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stepladder [879]
3 years ago
5

Solving a physics problem is . . .

Physics
2 answers:
denis23 [38]3 years ago
5 0
I think its c cuz its step by step processes
Vsevolod [243]3 years ago
3 0
C. A step-by-step process that takes time, and is essential for learning physics concepts.
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Three +3.0-μC point charges are at the three corners of a square of side 0.50 m. The last corner is occupied by a −3.0-μC charge
kramer

Answer:

E = 440816.32 N/C

Explanation:

Given data:

Three point charge of charge equal to +3.0 micro coulomb

fourth point charge = - 3.0 micro coulomb

side of square = 0.50 m

K =1/4 \pi \epsilon_0 = 8.99 \times 10^9 N.m^2/c^2

Due to having equal charge on center of square, 2 charge produce equal electric field at center and other two also produce electric field at center of same value

So we have

E_1 + E_3 = 0

E =E_2 + E_4

E = 2 E_2

[E_2 =\frac{2\times k \times q}{r^2}

[r= \frac{(0.5^2 + 0.5^2)^2}{2} = 0.35 m]

plugging all value

E = 2 E_2

E = 2 E_2 =\frac{2\times k \times q}{r^2}

E = \frac{2 \times 8.99 \times 10^93\times 10^{-6}}{0.35^2}

E = 440816.32 N/C

3 0
3 years ago
Read 2 more answers
Albert uses as his unit of length (for walking to visit his neighbors or plowing his fields) the albert (a), the distance albert
True [87]

To solve this problem, we know that:

1 Albert = 88 meters

1 A = 88 m

The first thing we have to do is to square both sides of the equation:

(1 A)^2 = (88 m)^2

1 A^2 = 7,744 m^2

<span>Since it is given that 1 acre = 4,050 m^2, so to reach that value, 1st let us divide both sides by 7,744:</span>

1 A^2 / 7,744 = 7,744 m^2 / 7,744

(1 / 7,744) A^2 = 1 m^2

Then we multiply both sides by 4,050.

(4050 / 7744) A^2 = 4050 m^2

0.523 A^2 = 4050 m^2

<span>Therefore 1 acre is equivalent to about 0.52 square alberts.</span>

7 0
3 years ago
When an 8 V battery is connected to a resistor, a 2 A current flows in the resistor. What is the resistor's value?
Vinvika [58]

Answer:

B

Explanation:

V=IR    I= curren V=volt R=resistor

8=2.R   8/2=R R=4

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3 years ago
What are the conditions necessary for a terrestrial planet to have a strong magnetic field?.
Stells [14]
Both a molten metallic core and reasonably fast rotation.
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What are all the<br> invertebrates with a<br> large foot
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