit width

and,

⇒ dSin \theta=m \lambda

Where, d = \frac{1}{N}

Hence,

⇒ \Delta\gamma=\frac{\lambda x}{d}

          =\lambda x\times N

On substituting the values, we get

⇒ \Delta\gamma=575\times 10^{-9}\times 1.25\times \frac{125}{10^{-2}}

⇒ \Delta\gamma=9\times 10^{-3} \ m

You might be interested in
A 200. kg object is pushed 12.0 m to the top of an incline to a height of 6.0 m. If the force applied along the incline is 3000.
Nataliya [291]

Answer:

Approximately 1.2 \times 10^{4}\; {\rm J} (assuming that g = 9.81\; {\rm N \cdot kg^{-1}}.)

Explanation:

The strength of the gravitational field near the surface of the earth is approximately constant: g = 9.81\; {\rm N \cdot kg^{-1}}.

The change in the gravitational potential energy ({\rm GPE}) of an object near the surface of the earth is proportional to the change in the height of this object. If the height of an object of mass m increased by \Delta h, the {\rm GPE} of that object would have increased by m\, g\, \Delta h.

In this question, the height of this object increased by \Delta h = 6.0\; {\rm m}. The mass of this object is m = 200\; {\rm kg}. Thus, the {\rm GPE} of this object would have increased by:

\begin{aligned}& (\text{Change in GPE}) \\ =\; & m\, g\, \Delta h \\ =\; & 200\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 6.0\; {\rm m} \\ \approx\; & 1.2 \times 10^{4}\; {\rm J}\end{aligned}.

(Note that 1\; {\rm N \cdot m} = 1\; {\rm J}.)

3 0
2 years ago
A wire that has a resistance of 1Ω is to be built from an aluminum block of volume 1 cm3
alina1380 [7]

Answer:

  L = 5,955 m

Explanation:

For this exercise we must use the relation

        R = ρ L / A

where R is the resistance that indicates that it is 1 Ω, the resistivity is taken from the tables ρ = 2.82 10⁻⁸ Ω m, L is the length of the wire and A is the cross section.

As it indicates to us in volume of aluminum to use we divide the two terms by the length

    R / L = ρ L / (A L)

the volume of a body is its area times its length, therefore

  R / L = ρ L / V

  R = ρ L² / V

we clear the length of the wire

   L = √ R V /ρ

we reduce the volume to SI units

   v = 1 cm³ (1m / 10² cm)³ = 1 10⁻⁶ m

let's calculate

   L = √ (1  1  10⁻⁶ / 2.82 10⁻⁸)

   L = √ (0.3546 10²)

   L = 5,955 m

5 0
3 years ago
55 POINTS!!!! What would happen if we removed the rabbit and the capybara from the food web?
Alekssandra [29.7K]
The food chain would be disturbed and the ecosystem would not be the same, without the consumers nothing would happen and it would lead to the downfall of the chain.
8 0
4 years ago
What is the magnitude of the electric field at the origin produced by a semi-circular arc of charge = 7.6 μc, twice the charge o
Anna [14]

Orient the semi-circle arc such that it is symmetric with respect to the y-axis. Now, by symmetry, the electric field in the x-direction cancels to zero. So the only thing of interest is the electric field in the y-direction.  


dEy=kp/r^2*sin(a) where k is coulombs constant p is the charge density r is the radius of the arc and a is the angular position of each point on the arc (ranging from 0 to pi. Integrating this renders 2kq/(pi*r^3). Where k is 9*10^9, q is 9.8 uC r is .093 m

I answeared your question can you answear my question pleas

6 0
3 years ago
Light is shining perpendicularly on the surface of the earth with an intensity of 680 W/m^2. Assuming that all the photons in th
andrew11 [14]

Answer:

3.066×10^21  photons/(s.m^2)

Explanation:

The power per area is:

Power/A = (# of photons /t /A)×(energy / photon)

E/photons = h×c/(λ)

photons /t /A = (Power/A)×λ /(h×c)  

photons /t /A = (P/A)×λ/(hc)

photons /t /A = (680)×(678×10^-9)/(6.63×10^-34)×(3×10^-8)

                      = 3.066×10^21

Therefore, the number of photons per second per square meter 3.066×10^21  photons/(s.m^2).

4 0
3 years ago
Other questions:
  • The work done to lift a 10kg object to a height of 1m above the ground is
    8·1 answer
  • Reading glasses with a power of 1.50 diopters make reading a book comfortable for you when you wear them 1.8 cmcm from your eye.
    5·1 answer
  • At its lowest setting a centrifuge rotates with an angular speed of calculate the angular velocity
    14·1 answer
  • The levels of organization are Cell, Tissue, Organs, ____________, and Organism. what goes in the blank?
    11·2 answers
  • Calculate the man’s mass. (Use PE = m × g × h, where g = 9.8 N/kg.)
    13·1 answer
  • Any two instruments based on Pascal's law​
    8·1 answer
  • CAN SOMEONE HELP ME PLEASE AND SHOW WORK
    7·1 answer
  • a father and his son are racing down a mountain. if they're traveling at the same velocity who has more kinetic energy​
    14·1 answer
  • Determine the distance above Earth's surface to a satellite that completes four orbits per day.
    14·1 answer
  • HELP ASAP
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!