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avanturin [10]
3 years ago
12

How can human health be affected by changes in air quality?

Physics
1 answer:
Anna11 [10]3 years ago
3 0
Answer: Air quality changes like pollution and can cause problems in humans like heart disease, lung cancer, and other diseases. Air is formed by a certain ratio of Oxygen, Carbon Dioxide, Argon, and others. When this proportion is disturbed, it can also cause humans health problems. EX: The ash clouds of volcanos can cause silicosis (lung impairment).
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A natural atom can be negatively charged by
svet-max [94.6K]

Answer:

By Gaining Electrons

Explanation:

A nuetral atom is negative when it gains electrons, and it can be positive when it loses electrons.

4 0
3 years ago
Read 2 more answers
A Chef is looking for a new frying pan that will allow her to cook food quickly at temperatures above 500°C. Use the data provid
makvit [3.9K]

Answer:

Explanation:

1st one

What is your evidence?

Very heavy professional or restaurant pans will have iron handles, while those for home use will be made of brass or stainless steel. All are perfectly safe for oven use.

6 0
2 years ago
A pebble is thrown into the air with a velocity of 19/m at an angle of 36 with respect to the horizontal.
kow [346]

Answer:

The maximum height the pebble reaches is approximately;

A. 6.4 m

Explanation:

The question is with regards to projectile motion of an object

The given parameters are;

The initial velocity of the pebble, u = 19 m/s

The angle the projectile path of the pebble makes with the horizontal, θ = 36°

The maximum height of a projectile, h_{max}, is given by the following equation;

h_{max} = \dfrac{\left (u \times sin(\theta) \right)^2}{2 \cdot g}

Therefore, substituting the known values for the pebble, we have;

h_{max} = \dfrac{\left (19 \times sin(36 ^{\circ}) \right)^2}{2 \times 9.8} = 6.3633894140470403035477570509439

Therefore, the maximum height of the pebble projectile, h_{max} ≈ 6.4 m.

3 0
3 years ago
Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equ
Irina-Kira [14]

Answer:

Magnitude of the net force on q₁-

Fn₁=1403 N

Magnitude of the net force on q₂+

Fn₂= 810 N

Magnitude of the net force on q₃+

Fn₃= 810 N

Explanation:

Look at the attached graphic:

The charges of the same sign exert forces of repulsion and the charges of   opposite sign exert forces of attraction.

Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:

F= (k*q*q)/(d)²

F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N

Magnitude of the net force on q₁-

Fn₁x= 0

Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N

Fn₁=1403 N

Magnitude of the net force on q₃+

Fn₃x= 810- 810 cos 60° = 405 N

Fn₃y= 810*sin 60° = 701.5 N

Fn_{3} = \sqrt{405^{2}+701.5^{2}  }

Fn₃ = 810 N

Magnitude of the net force on q₂+

Fn₂ = Fn₃ = 810 N

6 0
3 years ago
Which amusement park ride utilizes periodic motion?
tatyana61 [14]
Any ride that oscillates back and forth or moves only in a complete circle utilizes periodic motion.
8 0
3 years ago
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