Ask question

Ask question

All categories

aleksklad [387]

3 years ago

6

Coherent light with wavelength 460 nm falls on a pair of slits. On a screen 1.86 m away, the distance between dark fringes is 3.

90 mm. What is the slit separation?

1 answer:

Jobisdone [24]3 years ago

7 0

Answer:

The split separation d = 2.12 * 10^-4 m

Explanation:

Mathematically, for a double split interference, the relation between separation of slit and fringe width is given by the formula;

d = Rλ/Δy

Where R represents the distance of screen from the slit given as 1.86m , λ is the wavelength of incident light given as 460 nm ,

Δy is the fringe width given as 3.90 nm

plugging these values into the equation, we have;

d = (1.86)(460 * 10^-9)/(3.9 * 10^-3)

d = 2.12 * 10^-4 m

You might be interested in

How can an object overcome static friction?

larisa86 [58]

Answer:

Applying enough force in one direction to move the object, making kinetic energy.

Explanation:

Simpleness

4 0

3 years ago

The membrane that surrounds a certain type of living cell has a surface area of 5.3 x 10-9 m2 and a thickness of 1.1 x 10-8 m. A

kotykmax [81]

Answer:

$2.1\times 10^{-12} c$

Explanation:

We are given that

Surface area of membrane= $5.3\times 10^{-9} m^2$

Thickness of membrane= $1.1\times 10^{-8} m$

Assume that membrane behave like a parallel plate capacitor.

Dielectric constant=5.9

Potential difference between surfaces=85.9 mV

We have to find the charge resides on the outer surface of membrane.

Capacitance between parallel plate capacitor is given by

$C=\frac{k\epsilon_0 A}{d}$

Substitute the values then we get

Capacitance between parallel plate capacitor= $\frac{5.9\times 8.85\times 10^{-12}\times 5.3\times 10^{-9}}{1.1\times 10^{-8}}$

$C=0.25\times 10^{-12}F$

V= $85.9 mV=85.9\times 10^{-3}$

$Q=CV$

$Q=0.25\times 10^{-12}\times 85.9\times 10^{3}=2.1\times 10^{-12} c$

Hence, the charge resides on the outer surface= $2.1\times 10^{-12} c$

5 0

3 years ago

Read 2 more answers

A truck using a rope to tow a 2230-kg car accelerates from rest to 13.0 m/s in a time of 15.0s. How strong must the rope be? μk

Leokris [45]

Answer:

The rope must have a force of 10084,21 N

Explanation

Acceleration calculation

The car acceleration is equal to the acceleration of the truck

ac: car acceleration $\frac{m}{s^{2} }$

at: truck acceleration $\frac{m}{s^{2} }$ )

$ac = at= \frac{vf-vi}{t-ti}$ equation(1)

Known information:

vi = Initial speed = 0, ti = initial time = 0

vf = Final speed = 13 $\frac{m}{s}$ , t = final time =5 s

We replaced the known information in the equation(1):

$ac = at = \frac{13-0}{15-0}$

$ac=ac=\frac{13}{15} \frac{m}{s}$

Dynamic analysis

The forces acting on the car are the following:

Wc: Car weight

N: normal force, road force on the car

Ff: Friction force

T: Force of tension

Car weight calculation:

$Wc=mc*g$

mc = Car mass = 2230kg

g = Gravity acceleration=9.8 $\frac{m}{s^{2} }$

$Wc= 2230*9.8$

$Wc=21854 N$

Normal force calculation:

Newton's first law

$sum Fy= 0$

$N-W=0$

$N=W$

$N=21854 N$

Friction force calculation (Ff):

We have the formula to calculate the friction force:

Ff = μk * N Equation (3)

μk kinetic coefficient of friction

We know that μk = 0.373and N= 21854N ,then:

$Ff=0.373*21854$

$Ff=8151.54 N$

Calculation of the tension force in the rope (T):

Newton's Second law

$sum Fx= mc*ac$

$T-Ff=mc*ac$

$T=2230(\frac{13}{15}) + 8151.54$

T=10084,21 N

Answer: The rope must have a force of 10084,21 N

8 0

3 years ago

A cart with a mass of 0.5 kg is at the top of the ramp. The height is 0.40m .

Tju [1.3M]

A=0.05.0M.

B=68.9244GPE.34

C=0

D it would be 79%HIGHER

3 0

3 years ago

How many days are in 345000 minutes? (Set up by dimensional analysis)

Komok [63]

239.583 days is the answer

3 0

2 years ago