Ask question

Ask question

All categories

3 years ago

15

Light from a laser of wavelength λ1 shines normally on a pair of narrow slits separated by distance D. This results in a differe

nce in angle between the central and first intensity maxima of 0.02 radians. A second laser of wavelength λ2=2λ1 then shines on the same slit system. The new angular difference between the central first intensity maxima is closest to:

1 answer:

lana66690 [7]3 years ago

6 0

Answer:

Explanation:

The angular position of first intensity maxima is given by the expression

= λ / d where λ is wave length of light and d is slit separation

putting the values given in the problem. ,

.02 = λ₁ / D

Now wavelength has been increased to 2λ₁

angular separation of first maxima

= 2λ₁ /D

= 2 x .02

= .04 radians.

You might be interested in

As a glacier melts, the volume V of the ice, measured in cubic kilometers, decreases at a rate modeled by the differential equat

DaniilM [7]

Answer:

the volume in terms of time t ⇒ $\\V = e^{0.05t} + 399\\$

Explanation:

Given that :

$\frac{dv}{dt}= kv\\\\\int\limits \, \frac{dv}{v}= \int\limits \, kdt\\In v = kt + C_1\\v = e^{kt} + C\\400 = e^{k*0} + C\\400 = 1 + C\\C = 400 -1\\C = 399$

$V = e^{kt} + 399$

When v = 300 ; $\frac{dv}{dt}= - 15$

then

$\frac{dv}{dt}= kv\\\\-15 = 300*k\\\\k = \frac{-15}{300}\\\\\\k = \frac{-1}{20}\\\\k = -0.05$

∴ $\\V = e^{0.05t} + 399\\$

Therefore, the volume in terms of time t ⇒ $\\V = e^{0.05t} + 399\\$

6 0

3 years ago

Squids propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracting the cavity

Musya8 [376]

Answer:

v_squid = - 2,286 m / s

Explanation:

This exercise can be solved using conservation of the moment, the system is made up of the squid plus the water inside, therefore the force to expel the water is an internal force and the moment is conserved.

Initial moment. Before expelling the water

p₀ = 0

the squid is at rest

Final moment. After expelling the water

$p_{f}$ = M V_squid + m v_water

p₀ = p_{f}

0 = M V_squid + m v_water

c_squid = -m v_water / M

The mass of the squid without water is

M = 9 -2 = 7 kg

let's calculate

v_squid = 2 8/7

v_squid = - 2,286 m / s

The negative sign indicates that the squid is moving in the opposite direction of the water

8 0

3 years ago

A solid sphere of mass 8.6 kg, made of metal whose density is 3,400 kg/, hangs by a cord. When the sphere is immersed in a liqui

creativ13 [48]

Answer:

A. $1900 kg/m^3$

Explanation:

We are given that

$m=8.6 kg$

Density, $\rho_s=3400 kg/m^3$

Tension,T=38 N

We have to find the density of liquid.

$T=mg-\rho_l Vg$

$g=9.8 m/s^2$

Volume,V= $\frac{m}{\rho_s}$

$38=8.6\times 9.8-\rho_l\times \frac{8.6}{3400}\times 9.8$

$\rho_l\times \frac{8.6}{3400}\times 9.8=8.6\times 9.8-38$

$\rho_l=\frac{(8.6\times 9.8-38)\times 3400}{8.6\times 9.8}$

$\rho_l=1867kg/m^3\approx 1900 kg/m^3$

Option A is true.

4 0

3 years ago

Suppose that in a lightning flash the potential difference between a cloud and the ground is 0.96×109 V and the quantity of char

Dvinal [7]

(a) $2.98\cdot 10^{10} J$

The change in energy of the transferred charge is given by:

$\Delta U = q \Delta V$

where

q is the charge transferred

$\Delta V$ is the potential difference between the ground and the clouds

Here we have

$q=31 C$

$\Delta V = 0.96\cdot 10^9 V$

So the change in energy is

$\Delta U = (31 C)(0.96\cdot 10^9 V)=2.98\cdot 10^{10} J$

(b) 7921 m/s

If the energy released is used to accelerate the car from rest, than its final kinetic energy would be

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

where

m = 950 kg is the mass of the car

v is the final speed of the car

Here the energy given to the car is

$K=2.98\cdot 10^{10} J$

Therefore by re-arranging the equation, we find the final speed of the car:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(2.98\cdot 10^{10})}{950}}=7921 m/s$

5 0

3 years ago

A speedy tortoise can run with a speed of <img src="https://tex.z-dn.net/?f=v_T" id="TexFormula1" title="v_T" alt="v_T" align="a

Vinil7 [7]

Answer:

$t_{H} = t_{T} - 60$

Explanation:

Thinking process:

Let:

distance = speed × time

for the hare:

$d_{H} = s_{H} t_{H}$

for the tortoise:

$d_{T} = s_{T} t_{T}$

Let the length of the shell be s. Then the two are related. So:

$d_{H} = d_{T} - s$

Let's say the hare ran at x times the speed of the tortoise:

$s_{H} = xs_{T}$

So, the equation still holds : $t_{H} = t_{T} - 60$

8 0

3 years ago