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kenny6666 [7]
3 years ago
12

The electrical charge of an atom as a whole is?

Physics
2 answers:
gayaneshka [121]3 years ago
8 0
<span>The electrical charge of an atom as a whole is?

Answer:</span><span>C. Neutral</span>

Natali5045456 [20]3 years ago
5 0
Im pretty sure its <span>C. Neutral.</span>
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Choose all the answers that apply.
slava [35]
<h2>So, the correct answers are:</h2>

Travels in longitudinal waves

Travels most slowly through a gas

Speeds up when temperature is increased

Is caused by vibration

Explanation for correct answers:

Yes, it does travel in longitudinal waves

Yes, sounds weird, but travels faster in the water

Yes, does speed up when temperature is increased

And yes, Is caused by vibration.

<h2>Wrong answer is:</h2>

Can travel through a vacuum

Explanation for wrong answer:

actually, in space, there is NO sound, because there are no particals for the sound to vibrate with, there's just empty SPACE.

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2 years ago
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3 years ago
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What does fossil fuels do?
katrin [286]
Fossil fuel are collected and turned into oils like car gas and many other products have fossil fuels in them and we don't even realize it idk oil was made out of fossils till i studied fossils in class Hope this helps:)
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3 years ago
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The pressure drop needed to force water through a horizontal 1-in diameter pipe if 0.60 psi for every 12-ft length of pipe. Dete
oksian1 [2.3K]

Answer:

The shear stress at a distance 0.3-in away from the pipe wall is 0.06012lb/ft²

The shear stress at a distance 0.5-in away from the pipe wall is 0

Explanation:

Given;

pressure drop per unit length of pipe = 0.6 psi/ft

length of the pipe = 12 feet

diameter of the pipe = 1 -in

Pressure drop per unit length in a circular pipe is given as;

\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\

make shear stress (τ) the subject of the formula

\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\\tau = \frac{\delta P *r}{2L}

Where;

τ is the shear stress on the pipe wall.

ΔP is the pressure drop

L is the length of the pipe

r is the distance from the pipe wall

Part (a) shear stress at a distance of  0.3-in away from the pipe wall

Radius of the pipe = 0.5 -in

r = 0.5 - 0.3 = 0.2-in = 0.0167 ft

ΔP = 0.6 psi/ft

ΔP, in lb/ft² = 0.6 x 144 = 86.4 lb/ft²

\tau = \frac{\delta P *r}{2L}  = \frac{86.4 *0.0167}{2*12} =0.06012 \ lb/ft^2

Part (b) shear stress at a distance of  0.5-in away from the pipe wall

r = 0.5 - 0.5 = 0

\tau = \frac{\delta P *r}{2L}  = \frac{86.4 *0}{2*12} =0

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4 years ago
How much physical activity do adults need?
serg [7]
30? Physical activity
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3 years ago
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