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1 year ago

Which of the following is a form of pollution created when vehicle exhaust interacts with sunlight?

Physics

1 answer:

SVEN [57.7K]1 year ago

8 0

Photochemical smog is formed when primary air pollutants interact with sunlight.

Photochemical smog is the result of the reaction between pollutants like nitrogen oxides (NO), sunlight and volatile organic compound (VOC) in the atmosphere. The sources of NO are car exhaust, coal power plants, factory emissions, etc. This type of smog is also known by the name Los Angeles smog.

Air pollutants are the particles present dissolved in the air, which when inhaled by the organisms can cause serious health issues. These pollutants are :ozone, particulate matter, gaseous oxides, etc. These pollutants majorly affect the respiratory system of the humans.

Therefore, photochemical smog is a form of pollution created when vehicle exhaust interacts with sunlight.

To know more about photochemical smog, here: brainly.com/question/15728274

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Consider that a ray of light is travelling from glass to water. The refractive index of water is 1.30 (i e n . ., 1.30 w = ) and

Bess [88]

Answer:

$\theta_i=49.88^{\circ}$

Explanation:

Total internal reflection can happen when light goes from a medium with higher refractive index (in this case, glass) to a medium with lower refractive index (in this case, water).

Snell's Law tells us that $n_isin\theta_i=n_rsin\theta_r$ , where the <em>i</em> stands for incident (in this case, glass) and the <em>r</em> for refracted (in this case, water). We want to know when $\theta_r=90^{\circ}$ , that is, when $n_isin\theta_i=n_r$ , and this happens when the incident angle is:

$\theta_i=arcsin(\frac{n_r}{n_i})$

Which for our values means:

$\theta_i=arcsin(\frac{1.3}{1.7})=arcsin(0.76470588235)=49.88^{\circ}$

6 0

3 years ago

An 8.0 m, 240 N uniform ladder rests against a smooth wall. The coefficient of static friction between the ladder and the ground

sweet [91]

Answer:

5.7 m

Explanation:

AD = length of the ladder = L = 8 m

AB = distance of the center of mass of the ladder = (0.5) L = (0.5) 8 = 4 m

AC = distance of person on the ladder from the bottom end = x

W = weight of the ladder = 240 N

$F_{g}$ = weight of the person = 710 N

F = force by the wall on the ladder

N = normal force by ground on the ladder = ?

Using equilibrium of force along the vertical direction

N = $F_{g}$ + W

N = 710 + 240

N = 950 N

μ = Coefficient of static friction = 0.55

f =static frictional force on the ladder

Static frictional force is given as

f = μ N

f = (0.55) (950)

f = 522.5 N

Force equation along the horizontal direction is given as

F = f

F = 522.5 N

using equilibrium of torque about point A

F Sin50 (AD) = W Cos50 (AB) + ( $F_{g}$ Cos50 (AC))

(522.5) Sin50 (8) = (240) Cos50 (4) + (710) Cos50 (x)

x = 5.7 m

7 0

3 years ago

Why are wheels and steam engine important in the history of transport

STALIN [3.7K]

Explanation:

Because it's a power to turn the wheels of industry

5 0

3 years ago

The pressure on a balloon holding 356 mL of an ideal gas is increased from 267 torr to 1.00 atm. What is the new volume of the b

Natalija [7]

Answer:

124.52 mL

Explanation:

from Boyle's Law,

PV = P'V' ................... Equation 1

Where P = Initial pressure of the gas, V = Initial volume of the gas, P' = Final pressure of the gas, V' = Final volume of the gas.

make V' the subject of the equation.

V' = PV/P'............. Equation 2

Given: P = 267 torr = (267×0.00131) = 0.34977 atm, V = 356 mL, P' = 1 atm

Substitute into equation 2

V' = (0.34977×356)/1

V' = 124.52 mL.

Hence the new volume of the balloon = 124.52 mL

4 0

3 years ago

A small object carrying a charge of -4.00 nC is acted upon by a downward force of 19.0 nN when placed at a certain point in an e

Gala2k [10]

Explanation:

Given that,

Charge acting on the object, $q=-4\ nC=-4\times 10^{-9}\ C$

Force acting on the object, $F=19\ nC=19\times 10^{-9}\ C$ (in downward direction)

(a) The electric force acting in the electric field is given by :

$F=qE$

E is the electric field

$E=\dfrac{F}{q}$

$E=\dfrac{19\times 10^{-9}\ N}{4\times 10^{-9}\ C}$

E = 4.75 N/C

The direction of electric field is as same as electric force. But it is negative charge. So, the direction of electric field is in upward direction.

(b) The charge on the proton is, $q=1.6\times 10^{-19}\ C$

The force acting on the proton is :

$F=qE$

$F=1.6\times 10^{-19}\times 4.75$

$F=7.6\times 10^{-19}\ N$

If the charge on the proton is positive, the force on the proton is in upward direction.

Hence, this is the required solution.

8 0

3 years ago