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3 years ago

What are the principals of electric arc welding

1 answer:

7 0

Principles of arc welding. Arc welding is a welding process, in which heat is generated by an electric arc struck between an electrode and the work piece. Electric arc is luminouselectrical discharge between two electrodes through ionized gas.

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How much force is needed to accelerate a 15kg bowling ball at 2 m/s^2

olga_2 [115]

Golf no doing what I iuroeurir to do my homework and my homework

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2 years ago

Read 2 more answers

*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude

kvv77 [185]

Answer:

Approximately $3.86\; {\rm N}$ (given that the magnitude of this charge is $-7.35\; {\rm \mu C}$ .)

Explanation:

If a charge of magnitude $q$ is placed in an electric field of magnitude $E$ , the magnitude of the electrostatic force on that charge would be $F = E\, q$ .

The magnitude of this charge is $q = 7.35\; {\rm \mu C}$ . Apply the unit conversion $1\; {\rm \mu C} = 10^{-6}\; {\rm C}$ :

$\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}$ .

An electric field of magnitude $E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}}$ would exert on this charge a force with a magnitude of:

$\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}$ .

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

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1 year ago

A star is held together by:gravitynuclear fissionnuclear fusionacceleration

alina1380 [7]

Nuclear fusion in the core tries to blow the star apart. Gravity holds it together. Whoever designed that system really knew what he was doing. I'm kinda grateful to him.

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2 years ago

Determine the magnitude of the effective value of g⃗ at a latitude of 60 ∘ on the earth. assume the earth is a rotating sphere.

dezoksy [38]

In addition to acceleration of gravity we experience centrifugal acceleration away from the axis of rotation of the earth. this additional acceleration has value ac = r w^2 where w = angular velocity and r is distance from your spot on earth to the earth's axis of rotation so r = R cos(l) where l = 60 deg is the lattitude and R the earth's radius and w = 1 / (24hr x 3600sec/hr)
now you look up R and calculate ac then you combine the centrifugal acc. vector ac with the gravitational acceleration vector ag = G Me/R^2 to get effective ag' = ag -

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2 years ago

Solve for the unknown in each of these circuits

Tamiku [17]

Ohms law: This law relates voltage difference between two points. Mathematically, the law states that V=IR;

Where

V = voltage difference ; in volts

I = Current ; in Amperes

R = Resistance ; in ohms

1. Answer : given that R = 10 ; V= 12 V ; I = ?

From ohms law, I = V/R

= 12/10

= 1.2 Amp.

2. Answer: given that R = 10 ; V= ? ; I = 5

From ohms law, V = IR

= 10×5 = 50 V

3 . Answer: given that R = ? ; V= 120 ; I = 5

From ohms law, R = V/I

= 120/5

= 24 Ω

4 . Answer: given that R = ? ; V= 10 ; I = 20

From ohms law, R = V/I

= 10/20

= 0.5 Ω

5 . Answer: given that R = 480 ; V= 24 ; I = ?

From ohms law, I = V/R

= 24/480

= 0.05 A

6. Answer: given that R = 150 ; V= ? ; I = 1

From ohms law, V = IR

= 1 × 150

= 150 V

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3 years ago