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Jet001 [13]
3 years ago
10

Which of the following would be useful to show the growth rate of plants at varying temperatures?

Physics
1 answer:
Sergeu [11.5K]3 years ago
5 0

Answer:

the correct  answer my dude its the last choice

Explanation:

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When soccer players run they are using friction to propell themselves
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<span>If you put a magnet right next to a USB drive, depending on the strength of the magnet and the amount of steel, nickel or cobalt used in the construction of that particular model of USB drive, the drive would either adhere to, or not adhere to, the magnet. This would cause no other significant effects. The storage of data in solid state form (as in USB drives) is not magnetic in nature, so no deletion or any other damage of the stored data would occur.</span>
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7. Using Analogies Explain how the behavior
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The relative motion of gaseous particles increases with increase in the temperature of the gas molecules just like the motion of popcorn in a popper increases when heat is applied to the popper.

<h3>What is kinetic theory of gas?</h3>

The kinetic theory of gases or matter states that matter consists of tiny particles which are constant motion, colliding with one another and with walls of the containing vessels.

Just like a popcorn in a popcorn popper pops when heat is applied to the popper, gases contained in a cylinder increases their speed when they acquire more kinetic energy as the temperature of the cylinder increases.

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9 months ago
What is an example of convection currents? a. marshmallows toasting over a campfire b. a pot being heated by an electric burner,
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Answer:

B

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3 years ago
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A +71 nC charge is positioned 1.9 m from a +42 nC charge. What is the magnitude of the electric field at the midpoint of these c
viva [34]

Answer:

The net Electric field at the mid point is 289.19 N/C

Given:

Q = + 71 nC = 71\times 10^{- 9} C

Q' = + 42 nC = 42\times 10^{- 9} C

Separation distance, d = 1.9 m

Solution:

To find the magnitude of electric field at the mid point,

Electric field at the mid-point due to charge Q is given by:

\vec{E} = \frac{Q}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E} = \frac{71\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

\vec{E} = 708.03 N/C

Now,

Electric field at the mid-point due to charge Q' is given by:

\vec{E'} = \frac{Q'}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E'} = \frac{42\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

\vec{E'} = 418.84 N/C

Now,

The net Electric field is given by:

\vec{E_{net}} = \vec{E} - \vec{E'}

\vec{E_{net}} = 708.03 - 418.84 = 289.19 N/C

5 0
3 years ago
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