Which statement about the carbon cycle is most true?

What is a proper way and a safe way to dispose batteries

gregori [183]

Answer:

throw them away............

4 0

3 years ago

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Amy gets up from her seat on a train to move closer to the front. Just as she begins to walk forward, the train stops at a stati

Ivahew [28]

Have you ever been in a bus, a train, or a car ?
What happens to you when it stops suddenly ?

A body in motion continues in motion unless
an external force acts on it and makes it stop.

Amy's body keeps moving forward when the train stops.
She pitches forward, and if she doesn't reach out and grab
a seat or a seated person, she may lose her footing and fall
on her face.

Choice - 'A' is a very good explanation.
The other choices aren't.

'C' is a good hunch, but it only applies to her feet.
The rest of her keeps going.

'D' is nonsense. There are no mysterious forces of
'repulsion' or 'attraction' on the train.

5 0

3 years ago

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A uniform line charge of density λ lies on the x axis between x = 0 and x = L. Its total charge is 7 nC. The electric field at x

DedPeter [7]

Answer:

The electric field at x = 3L is 166.67 N/C

Solution:

As per the question:

The uniform line charge density on the x-axis for x, 0< x< L is $\lambda$

Total charge, Q = 7 nC = $7\times 10^{- 9} C$

At x = 2L,

Electric field, $\vec{E_{2L}} = 500N/C$

Coulomb constant, K = $8.99\times 10^{9} N.m^{2}/C^{2}$

Now, we know that:

$\vec{E} = K\frac{Q}{x^{2}}$

Also the line charge density:

$\lambda = \frac{Q}{L}$

Thus

Q = $\lambda L$

Now, for small element:

$d\vec{E} = K\frac{dq}{x^{2}}$

$d\vec{E} = K\frac{\lambda }{x^{2}}dx$

Integrating both the sides from x = L to x = 2L

$\int_{0}^{E}d\vec{E_{2L}} = K\lambda \int_{L}^{2L}\frac{1}{x^{2}}dx$

$\vec{E_{2L}} = K\lambda[\frac{- 1}{x}]_{L}^{2L}] = K\frac{Q}{L}[frac{1}{2L}]$

$\vec{E_{2L}} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{2L}] = \frac{63}{L^{2}}$

Similarly,

For the field in between the range 2L< x < 3L:

$\int_{0}^{E}d\vec{E} = K\lambda \int_{2L}^{3L}\frac{1}{x^{2}}dx$

$\vec{E} = K\lambda[\frac{- 1}{x}]_{2L}^{3L}] = K\frac{Q}{L}[frac{1}{6L}]$

$\vec{E} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{6L}] = \frac{63}{6L^{2}}$

Now,

If at x = 2L,

$\vec{E_{2L}} = 500 N/C$

Then at x = 3L:

$\frac{\vec{E_{2L}}}{3} = \frac{500}{3} = 166.67 N/C$

4 0

4 years ago

In a generator, as the magnet spins, opposite poles of the magnet push the electrons in opposite directions. This back-and-forth

Natali [406]

C) alternating current .
<span>
B)direct current </span>

6 0

4 years ago

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Which describes the average velocity of an ant traveling at a constant speed

yanalaym [24]

Answer: A. The total displacement divided by the time and C. The slope of the ant's displacement vs. time graph.

Explanation:

Hi! The question seems incomplete, but I found the options on the internt:

A. The total displacement divided by the time.

B. The slope of the ant's acceleration vs. time graph.

C. The slope of the ant's displacement vs. time graph.

D. The average acceleration divided by the time.

Now, since we know the ant is travelling at a constant speed, its average velocity $V$ will be expressed by the following equation:

$V=\frac{d}{t}$

Where:

$d$ is the ant's total displacement

$t$ is the time it took to the ant to travel to the kitchen

Hence one of the correct options is: A. The total displacement divided by the time

On the other hand, this can be expressed by a displacement vs. time graph graph, where the slope of that line leads to the equation written above. So, the other correct option is: C. The slope of the ant's displacement vs. time graph.

3 0

3 years ago