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3 years ago

The Clean Air Act emphasizes that one way to prevent and reduce air pollution is to involve public participation true or false

Physics

2 answers:

Anna35 [415]3 years ago

7 0

True hope this helped

oksian1 [2.3K]3 years ago

3 0

Answer:

The answer is: True.

Explanation:

The Clean Air Act considers that, in order to protect the environment as a whole as well as human health, it is necessary for member states to take measures, the concentrations of harmful air pollutants must be avoided, prevented or reduced, and limit values established, or alert thresholds for air pollution levels.

As well as having adequate information on the quality of the ambient air, and ensuring that the public has knowledge of it, among other things through alert thresholds.

The answer is: True.

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Can there be a virus that can turn humans into phsyical zombies or something like that?...

seraphim [82]

Answer:

Well for humans, most people think a rabies virus could do it (if it mutates fast). The problem is, rabies isn't really contagious enough to start a pandemic per se, but it would basically turn you into a "zombie". There are also a ton of other rare viruses and fungi that could trigger that, but they aren't proven on humans.

5 0

3 years ago

A mysterious crate has shown up at your place of work, Firecracker Company, and you are told to measure its inertia. It is too h

LUCKY_DIMON [66]

Answer:

the inertia of the crate is (49.67 kg)r²

Explanation:

Given the data in the question;

First; we will use the law of conservation of momentum to determine the mass of the crate;

m₁v₁ - m₂v₂ = 0

given that; m₁ = 0.60 kg and v₂ = 0.058 m/s

we substitute

0.60 × v₁ = m₂ × 0.058 = 0

m₂ = 0.60v₁ / 0.058 ----------- EQU 1

Next, we use the energy conservation relation to find the velocity

According to conservation of energy;

1/2m₁v₁² + 1/2m₂v₂² = 7 J

we substitute

1/2×0.60×v₁² + 1/2×m₂×(0.058)² = 7 J

0.3v₁² + 0.001682m₂ = 7 J ----- EQU 2

substitute value of m₂ form equ 1 into equ 2

0.3v₁² + 0.001682(0.60v₁ / 0.058) = 7 J

0.3v₁² + 0.0174v₁ = 7 J

0.3v₁² + 0.0174v₁ - 7 J = 0

we solve the quadratic equation;

{ x = [-b±√( b² - 4ac)] / 2a }

v₁ = [-0.0174 ±√( 0.0174² - 4×0.3×-7)] / 2×0.3

= [-0.0174 ±√(8.4003)] / 0.6

= [-0.0174 ± 2.8983 ] / 0.6

= -4.8595 or 4.8015 but{ v₁ ≠ - }

so v₁ = 4.8015 m/s ≈ 4.802 m/s

next we input value of v₁ into equation 1

m₂ = (0.60×4.8015) / 0.058

m₂ = 2.8809 / 0.058

m₂ = 49.67 kg

So, the moment of inertia of the crate will be;

I₂ = m₂r²

we substitute value of m₂

I₂ = (49.67 kg)r²

Therefore, the inertia of the crate is (49.67 kg)r²

5 0

3 years ago

A student pulls on a rope attached to a box of books and moves the box down the hall. The student pulls with a force of 185N at

Vitek1552 [10]

Answer:

$2.75 m/s^2$

Explanation:

We can solve the problem by writing the equations of motion along the horizontal and vertical direction.

Along the horizontal direction we have:

$T cos \theta - \mu N = ma$ (1)

where

$T cos \theta$ is the horizontal component of the tension, where

T = 185 N is the tension in the rope

$\theta=25^{\circ}$ is the angle between the rope and the horizontal

$\mu N$ is the force of friction, where

$\mu=0.27$ is the coefficient of friction

N is the normal reaction of the floor on the box

m = 35.0 kg is the mass of the box

a is the acceleration

Along the vertical direction we have:

$N+T sin \theta-mg=0$ (2)

where

N is the normal force (upward direction)

$T sin \theta$ is the vertical component of the tension in the rope (upward direction)

$mg$ is the weight of the box (downward direction), where

m = 35.0 kg is the mass of the box

$g=9.8 m/s^2$ is the acceleration due to gravity

From eq.(2) we get:

$N=mg-T sin \theta$

And substituting into (1), we can find the acceleration:

$T cos \theta - \mu (mg-T sin \theta) = ma\\Tcos \theta -\mu mg + \mu T sin \theta = ma\\a=\frac{T cos \theta- \mu mg + \mu T sin \theta}{m}=\\=\frac{(185)(cos 25^{\circ})-(0.27)(35.0)(9.8)+(0.27)(185)(sin 25^{\circ})}{35.0}=2.75 m/s^2$

8 0

2 years ago

Consider a thin circular disk that has been heated up to 400 °C and then left inside a chamber to cool down. The chamber surface

kogti [31]

Answer:

hello your question lacks the required diagram attached below is the complete question with the required diagram

answer : Qtotal = 807.4 Mw

Explanation:

Given Data :

disk properties :

∈ = 0.65

D = 200 mm

Ts = 400⁰c

attached below is the detailed solution

The total rate of Heat transferred from the disk

Qtotal = 807.4 Mw

7 0

3 years ago

A lightbulb is rated by the power that it dissipates when connected to a given voltage. For a lightbulb connected to 120 V house

zhenek [66]

1. increase, 2. decrease

3 0

3 years ago