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Likurg_2 [28]
3 years ago
7

At what location in a circuit is the electrical potential energy the greatest

Physics
1 answer:
Eduardwww [97]3 years ago
4 0

It's not the potential energy.  It's just the potential.

It's greatest at the positive terminal of the battery or power supply.


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In addition to observing wind speed, wind direction, temperature, air pressure, and humidity, meteorologists observe what?
vagabundo [1.1K]

Answer:

Precipitation amounts

Explanation:

Meteorologists are scientists whose job is to predict different weather weather conditions. They study the atmosphere of the earth thoroughly so as to make a clear and concise weather forecast.

Meteorologists makes it possible for people to get information on possible rainfall or sunshine, it also enable people to know about a possible hurricance occurrence.

Meteorologists make use of a variety of tools to predict the weather, they include: wind vanes, barometers, anemometers, rain gauges, thermometer, hydrometer, Campbell Stokes Recorder,Transmissometer.

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3 years ago
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Ted wants to hang a wall clock on the wall by using a string. If the mass of the wall clock is 0. 250 kilograms, what should be
antoniya [11.8K]
Sum of all forces = mass * acceleration

Ft= tension force
Fw= force of gravity (Fw= mass* acceleration of gravity which is 9.8 this only applies to force of gravity)

Ft- Fw = 0 (there is no acceleration)
Ft = Fw
Ft= m*g
Ft= 0.250kg*9.8m/s
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7 0
2 years ago
The peak value of an alternating current in a 1500-W device is 6.4 A. What is the rms voltage across it
horrorfan [7]

Answer:

6787.5 V

Explanation:

From the question,

P = IV..................... Equation 1

Where P = Power, I = rms current, V = rms voltage.

make V the subject of the equation

V = P/I................. Equation 2

Given: P = 1500 W, I = 6.4/√2 = 4.525 A

Substitute these values into equation 2

V = 1500(4.525)

V = 6787.5 V

Hence the rms voltage = 6787.5 V

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3 years ago
Which is NOT one of the four principals of Dalton’s atomic theory?
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Answer:

Your answer would be letter <em><u>B</u></em><em><u>.</u></em><em><u> </u></em><em><u>Electrons</u></em><em><u> </u></em><em><u>orbit</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>nucleus</u></em><em><u> </u></em><em><u>in</u></em><em><u> </u></em><em><u>energy</u></em><em><u> </u></em><em><u>level</u></em><em><u>.</u></em>

Explanation:

Hope it helps..

Just correct me if I'm wrong, okay?

But ur welcome!!

(;ŏ﹏ŏ)(◕ᴗ◕✿)

8 0
2 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

3 0
3 years ago
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