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3 years ago

A source charge generates an electric field of 1236 N/C at a distance of 4 m. What is the magnitude of the source charge?

Physics

2 answers:

SashulF [63]3 years ago

8 0

Answer:

Explanation:

A region around a charged particle where another charged particle experiences a force is called electric field.

Electric field strength is defined as the force experienced by unit positive test cahrge when placed in an electric field.

The formula of electric field due to a point cahrge is given by

E = k q / r^2

Where k is a constant and it's value is

9 × 10^9 Nm^2/C^2.

1236 = 9 × 10^9 × q / 16

q = 1.48 × 10^-3 Coulomb

lara [203]3 years ago

6 0

The magnitude of the source is 2.2 C

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Akimi4 [234]

Answer:

Approximately $122.625\; {\rm m}$ (assuming that $g = 9.81\; {\rm m\cdot s^{-2}}$ , the ball was launched from ground level, and that the drag on the ball is negligible.)

Explanation:

Let $v_{0}$ denote the velocity at which the ball was thrown upward.

If the drag (air friction) on the ball is negligible, the ball would land with a velocity of exactly $(-v_{0})$ . The velocity of the ball would be changed from $v$ to $(-v_{0})\!$ (such that $\Delta v = (-v_{0}) - v_{0} = (-2\, v_{0})$ ) within $t = 10\; {\rm s}$ .

Also because the drag on the ball is negligible, the acceleration of the ball would be $a = -g = -9.81\; {\rm m\cdot s^{-2}}$ . Thus:

$\Delta v = a\, t = 10\; {\rm s} \times (-9.81\; {\rm m\cdot s^{-2}}) = -98.1\; {\rm m\cdot s^{-1}}$ .

Since $\Delta v = (-2\, v_{0})$ :

$-2\, v_{0} = \Delta v = -98.1\; {\rm m\cdot s^{-1}$ .

$\begin{aligned}v_{0} &= \frac{-98.1\; {\rm m\cdot s^{-1}}}{-2}= 49.05\; {\rm m \cdot s^{-1}}\end{aligned}$ .

The ball reaches maximum height when its velocity is $v_{1} = 0\; {\rm m\cdot s^{-1}}$ . Apply the SUVAT equation $x = ({v_{1}}^{2} - {v_{0}}^{2}) / (2\, a)$ to find the displacement $x$ between the original position (ground level, where $v_{0} = 49.05\; {\rm m\cdot s^{-1}}$ ) and the max-height position of the ball (where $v_{1} = 0\; {\rm m\cdot s^{-1}}$ .)

$\begin{aligned}x &= \frac{(0\; {\rm m\cdot s^{-1}})^{2} - (49.05\; {\rm m\cdot s^{-1}})^{2}}{2 \times (-9.81\; {\rm m\cdot s^{-2}})} \\ &\approx 122.625\; {\rm m\cdot s^{-1}}\end{aligned}$ .

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3 years ago

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alekssr [168]

Answer:

Choice a. 1 kg, assuming that all other forces on the object (if any) are balanced.

Explanation:

By Newton's Second Law,

$\displaystyle a = \frac{\Sigma F}{m}$ ,

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$\displaystyle m = \frac{\Sigma F}{a}$ .

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<h3>Apply Newton's Second Law to find the mass of the object.</h3>

$\displaystyle m = \frac{\Sigma F}{a} = \frac{100\;\text{N}}{100\;\text{m}\cdot\text{s}^{-2}} = 1\;\text{kg}$ .

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Answer:

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