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svlad2 [7]
3 years ago
6

Environmental factors that can affect activity selection include

Physics
2 answers:
lord [1]3 years ago
8 0

Answer:A

Explanation:

Yuki888 [10]3 years ago
6 0

Answer: A.

Explanation: I hope this helps :) -

Environmental factors are diverse and far reaching.

...

They include:

Exposure to hazardous substances in the air, water, soil, and food.

Natural and technological disasters.

Climate change.

Occupational hazards.

The built environment. Goal

Promote health for all through a healthy environment.

Overview

Humans interact with the environment constantly. These interactions affect quality of life, years of healthy life lived, and health disparities. The World Health Organization (WHO) defines environment, as it relates to health, as “all the physical, chemical, and biological factors external to a person, and all the related behaviors.”1 Environmental health consists of preventing or controlling disease, injury, and disability related to the interactions between people and their environment.

The Healthy People 2020 Environmental Health objectives focus on 6 themes, each of which highlights an element of environmental health:

Outdoor air quality

Surface and ground water quality

Toxic substances and hazardous wastes

Homes and communities

Infrastructure and surveillance

Global environmental health

Creating healthy environments can be complex and relies on continuing research to better understand the effects of exposure to environmental hazards on people’s health.

Why Is Environmental Health Important?

Maintaining a healthy environment is central to increasing quality of life and years of healthy life. Globally, 23% of all deaths and 26% of deaths among children under age 5 are due to preventable environmental factors.1 Environmental factors are diverse and far reaching. They include:

Exposure to hazardous substances in the air, water, soil, and food

Natural and technological disasters

Climate change

Occupational hazards

The built environment

Related Topic Areas

Cancer

Heart Disease and Stroke

Occupational Safety and Health

Physical Activity

Respiratory Diseases

Poor environmental quality has its greatest impact on people whose health status is already at risk. Therefore, environmental health must address the societal and environmental factors that increase the likelihood of exposure and disease.

 -Have a good day :)

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What's the equivalent resistance to the resistances shown?<br><br>A) 5<br>B) 6<br>C) 2<br>D) 3
PIT_PIT [208]
From A to B its 5 ohm.
above shown 6 and 12 ohm resistors are in parallel to short circuit hence their equivalent resistance is zero.
(Current doesnt flow through a resisstor if there is a Short circuit alternate.
6 0
3 years ago
Please help this is for my final exam
ki77a [65]

Answer:

A

Explanation:

5 0
3 years ago
You measure the velocity of a drag racer that accelerates with constant acceleration. You want to plot the data and determine th
Ivenika [448]

Answer:

a) Linear equation

Explanation:

Definition of acceleration

a=\frac{dv}{dt}\\

if a=constant and we integrate the last equation

v(t)=v_{o}+a*t

So the relation between the time and the velocity is linear. If we plot the velocity in function of time, the plot is a line, and the acceleration is the slope of this line.

8 0
3 years ago
What eccentricity value results in a circular orbit?
Oxana [17]

Answer:

Zero

Explanation:

Given the equation of an ellipse:

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1

The eccentrity of an ellipse is given by:

e=\sqrt{1-\frac{b^2}{a^2}}

For a circle, we have

a=b

Therefore the eccentricity of a circle is

e=\sqrt{1-\frac{1^2}{1^2}}=0

7 0
3 years ago
A charge of 50 µC is placed on the y axis at y = 3.0 cm and a 77-µC charge is placed on the x axis at x = 4.0 cm. If both charge
zheka24 [161]

Answer:

The acceleration of an electron is 1.2\times10^{20}\ m/s^2

Explanation:

Given that,

One Charge = 50 μC

Distance on y axis = 3.0 cm

Second charge = 77 μC

Distance on x axis = 4.0 cm

We need to calculate the force on electron due to q₁

Using formula of force

F_{1}=\dfrac{kq_{1}q}{r^2}

Here, q = charge of electron

Put the value into the formula

F_{1}=\dfrac{9\times10^{9}\times50\times10^{-6}\times1.6\times10^{-19}}{(3\times10^{-2})^2}

F_{1}=8\times10^{-11}j\ \ N

We need to calculate the force on electron due to q₂

Using formula of force

F_{2}=\dfrac{kq_{2}q}{r^2}

Here, q = charge of electron

Put the value into the formula

F_{2}=\dfrac{9\times10^{9}\times77\times10^{-6}\times1.6\times10^{-19}}{(4\times10^{-2})^2}

F_{2}=6.93\times10^{-11}i\ \ N

We need to calculate the net force

Using formula of net force

F=F_{1}+F_{2}

Put the value into the formula

F=8\times10^{-11}j+6.93\times10^{-11}i

The magnitude of the net force

F=\sqrt{(8\times10^{-11})^2+(6.93\times10^{-11})^2}

F=1.058\times10^{-10}\ N

We need to calculate the acceleration of an electron

Using formula of force

F = ma

a=\dfrac{F}{m}

Put the value into the formula

a=\dfrac{1.058\times10^{-10}}{9.1\times10^{-31}}

a=1.2\times10^{20}\ m/s^2

Hence, The acceleration of an electron is 1.2\times10^{20}\ m/s^2

3 0
3 years ago
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