Which material(s) listed below is an example of a persistent organic pollutant?

A 0.50 mm-wide slit is illuminated by light of wavelength 500 nm.What is the width of the central maximum on a screen 2.0m behin

Novay_Z [31]

Answer:

0.004 m

Explanation:

For light passing through a single slit, the position of the nth-minimum in the diffraction pattern is given by

$y=\frac{n\lambda D}{d}$

where

$\lambda$ is the wavelength

D is the distance of the screen from the slit

d is the width of the slit

Therefore, the width of the central maximum is equal to twice the value of y for n=1 (first minimum):

$w=2\frac{\lambda D}{d}$

where we have

$\lambda=500 nm = 5\cdot 10^{-7}m$ is the wavelength

D = 2.0 m is the distance of the screen

$d=0.50 mm=5\cdot 10^{-4}m$ is the width of the slit

Substituting, we find

$w=2\frac{(5\cdot 10^{-7} m)(2.0 m)}{5\cdot 10^{-4} m}=0.004 m$

5 0

3 years ago

Which is the correctly balanced equation for the reaction of rust, Fe2O3, and hydrochloric acid, HCl?

Mariana [72]

Answer: Option (c) is correct answer.

Explanation:

The given reaction equation is as follows.

$Fe_{2}O_{3} + HCl \rightarrow FeCl_{3} + H_{2}O$

The number of atoms on reactant side are as follows.

Fe = 2
O = 3
H = 1
Cl = 1

The number of atoms on product side are as follows.

Fe = 1
O = 1
H = 2
Cl = 3

Therefore, in order to balance the equation we will balance the number of atoms in both reactant and product side.

Since number of Fe atoms in reactant side is 2, therefore, multiply $FeCl_{3}$ by 2 in the product side.

Number of O atoms in reactant side is 3, therefore, multiply $H_{2}O$ by 3 in the product side.

Now, the number of H atoms on product side is 6, therefore, multiply HCl by 6 in the reactant side.

The number of chlorine atoms on both reactant and product side equals 6.

Thus, the equation will become as follows.

$Fe_{2}O_{3} + 6HCl \rightarrow 2FeCl_{3} + 3H_{2}O$

Hence, the equation is balanced.

5 0

3 years ago

Read 2 more answers

One object is at rest, and another is moving. The two collide in a one-dimensional, completely inelastic collision. In other wor

zhannawk [14.2K]

Answer:

Part a)

v = 16.52 m/s

Part b)

v = 7.47 m/s

Explanation:

Part a)

(a) when the large-mass object is the one moving initially

So here we can use momentum conservation as the net force on the system of two masses will be zero

so here we can say

$m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v$

since this is a perfect inelastic collision so after collision both balls will move together with same speed

so here we can say

$v = \frac{(m_1v_{1i} + m_2v_{2i})}{(m_1 + m_2)}$

$v = \frac{(8.4\times 24 + 3.8\times 0)}{3.8 + 8.4}$

$v = 16.52 m/s$

Part b)

(b) when the small-mass object is the one moving initially

here also we can use momentum conservation as the net force on the system of two masses will be zero

so here we can say

$m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v$

Again this is a perfect inelastic collision so after collision both balls will move together with same speed

so here we can say

$v = \frac{(m_1v_{1i} + m_2v_{2i})}{(m_1 + m_2)}$

$v = \frac{(8.4\times 0 + 3.8\times 24)}{3.8 + 8.4}$

$v = 7.47 m/s$

4 0

3 years ago

Two objects with masses of 2.75 kg and 4.45 kg are connected by a light string that passes over a light frictionless pulley to f

Llana [10]

Answer:

Explanation:

F = ma

4.45g - 2.75g = (4.45 + 2.75)a

a = 9.81(4.45 - 2.75) / (4.45 + 2.75) = 2.31625 ≈ 2.32 m/s²

a)

T = 2.75(9.81 + 2.32) = 33.3 N

or

T = 4.45(9.81 - 2.32) = 33.3 N

b) 2.32 m/s² upward for 2.75 kg mass

2.32 m/s² downward for 4.45 kg mass

c) y = ½at² = ½(2.31625/3)1² = 1.158125 ≈ 1.16 m

4 0

2 years ago

Bob is threatening Tom’s life with a giant laser with wavelength (650 nm), a distance (D = 10 m) from the wall James is shackled

Fittoniya [83]

Answer:

He should stand from the center of laser pointed on the wall at 1.3 m.

Explanation:

Given that,

Wave length = 650 nm

Distance =10 m

Double slit separation d = 5 μm

We need to find the position of fringe

Using formula of distance

$d\sin\theta=n\lambda$

$d\dfrac{y}{D}=n\lambda$

$y=\dfrac{\lambda D}{d}$

Put the value into the formula

$y=\dfrac{650\times10^{-9}\times10}{5\times10^{-6}}$

$y=1.3\ m$

Hence, He should stand from the center of laser pointed on the wall at 1.3 m.

8 0

2 years ago