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3 years ago

two negative charges that are both -3.0 C push each other apart with a force of 19.2 N how far apart are the charges

1 answer:

5 0

The electrostatic force between two charges q1 and q2 is given by
$F=k_e \frac{q_1 q_2}{r^2}$
where $k_e =8.99 \cdot 10^9 N m^2 C^{-2}$ is the Coulomb's constant and r is the distance between the two charges.

If we use F=19.2 N and q1=q2=-3.0 C, we can find the value of r, the distance between the two charges by re-arranging the previous formula:
$r= \sqrt{k_e \frac{q_1 q_2}{F} }= \sqrt{ 8.99 \cdot 10^9 N m^2 C^{-2} \frac{(-3.0C)^2}{19.2 N} } =6.49 \cdot 10^4 m=64.9 km$

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3 years ago

A combination of two identical resistors connected in series has an equivalent resistance of 12. ohms. What is the equivalent re

SIZIF [17.4K]

Answer:

R1 + R2 = R = 12 for resistors in series - so R1 = R2 if they are identical

2 R1 = 12 and R1 = R2 = 6 ohms

1 / R = 1 / R1 + 1 / R2 for resistors in parallel

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2 years ago

A dockworker applies a constant horizontal force of 80.0 N to a block of ice on a smooth horizontal floor. The frictional force

Tamiku [17]

Answer:

(a) 91 kg (2 s.f.) (b) 22 m

Explanation:

Since it is stated that a constant horizontal force is applied to the block of ice, we know that the block of ice travels with a constant acceleration and but not with a constant velocity.

(a)

$s \ = \ ut \ + \ \displaystyle\frac{1}{2} at^{2} \\ \\ a \ = \ \displaystyle\frac{2(s \ - \ ut)}{t^{2}} \\ \\ a \ = \ \displaystyle\frac{2(11 \ - \ 0)}{5^{2}} \\ \\ a \ = \ \displaystyle\frac{22}{25} \\ \\ a \ = \ 0.88 \ \mathrm{m \ s^{-2}}$

Subsequently,

$F \ = \ ma \\ \\ m \ = \ \displaystyle\frac{F}{a} \\ \\ m \ = \ \displaystyle\frac{80 \ \mathrm{kg \ m \ s^{-2}}}{0.88 \ \mathrm{m \ s^{-2}}} \\ \\ m \ = \ 91 \mathrm{kg} \ \ \ \ \ \ (2 \ \mathrm{s.f.})$

*Note that the equations used above assume constant acceleration is being applied to the system. However, in the case of non-uniform motion, these equations will no longer be valid and in turn, calculus will be used to analyze such motions.

(b) To find the final velocity of the ice block at the end of the first 5 seconds,

$v \ = \ u \ + \ at \\ \\ v \ = \ 0 \ + \ (0.88 \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \\ \\ v \ = \ 4.4 \ \mathrm{m \ s^{-1}}$

According to Newton's First Law which states objects will remain at rest

or in uniform motion (moving at constant velocity) unless acted upon by

an external force. Hence, the block of ice by the end of the first 5

seconds, experiences no acceleration (a = 0) but travels with a constant

velocity of 4.4 $m \ s^{-1}$ .

$s \ = \ ut \ + \ \displaystyle\frac{1}{2}at^{2} \\ \\ s \ = \ (4.4 \ \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \ + \ \displaystyle\frac{1}{2}(0)(5^{2}) \\ \\ s \ = \ 22 \ \mathrm{m}$

Therefore, the ice block traveled 22 m in the next 5 seconds after the

worker stops pushing it.

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2 years ago

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It was Niels Bohr who proposed it

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3 years ago

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babunello [35]

Answer:

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Explanation:

"the problem of soil erosion"

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are examples of how it could be used.

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3 years ago

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