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creativ13 [48]
3 years ago
14

A 15 C point charge is located on the yaxis at (0, 0.25). A second charge of 10 C is located on the x-axis at (0.25, 0). If th

e two charges are separated by air, what is most nearly the force between them?

Physics
1 answer:
igomit [66]3 years ago
7 0

Answer:

10.8 N

Explanation:

The question requires the force between them, hence, we only need the magnitude of the force without considering what direction it's acting.

Parameters given:

Q1 = 15 * 10^(-6) C

Q2 = 10 * 10^(-6) C

The diagram explains better.

The electrostatic force BETWEEN Q1 and Q2 is:

F = (k * Q1 * Q2)/r²

Using Pythagoras theorem:

r² = 0.25² + 0.25² = 0.0625 + 0.0625

r² = 0.125

=> F = [9 * 10^9 * 15 * 10^(-6) * 10 * 10^(-6)]/0.125

F = 1.35/0.125

F = 10.8 N

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abruzzese [7]

Answer:

\boxed{\sf Kinetic \ energy \ (KE) = 85 \ J}

Given:

Mass (m) = 6.8 kg

Speed (v) = 5.0 m/s

To Find:

Kinetic energy (KE)

Explanation:

Formula:

\boxed{ \bold{\sf KE =  \frac{1}{2} m {v}^{2} }}

Substituting values of m & v in the equation:

\sf \implies KE =  \frac{1}{2}  \times 6.8 \times  {5}^{2}

\sf \implies KE = \frac{1}{ \cancel{2}}  \times  \cancel{2} \times 3.4 \times 25

\sf \implies KE =3.4 \times 25

\sf \implies KE = 85 \: J

8 0
4 years ago
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3 years ago
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The sound from a bolt of lightning travelled 4.08 km in 12.0 s. What was the speed
LenaWriter [7]

Answer:

83.67 m/s

Explanation:

Set up a calculation to convert units of measure to what you need.

You have km/s and you need m/s.

4.08km     1000 m         83.67m

-----------  X ----------  =  ---------------   the km will cancel out and you are left

 12.0 s          1 km              s              with m/s

6 0
3 years ago
A confined aquifer with a porosity of 0.15 is 30 m thick. The potentiometric surface elevation at two observation wells 1000 m a
AlekseyPX

Answer:

Part (a) The flow rate per unit width of the aquifer is 1.0875 m³/day

Part (b) The specific discharge of the flow is 0.0363 m/day

Part (c) The average linear velocity of the flow is 0.242 m/day

Part (d) The time taken for a tracer to travel the distance between the observation wells is 4132.23 days = 99173.52 hours

Explanation:

Part (a) the flow rate per unit width of the aquifer

From Darcy's law;

q = -Kb\frac{dh}{dl}

where;

q is the flow rate

K is the permeability or conductivity of the aquifer = 25  m/day

b is the aquifer thickness

dh is the change in th vertical hight = 50.9m - 52.35m = -1.45 m

dl is the change in the horizontal hight = 1000 m

q = -(25*30)*(-1.45/1000)

q = 1.0875 m³/day

Part (b) the specific discharge of the flow

V = \frac{Q}{A} = \frac{q}{b} = -K\frac{dh}{dl}\\\\V = -(25 m/d).(\frac{-1.45 m}{1000 m}) = 0.0363 m/day

V = 0.0363 m/day

Part (c) the average linear velocity of the flow assuming steady unidirectional flow

Va = V/Φ

Φ is the porosity = 0.15

Va = 0.0363 / 0.15

Va = 0.242 m/day

Part (d) the time taken for a tracer to travel the distance between the observation wells

The distance between the two wells = 1000 m

average linear velocity = 0.242 m/day

Time = distance / speed

Time = (1000 m) / (0.242 m/day)

Time = 4132.23 days

        = 4132.23 days *\frac{24 .hrs}{1.day} = 99173.52, hours

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3 years ago
What force holds the earth-moon system together?
Verdich [7]
Gravity holds the system together
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