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Flura [38]
3 years ago
12

What are some actions that can be taken to remedy air pollution? Select all that apply. Run your air conditioner at a higher set

ting Using humidifiers everywhere Government passing regulatory laws Car emissions testing Burning wood instead of charcoal in your grill Carpooling and less driving overall
Physics
1 answer:
Sati [7]3 years ago
4 0

Answer:  Government passing regulatory laws

Car emissions testing

Carpooling and less driving overall

Air pollution results when a mixture of particles and gases that can reach in harmful concentrations both outdoors as well as indoors. This increase in concentration can lead to rise in the temperatures and global warming it is also related with the health hazards. Soot, smoke, pollen, carbon dioxide, carbon monoxide and methane are the examples of the common pollutants.

Government passing regulatory laws: The government can ban some of the fossil fuels to prevent air pollution, and some vehicles which are likely to emit more partially combustible gases.

Car emission testing: Nowadays, cars and other vehicles are equipped with catalytic converters which converts the uncombustible gaseous products into less harmful gases which are emitted from the vehicle. The car gas emission testing should be performed to either allow a vehicle or ban the movement of the vehicle.

Carpooling and less driving overall: Carpooling is the sharing of one vehicle with more than one individual. This will prevent the need of others to drive another vehicle. The less the number of vehicles on the road the lesser will be the emission of gases and less pollution will be caused.

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Use technology and the given confidence level and sample data to find the confidence interval for the population mean muμ. Assum
Ronch [10]

Answer:

a.  μ_{95%} = 3 ± 1.8 = [1.2,4.8]

b. The correct answer is option D. No, because the sample size is large enough.

Explanation:

a. The population mean can be determined using a confidence interval which is made up of a point estimate from a given sample and the calculation error margin. Thus:

μ_{95%} = x_±(t*s)/sqrt(n)

where:

μ_{95%} = = is the 95% confidence interval estimate

x_ = mean of the sample = 3

s = standard deviation of the sample = 5.8

n = size of the sample = 41

t = the t statistic for 95% confidence and 40 (n-1) degrees of freedom = 2.021

substituting all the variable, we have:

μ_{95%} = 3 ± (2.021*5.8)/sqrt(41) = 3 ± 1.8 = [1.2,4.8]

b. The correct answer is option D. No, because the sample size is large enough.

Using the the Central Limit Theorem which states that regardless of the distribution shape of the underlying population, a sampling distribution of size which is ≥ 30 is normally distributed.

4 0
3 years ago
In a mail-sorting facility, a 2.50-kg package slides down an inclined plane that makes an angle of 20.0° with the horizontal. Th
lawyer [7]

Answer:

The coefficient of kinetic friction is 0.382.

Explanation:

Given:

Angle of inclination is, \theta=20.0°

Mass of package is, m=2.50\ kg

Initial speed of package is, u=2.00\ m/s

Final speed of the package at the bottom is, v=0\ m/s

Distance of travel along the incline is, d=12.0\ m

Acceleration due to gravity is, g=9.8\ m/s^2

Let the coefficient of kinetic friction be \mu.

Now, the frictional force will be acting along the incline but in the direction opposite to the direction of motion.

So, the net acceleration acting on the package will be up the incline and is equal to:

a=\mu g\cos\theta-g\sin\theta ----------------- 1

Now, using equation of motion, we have:

v^2-u^2=2ad\\\\0-(2.00)^2=2a(12.0)

Solving for 'a', we get:

-4.00=24.0a\\\\a=-\frac{4}{24}=-\frac{1}{6}\ m/s^2

Now, plug in the value of 'a' in equation (1). This gives,

\mu g\cos\theta-g\sin\theta=\frac{1}{6} ( Neglecting negative sign)

Plug in all the given values and solve for \mu. This gives,

9.8(-sin(20)+\mu cos(20))=\frac{1}{6}\\\\-0.342+\mu\times 0.94=0.017\\\\0.94\mu=0.342+0.017\\\\0.94\mu=0.359\\\\\mu=\frac{0.359}{0.94}=0.382

Therefore, the coefficient of kinetic friction is 0.382.

5 0
4 years ago
An object located near the surface of Earth has a weight of a 245 N
marusya05 [52]

Answer:

The mass of the object is 24.5 kg and weight of the object on Mars is 91.14 N.

Explanation:

Weight of the object on the surface of Earth, W = 245 N

On the surface of Earth, acceleration due to gravity, g = 10 m/s²

Weight of an object is given by :

W = mg

m is mass

m=\dfrac{W}{g}\\\\m=\dfrac{245\ N}{10\ m/s^2}\\\\=24.5\ kg

So, the mass of the object is 24.5 kg

Acceleration due to gravity on Mars, g' = 3.72 m/s²

Weight of the object on Mars,

W' =mg'

W' = 24.5 kg × 3.72 m/s²

= 91.14 N

So, the weight of the object on Mars is 91.14 N.

4 0
3 years ago
Find a numerical value for rhoearth, the average density of the earth in kilograms per cubic meter. Use 6378km for the radius of
mylen [45]

Answer:

density = 5520 kg/m^3

Explanation:

given that

radius of earth = 6378 km

G = 6.67 x 10⁻¹¹ m³/kg.s²

g = 9.80 m/s²

we know,

g = \dfrac{GM}{r^2}

mass of earth

M = \dfrac{gr^2}{G}

M = \dfrac{9.8 \times (6378 \times 10^3)^2}{6.67 \times 10^{-11}}

M = 5.972 x 10²⁴ kg

density =\dfrac{mass}{volume}

V = volume of the earth = 4/3πr³

V = 4/3 x 3.14 x (6378  x 10³)³

V = 1.08 x 10²¹ m³

density = \dfrac{5.972\times 10^{24}}{1.08\times 10^{21}}

density = 5.52 x 10³  kg/m^3

density = 5520 kg/m^3

8 0
3 years ago
A vector is 14.4 m long and
MaRussiya [10]

Answer:

Explanation:

The x-component is found in the magnitude of the vector times the cosine of the angle.

A_x=14.4cos133 and, to 3 sig dig,

A_x=-9.82m

3 0
3 years ago
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