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NikAS [45]
4 years ago
15

In a double-slit experiment the distance between slits is 5.0 mm and the slits are 1.4 m from the screen. Two interference patte

rns can be seen on the screen: one due to light of wavelength 460 nm, and the other due to light of wavelength 650 nm. What is the separation in meters on the screen between the m = 2 bright fringes of the two interference patterns?
Physics
1 answer:
My name is Ann [436]4 years ago
5 0

Explanation:

It is given that,

Distance between the slits, d = 5 mm = 0.005 m

Distance between slit and screen, D = 1.4 m

For m = 2, and \lambda_1=460\ nm=460\times 10^{-9}\ m

x_1=\dfrac{m\lambda_1 D}{d}

x_1=\dfrac{2\times 460\times 10^{-9}\times 1.4}{0.005}

x_1=0.0002576\ m

For m = 2, and \lambda_2=650\ nm=650\times 10^{-9}\ m

x_2=\dfrac{m\lambda_2 D}{d}

x_2=\dfrac{2\times 650\times 10^{-9}\times 1.4}{0.005}

x_2=0.000364\ m

Separation between two fringes is :

\Delta x=x_2-x_1

\Delta x=0.000364-0.0002576

\Delta x=0.0001064

\Delta x=1.064\times 10^{-4}\ m

Hence, this is the required solution.

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An alert driver can apply the brakes fully in about 0.5 seconds. How far would the car travel if it
dybincka [34]

Answer:

The car would travel after applying brakes is, d = 14.53 m

Explanation:

Given that,

The time taken to apply brakes fully is, t = 0.5 s

The velocity of the car, v = 29.06 m/s

The distance traveled by the car in 0.5 s, d = ?

The relation between the velocity, displacement, and time is given by the formula                

                                d = v x t    m

Substituting the values in the above equation,

                                  d = 29.06 m/s x 0.5 s

                                     = 14.53 m

Therefore, the car would travel after applying brakes is, d = 14.53 m

8 0
4 years ago
Calculate the pressure on the bottom of a swimming pool 3. 5 m deep. How does the pressure compare with atmospheric pressure, 10
Nostrana [21]

Answer:

  1.343 atm

Explanation:

The mass of water above 1 square meter of swimming pool bottom is ...

  M = (3.5 m)·(1000 kg/m^3) = 3500 kg/m^2

Then the force exerted by the water on the pool bottom is ...

  F = Mg = (3500 kg/m^2)(9.8 m/s^2) = 34300 N/m^2 = 34300 Pa

Compared with atmospheric pressure, this is ...

  34,300/10^5 = 0.343 . . . . atmospheres

Added to the atmospheric pressure on the water's surface, the total pressure on the pool bottom is 1.343 atmospheres.

5 0
4 years ago
A 72 kg skydiver is descending on a parachute. His speed is still increasing at 1.2 m/s2. What are the magnitude and direction o
frozen [14]

Answer:

86.4 N downward

Explanation:

Force: This can be defined as the product of mass and acceleration of a body.

The S.I unit of Force is Newton(N).

The Expression of force is given as,

F = ma ................ Equation 1

Where F = force of the parachute harness, m = mass of the skydiver, a = acceleration of the skydiver.

Given: m = 72 kg, a = 1.2 m/s²

Substitute into equation 1

F = 72(1.2)

F = 86.4 N down ward.

Hence the force on the parachute harness = 86.4 N downward

8 0
3 years ago
All of the following shows density and buoyancy except:
Evgesh-ka [11]
The answer is <span>B. the amount of hot soup contained in a bowl. Buoyancy is defined as the upward force exerted by a fluid on an object immersed in that fluid. Buoyancy and density are two factors that affect the downward and upward forces exerted on object that affect its ability to "float" or "sink". Only B does not have anything to do with such forces.</span>
4 0
3 years ago
A circular wire loop of radius LaTeX: RR lies in the xy-plane with the z-axis running through its center. There is initially no
frosja888 [35]

Answer:

Explanation:

Given a circular loop of radius R

r = R

Note: the radius lies in the xy plane

Area is given as

A = πr² = πR²

At t = 0, no magnetic field B=0

The magnetic field is given as a function of time

B = C•exp(t) •i + D•t² •k

Where C and D are constant

We want to find the magnitude of EMF in the circular loop.

EMF is given as

ε = - N•dΦ/dt

Where,

N is number of turn and in this case we will assume N = 1.

Φ is magnetic flux and it is given as

Φ = BA

ε = - N•d(BA)/dt

Where A is a constant, then we have

ε = - N•A•dB/dt

B = C•exp(t) •i + D•t² •k

dB/dt = C•exp(t) •i + 2D•t •k

Then,

ε = - N•A•dB/dt

ε = - 1•πR²•(C•exp(t) •i + 2D•t •k)

ε = -πR²•(C•exp(t) •i + 2D•t •k)

So, let find the magnitude of EMF

Generally finding magnitude of two vectors R = a•i + b•j

Then, |R| = √a² + b²

So, applying this we have,

ε = πR² (√(C²•exp(2t) + 4D²t²))

From the given magnetic field, we are given that,

B = 0 at t = 0

B = C•exp(t) •i + D•t² •k

B = 0 = C•exp(0) •i + D•0² •k

0 = C

Then, C = 0.

So, substituting this into the EMF.

ε = πR² (√(0²•exp(2t) + 4D²t²))

ε = πR² (√4D²t²)

ε = πR² × 2Dt

ε = 2πDR²t

So, the EMF is also a function of time

ε = 2πDR²t

4 0
4 years ago
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