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mixas84 [53]
4 years ago
15

PLEASE HELP!!!!

Physics
1 answer:
andrew11 [14]4 years ago
6 0
0.304 cm I think - let me check
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What is system thinking ?
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Answer: Systems thinking is a holistic approach to analysis that focuses on the way that a system's constituent parts interrelate and how systems work over time and within the context of larger systems.

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In root pass welding, that electrode we are going to use?​
postnew [5]

Answer:

The root pass is made with a 5/32” (4.0mm) diameter electrode. A cellulosic electrode (E-XX10) is being used. The root pass is welded with reverse (DC+) polarity.

Explanation:

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Is a charge of 5.8×10^-18 is possible
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I think yes. possible. Because electron charge is much smaller than that..
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3 years ago
The velocity of an ocean wave is 5m/s. The distance between crests is 3m. What is the frequency of the wave
alexgriva [62]

Answer:

1.67m

5m/s

Explanation:

Wavelength of the wave = 3m

Speed of the wave  = 5m/s

The distance between crest and the adjacent trough of water waves is known as the wavelength of a wave.

 To find the frequency ;

        V = f∧

V is the speed of the wave

f is the frequency

∧ is the wavelength

  Insert the parameters and find the frequency;

        f  = V/ ∧   = 5 / 3  = 1.67Hz

The rate at which the wave passed a given point is the speed of the wave and it is 5m/s

4 0
3 years ago
Consider a positive charge Q and a point B twice as far away from Q as point A. What is the ratio of the electric field strength
Vikentia [17]

Answer:

\frac{E_{A}}{E_{B}}=4

Explanation:

The electric field is defined as the electric force per unit of charge, this is:

E=\frac{F}{q}.

The electric force can be obtained through Coulomb's law, which states that the electric force between to electrically charged particles is inversely proportional to the square of the distance between them and directly proportional to the product of their charges. The electric force can be expressed as

F=\frac{kQq}{r^{2}}.

By substitution we get that

E=\frac{kQq}{qr^{2}}\\\\E=\frac{kQ}{r^{2}}

Now, letting E_{A} be the electric field at point A, letting E_{B} be the electric field at point B, and letting R be the distance from the charge to A:

E_{A}=\frac{kQ}{R^{2}}\\\\E_{B}=\frac{kQ}{(2R)^{2}}.

The ration of the electric fields is

\frac{E_{A}}{E_{B}}=\frac{\frac{kQ}{R^{2}}}{\frac{kQ}{(2R)^{2}}}\\\\\frac{E_{A}}{E_{B}}=\frac{\frac{1}{R^{2}}}{\frac{1}{(2R)^{2}}}\\\\\frac{E_{A}}{E_{B}}=\frac{\frac{1}{R^{2}}}{\frac{1}{(4)R^{2}}}\\\\\\\frac{E_{A}}{E_{B}}=\frac{1}{\frac{1}{(4)}}\\\\\frac{E_{A}}{E_{B}}=4

This means that at half the distance, the electric field is four times stronger.

4 0
3 years ago
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