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Tomtit [17]
3 years ago
11

If you were a scientist trying to separate pollution molecules from the air, how might you do it? (This question appears on page

23 of Break It Down.)
Physics
2 answers:
joja [24]3 years ago
8 0
Using physical means such as electrostatic filters or mechanical filters :)
Readme [11.4K]3 years ago
5 0
Advances in technology can clean the pollutants from emissions or keep pollutants from being made. Some of the ways to improve the quality of the air are by using physical means, such as electrostatic filters or mechanical filters, or chemical means, such as wet scrubbers, to separate pollution from the air.
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Which statements describe a situation in which work is being done? Check all that apply.
andreev551 [17]
You didn’t submit a photo
5 0
3 years ago
Which example has gravitational potential energy?
Lostsunrise [7]

B. An apple about to fall from a tree branch

This is because potential energy is the energy before its actuall motion, or before kinetic energy (energy in motion). And the apple ABOUT TO FALL mentions a gravitational example.

5 0
3 years ago
Read 2 more answers
A wire carries 3.7 A of current. A second wire is placed parallel to the first 8.0 cm away. What is the current flowing through
IgorC [24]

Answer:

The current in second wire is 5.0 A.

(B) is correct option.

Explanation:

Given that,

Current in first wire = 3.7 A

Distance = 8.0 cm

We need to calculate the magnetic field due to the current carrying wire

Using formula of magnetic field

B=\dfrac{\mu_{0}I}{2\pi r}

Where, I = current

r = distance

Put the value into the formula

For first wire

B = \dfrac{\mu_{0}\times3.7}{2\pi \times3.4\times10^{-2}}...(I)

For second wire,

The distance is 8-3.7 =  4.3 cm

B' = \dfrac{\mu_{0}\times I'}{2\pi \times(8-3.4)\times10^{-2}}...(II)

The magnetic field in both the wires,

From equation (I) and (II)

\dfrac{\mu_{0}\times3.7}{2\pi \times3.4\times10^{-2}}= \dfrac{\mu_{0}\times I'}{2\pi \times(8-3.4)\times10^{-2}}

I'=\dfrac{3.7\times4.3\times10^{-2}}{3.4\times10^{-2}}

I'=4.68\ A\ approx = 5.0\ A

Hence, The current in second wire is 5.0 A.

8 0
4 years ago
A long solenoid, of radius a, is driven by an alternating current, so that the field inside is sinusoidal: B(t) = B0 cos(ωt) ˆz.
Alexxandr [17]

Answer:

Explanation:

Given that,

B(t) = B0 cos(ωt) • k

Radius r = a

Inner radius r' = a/2 and resistance R.

Current in the loop as a function of time I(t) =?

Magnetic flux is given as

Φ = BA

And the Area is given as

A = πr², where r = a/2

A = πa²/4

Then,

Φ = ¼ Bπa²

Φ(t) = ¼πa²Bo•Cos(ωt)

Then, the EMF is given as

ε(t) = -dΦ/dt

ε(t) = -¼πa²Bo • -ωSin(ωt)

ε(t) = ¼ωπa²Bo•Sin(ωt)

From ohms law,

ε = iR

Then, i = ε/R

I(t) = ¼ωπa²Bo•Sin(ωt) /R

This is the current induced in the loop.

Check attachment for better understanding

7 0
3 years ago
A positively charged object is brought near but not in contact with the top of an uncharged gold leaf electroscope. The experime
Olin [163]

Answer:

The leaves of the electroscope move further apart.

Explanation:

This is what happens; when the positive object is brought near the top, negative charges migrating from the gold leaves to the top. This is because the negative charges in the gold are attracted by the positive charge. Thus, it leaves behind a net positive charge on the leaves, though the scope remains neutral overall. To that effect, the leaves repel each other and move apart. If a finger touches the top of the electroscope at the moment when the positive object remains near the top, it basically grounds the electroscope and thus the net positive charge in the leaves flows to the ground through the finger. However, the positive object continues to "hold" negative charges in place at the top. Ar this moment the gold leaves have lost their net positive charge, so they no longer repel, and they move closer together. If the positive object is moved away, the negative charges at the top are no longer attracted to the top, and they redistribute themselves throughout the electroscope, moving into the leaves and charging them negatively.

Thus, the leaves move apart from each other again and we now have a negatively charged electroscope. If a negatively charged object is now brought close to the top, but without touching, the negative charges already in the electroscope will be repelled down toward the leaves, thereby making them more negative, causing them to repel more, and hence move even further apart.

So, the leaves move further apart.

7 0
3 years ago
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