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Ira Lisetskai [31]
4 years ago
13

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

Physics
1 answer:
Marina CMI [18]4 years ago
7 0
I think the answer would be true that’s my opinion
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Amy uses 20N of force to push a lawn mower 10 meters. How much work does she do? *
babunello [35]

She does 200J .

We know she uses 20N of force and 10m is the distance. We multiply both numbers and we are given our answer of 200J. Hope this was helpful. :)

4 0
4 years ago
A potential difference of 3.00 nV is set up across a 2.00 cm length of copper wire that has a radius of 2.00 mm. How much charge
Anvisha [2.4K]

The number of charge drifts are 3.35 X 10⁻⁷C

<u>Explanation:</u>

Given:

Potential difference, V = 3 nV = 3 X 10⁻⁹m

Length of wire, L = 2 cm = 0.02 m

Radius of the wire, r = 2 mm = 2 X 10⁻³m

Cross section, 3 ms

charge drifts, q = ?

We know,

the charge drifts through the copper wire is given by

q = iΔt

where Δt = 3 X 10⁻³s

and i = \frac{V}{R}

where R is the resistance

R = \frac{pL}{r^{2} \pi }

ρ is the resistivity of the copper wire = 1.69 X 10⁻⁸Ωm

So, i = \frac{\pi(r)^{2}V  }{pL}

q = \frac{\pi(r^{2} )Vt }{pL}

Substituting the values,

q = 3.14 X (0.02)² X 3 X 10⁻⁹ X 3 X 10⁻³ / 1.69 X 10⁻⁸ X 0.02

q = 3.35 X 10⁻⁷C

Therefore, the number of charge drifts are 3.35 X 10⁻⁷C

3 0
3 years ago
Describe in brief the importance of measurement in our daily life?<br>​
Volgvan

Measurement tools make our lives safer and better? And they increase the quality and quantity of life. Arguably, the ability to measure calculating

physical properties accurately have tremendous survival value that gives humans adaptive, evolutionary advantages sharpened through many years of natural selection.

6 0
4 years ago
In lab, a radiation detector was used to calculate the background radiation. The
Keith_Richards [23]
Not sure but try a tutor!
8 0
3 years ago
A capacitor consists of two closely spaced metal conductors of large area, separated by a thin insulating foil. It has an electr
Nata [24]

Answer:

- E = 11.55J

- Q = 0.17C

- E' = (1/4)E

Explanation:

- To calculate the amount of energy stored in the capacitor, you use the following formula:

E=\frac{1}{2}CV^2

C: capacitance = 3800.0*10^-6F

V: potential difference = 78.0V

E=\frac{1}{2}(3800.0*10^{-6}C)(78.0V)^2=11.55J

The energy stored in the capacitor is 11.55J

- If the electrical energy stored in the capacitor is 6.84J, the charge on the capacitor is:

E=\frac{1}{2}QV\\\\Q=\frac{2E}{V}\\\\Q=\frac{2(6.84J)}{78.0V}=0.17C

The charge on the capacitor is 0.17C

- If you take the capacitor as a parallel plate capacitor, you have that the energy stored on the capacitor is:

E=\frac{1}{2}CV^2=\frac{1}{2}(\frac{\epsilon_oA}{d})V^2=\frac{1}{2}\frac{\epsilon_oAV^2}{d}\\\\

A: area of the plates

d: distance between plates

If the distance between plates is increased by a factor of 4, you have:

E'=\frac{1}{2}\frac{\epsilon_oAV^2}{(4d)}=\frac{1}{4}\frac{\epsilon_oAV^2}{2d}=\frac{1}{4}E

Then, the stored energy in the capacitor is decreased by a a factor of (1/4)

6 0
3 years ago
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