All categories

3 years ago

2. Human Impact on the Environment

1 answer:

8 0

Burning fossil fuels emits a number of air pollutants that are harmful to both the environment and public health. Sulfur dioxide (SO2) emissions, primarily the result of burning coal, contribute to acid rain and the formation of harmful particulate matter.

You might be interested in

A proton, initially traveling in the +x-direction with a speed of 5.05×10^5 m/s , enters a uniform electric field directed verti

Korvikt [17]

Answer:

The strength of the electric field is $1.35\times10^{4}\ N/C$ .

Explanation:

Given that,

Speed $v= 5.05\times10^{5}\ m/s$

Time $t= 3.90\times10^{-7}\ s$

Angle = 45°

We need to calculate the acceleration

Using equation of motion

$v = u+at$

$5.05\times10^{5}=0+a\times3.90\times10^{-7}$

$a =\dfrac{5.05\times10^{5}}{3.90\times10^{-7}}$

$a=1.29\times10^{12}\ m/s^2$

We need to calculate the strength of the electric field

Using relation of newton's second law and electric force

$F= ma=qE$

$ma = qE$

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

Put the value into the formula

$E=\dfrac{1.67\times10^{-27}\times1.29\times10^{12}}{1.6\times10^{-19}}$

$E=1.35\times10^{4}\ N/C$

Hence, The strength of the electric field is $1.35\times10^{4}\ N/C$ .

3 0

3 years ago

A light plane must reach a speed of 33 m/s per take off. How long a runway is needed if the constant acceleration is 3.0 m/s^2.

aleksley [76]

Given the final velocity (Vf) and the acceleration (a), the distance that should be traveled by the plane is calculated through the equation,
d = (Vf² - Vi²) / 2a
V1 should be zero because the light plane started the motion from rest. Substituting the given values,
d = ((33 m/s)² - 0)) / 2(3 m/s²)
The distance is therefore equal to 181.5 meters.

3 0

3 years ago

A beam of light has a wavelength of 650 nm in vacuum. (a) What is the speed of this light in a liquid whose index of refraction

Lady_Fox [76]

Answer:

The speed of this light and wavelength in a liquid are $2.04\times10^{8}\ m/s$ and 442 nm.

Explanation:

Given that,

Wavelength = 650 nm

Index refraction = 1.47

(a). We need to calculate the speed

Using formula of speed

$n = \dfrac{c}{v}$

Where, n = refraction index

c = speed of light in vacuum

v = speed of light in medium

Put the value into the formula

$1.47=\dfrac{3\times10^{8}}{v}$

$v=\dfrac{3\times10^{8}}{1.47}$

$v= 2.04\times10^{8}\ m/s$

(b). We need to calculate the wavelength

Using formula of wavelength

$n=\dfrac{\lambda_{0}}{\lambda}$

$\lambda=\dfrac{\lambda_{0}}{n}$

Where, $\lambda_{0}$ = wavelength in vacuum

$\lambda$ = wavelength in medium

Put the value into the formula

$\lambda=\dfrac{650\times10^{-9}}{1.47}$

$\lambda=442\times10^{-9}\ m$

Hence, The speed of this light and wavelength in a liquid are $2.04\times10^{8}\ m/s$ and 442 nm.

3 0

3 years ago

Two forces Upper FSubscript Upper A Baseline Overscript right-arrow EndScripts and Upper FSubscript Upper B Baseline Overscript

Pavel [41]

Answer:

Part a)

$F_A = 4.59 N$

Part B)

$F_B = 1.28 N$

Explanation:

As we know that when both the forces are acting on the object in same direction then we will have

$F_A + F_B = ma$

as we know that

$a = 0.554 m/s^2$

m = 10.6 kg

now we will have

$F_A + F_B = 10.6(0.554)$

$F_A + F_B = 5.87 N$

Now two forces are in opposite direction then we have

$F_A - F_B = 10.6(0.313)$

$F_A - F_B = 3.32 N$

Part A)

Now we will have from above two equation

$F_A = 4.59 N$

Part B)

Similarly for other force we have

$F_B = 1.28 N$

5 0

3 years ago

An insulating sphere is 8.00 cm in diameter and carries a 6.50 ÂµC charge uniformly distributed throughout its interior volume.

Kobotan [32]

Explanation:

(a) Formula to calculate the density is as follows.

$\rho = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.04)^{3}}$

= $2.42 \times 10^{-2} C/m^{3}$

Now, calculate the charge as follows.

$q_{in} = \rho(\frac{4}{3} \pi r^{3})$

= $2.42 \times 10^{-2} C/m^{3} \times 4.1762 \times (0.01)^{3}$

= $10.106 \times 10^{-8}$ C

or, = 101.06 nC

(b) For r = 6.50 cm, the value of charge will be calculated as follows.

$q_{in} = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.065)^{3}}$

= 7.454 $\mu C$

7 0

3 years ago