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2 years ago

What are the sign and magnitude in coulombs of a point charge that produces a potential of -3.50 V at a distance of 2.00 mm

Physics

1 answer:

DanielleElmas [232]2 years ago

5 0

The magnitude and sign of the charge are 0.8 MC and negative respectively.

To find the answer, we need to know about the electric potential of a point charge.

<h3>What's the mathematical expression of the electric potential of a point charge?</h3>

Mathematically, the electric potential at a distance 'r' from a point charge 'q' is given as (Kq)/r.
K is the electrical constant with value 9×10^(-9) in vaccum.

<h3>What will be the magnitude and sign of a charge, if potential is -3.50V at 2mm?</h3>

From the expression of electric potential, charge is

q= (potential ×r)/K

= (-3.5×0.002)/ (9×10^(-9))

= -0.8 mega coulomb.

Thus, we can conclude that the magnitude and sign of the charge are 0.8 MC and negative respectively.

Learn more about the electric potential here:

brainly.com/question/14306881

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An airplane undergoes the following displacements, all at the same altitude: First, it flies 59.0 km in a direction 30.0° east o

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Answer:

B) 71.5 [km]

Explanation:

To solve this problem we will decompose each of the directions in the x & y axes.

To solve this problem we will decompose each of the directions in the x & y axes. also for a greater understanding of the angles, you should look at the attached image, which contains the orientations for each angle (clockwise or counterclockwise).

<u>59.0 km in a direction 30.0° east of north</u>

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$d_{1x}= 59*sin(30) = 29.5[km]\\d_{1y}= 59*cos(30) = 51.09[km]$

<u>58.0 km due south</u>

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$d_{2y} = - 58 [km]\\$

<u>It flies 100 km 30.0° north of west</u>

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Now we sum algebraically the components

$d_{x}=29.5-86.6 = -57.1[km]\\d_{y}=51.09 -58+50=43.09[km]\\\\$

Using the Pythagorean theorem we can find the magnitude of the displacement.

$d = \sqrt{(57.1)^{2} +(43.09)^{2} } \\d= 71.53[km]$

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3 years ago

What work is the person doing doing to the box if the box is just being held and it is not moving?

kobusy [5.1K]

Allowing for a platform in which the box holds potential energy

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3 years ago

Gayle runs at a speed of 3.85 m/s and dives on a sled, initially at rest on the top of a frictionless snow-covered hill. After s

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Answer:

Final velocity at the bottom of hill is 15.56 m/s.

Explanation:

The given problem can be divided into four parts:

1. Use conservation of momentum to determine the speed of the combined mass (Gayle and sled)

From the law of conservation of momentum (perfectly inelastic collision), the combined velocity is given as:

$p_i = p_f$

$m_1u_1 + m_2v_2 = (m_1 + m_2)v$

$v = \frac{(m_1u_1 + m_2v_2)}{(m_1 + m_2)}$

$v=\frac{[50.0\ kg)(3.85\ m/s) + 0]}{(50.0\ kg + 5.00\ kg)}= 3.5\ m/s$

2. Use conservation of energy to determine the speed after traveling a vertical height of 5 m.

The velocity of Gayle and sled at the instant her brother jumps on is found from the law of conservation of energy:

$E(i) = E(f)$

$KE(i) + PE(i) = KE(f) + PE(f)$

$0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)$

$v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]}$

Here, initial velocity is the final velocity from the first stage. Therefore:

$v(f) = \sqrt{[(3.5)^2+2(9.8)(5.00-0)]}= 10.5\ m/s$

3. Use conservation of momentum to find the combined speed of Gayle and her brother.

Given:

Initial velocity of Gayle and sled is, $u_1(i)=10.5$ m/s

Initial velocity of her brother is, $u_2(i)=0$ m/s

Mass of Gayle and sled is, $m_1=55.0$ kg

Mass of her brother is, $m_2=30.0$ kg

Final combined velocity is given as:

$v(f) = \frac{[m_1u_1(i) + m_2u_2(i)]}{(m_1 + m_2)}$

$v(f)=\frac{[(55.0)(10.5) + 0]}{(55.0+30.0)}= 6.79$ m/s

4. Finally, use conservation of energy to determine the final speed at the bottom of the hill.

Using conservation of energy, the final velocity at the bottom of the hill is:

$E(i) = E(f)$

$KE(i) + PE(i) = KE(f) + PE(f)$

$0.5mv^2(i) + mgh(i) = 0.5mv^2(f) + mgh(f)$

$v(f) = \sqrt{[v^2(i) + 2g(h(i) - h(f))]} \\v(f)=\sqrt{[(6.79)^2 + 2(9.8)(15 - 5.00)]}\\v(f)= 15.56\ m/s$

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In particular, the direction of the refracted ray is determined by Snell's Law:

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