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NARA [144]
3 years ago
11

Point charge μC is located at x =, y = , point charge is located at x = 0m. What are (a)the magnitude and (b)direction of the to

tal electric force that these charges exert on a third point charge Q = μC at x = m, y = 0?
Physics
1 answer:
pishuonlain [190]3 years ago
6 0

Answer: The question has some details missing. here is the complete question ; Point charge 1.5 μC is located at x = 0, y = 0.30 m, point charge -1.5 μC is located at x = 0 y = -0.30m. What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 5.0 μC at x = 0.40 m, y = 0

Explanation:

  • a) First of all find the distance between the two charges;
  • x = 0, y = 0.30  and x = 0.40 m, y = 0
  • r = √( 0.4² + 0.3²)
  • = 0.5m

hence, the force F = 2Kq1q2cosθ /r²...............equation 1

but cosθ = y/r = 0.3/0.5

cosθ = 0.6

plugging back to equation 1;

F = 2 x 9 x 10^9 x 1.5 x 10^-6 x 5 x 10^-6 /0.5^2

F = 540 x 10^-3

Magnitude of Force = 0.54N

b) Direction is at angle 90

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SP1b.
nata0808 [166]

Answer:

2 m/s^2, west

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Vf=final velcoity

Vi=initial velocity

t=timw

a =  \frac{vf - vi}{t}

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\frac{15 - 25}{5}

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3 0
2 years ago
The iron nail’s mass is 16 grams and its temperature drops 650 C when dropped into the water. How much heat energy did the iron
Mice21 [21]

The heat energy transferred by the iron nail is 4680 J

Explanation:

The thermal energy transferred by a substance to another substance is given by the equation

Q=mC\Delta T

where

m is the mass of the substance

C is its specific heat capacity

\Delta T is its change in temperature

For the iron nail in this problem, we have:

m = 16 g

C=0.450 J/g^{\circ}C

\Delta T = -650^{\circ}C

So, the amount of heat energy given off by the nail is

Q=(16)(0.450)(-650)=-4680 J

where the negative sign indicates that the heat is given off.

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

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5 0
2 years ago
Need help solving this question.
MatroZZZ [7]

Answer:

See the answers below.

Explanation:

to solve this problem we must make a free body diagram, with the forces acting on the metal rod.

i)

The center of gravity of the rod is concentrated in half the distance, that is, from the end of the bar to the center there is 40 [cm]. This can be seen in the attached free body diagram.

We have only two equilibrium equations, a summation of forces on the Y-axis equal to zero, and a summation of moments on any point equal to zero.

For the summation of forces we will take the forces upwards as positive and the negative forces downwards.

ΣF = 0

-15+T-W=0\\T-W=15

Now we perform a sum of moments equal to zero around the point of attachment of the string with the metal bar. Let's take as a positive the moment of the force that rotates the metal bar counterclockwise.

ii) In the free body diagram we can see that the force acts at 18 [cm] of the string.

ΣM = 0

(15*9) - (18*W) = 0\\135 = 18*W\\W = 7.5 [N]

7 0
2 years ago
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