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4 years ago

What is the diameter of a circle

2 answers:

5 0

A diameter is a straight line that begins at one end of the circle and ends at the opposite end. The condition for a line to be a diameter is that it must pass through the centre of the circle. If a line that touches two points on the circle, does not pass from the centre of the circle, it cannot be called a diameter.
The length of the diameter is the measurement of the line that connects the two points of the circle.

Komok [63]4 years ago

3 0

The Diameter starts at one side of the circle, goes through the center and ends on the other side.

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One point of the circuit is grounded (V = 0). What are the (a) size and (b) direction (up or down) of the current through resist

Svetach [21]

<h3>Answer:</h3>

(a) i₁ = 0.03818 A = 38.18 mA

(b) downward

(c) i₂ = 0.01091 A = 10.91 mA

(d) rightward

(e) i₃ = 0.02727 A = 27.27 mA

(f) leftward

(g) Eₐ = 3.818 Volts

<h3>Question:</h3>

The complete question is stated below and the figure is provided in the attachment:

In Fig. 27-47, E 1 = 6.00 V, E 2 = 12.0 V, R1 = 100 Ω,

R2 = 200 Ω, and R3 = 300 Ω. One point of the circuit is grounded

(V = 0). What are the (a) size and (b) direction (up or down) of the

current through resistance 1, the (c) size and (d) direction

(left or right) of the current through resistance 2, and the

(e) size and (f) direction of the current through resistance 3?

(g) What is the electric potential at point A?

<h3></h3><h3>Explanation:</h3>

Applying Kirchoff's voltage law in the loops of botg E₁ and E₂, in the clockwise and anti clockwise direction:

E₁ - i₂R₂ - i₁R₁ = 0

E₂ - i₃R₃ - i₁R₁ = 0

If, we apply Kirchhoff's current law at junction A, we get:

i₁ = i₂ + i₃

Using these relations in loop equations, and re-arranging:

E₁ - i₂R₂ - (i₂ + i₃) R₁ = 0 ___________ eqn (1)

E₂ - i₃R₃ - (i₂ + i₃) R₁ = 0 ___________ eqn (2)

Eqn (1) implies:

6 - 200 i₂ - 100 i₂ - 100 i₃ = 0

i₂ = (6 - 100i₃)/300

Eqn (2) implies:

12 - 300 i₃ - 100 i₂ - 100 i₃ = 0

12 - 400 i₃ = 100 i₂

using value of i₃ from eqn (1)

12 - 400 i₃ = (1/3)(6 - 100 i₃)

36 - 1200 i₃ = 6 - 100 i₃

1100 i₃ = 30

i₃ = 0.02727 A

using this value in eqn of i₂:

i₂ = [6 - 100(0.02727)]/300

i₂ = (6 - 2.727)/300

i₂ = 0.01091 A

Since:

i₁ = i₂ + i₃

i₁ = 0.01091 A + 0.02727 A

i₁ = 0.03818 A

(a)

i₁ = 0.03818 A = 38.18 mA

(b)

Since, the value of current is positive, thus it will have the direction that was assumed.

Therefore, its direction will downward

(c)

i₂ = 0.01091 A = 10.91 mA

(d)

Since, the value of current is positive, thus it will have the direction that was assumed.

Therefore, its direction will rightward

(e)

i₃ = 0.02727 A = 27.27 mA

(f)

Since, the value of current is positive, thus it will have the direction that was assumed.

Therefore, its direction will leftward

(g)

With respect to the grounded portion, the potential drop at the resistance 1 will be equal to the potential at A Eₐ.

Therefore,

Eₐ = i₁R₁

Eₐ = (0.03818 A)(100 Ω)

Eₐ = 3.818 Volts

3 0

4 years ago

What does increasing the impulse on an object due to its momentum?

allochka39001 [22]

increaskng the impulse also increases momentum

7 0

2 years ago

Why does the amount of water that flows in a river change during the year?

lyudmila [28]

It mainly depends on the season of the year

4 0

4 years ago

Read 2 more answers

What two things affect the density of Water?

nikdorinn [45]

The temperature of the water and the and the salinity of water

5 0

3 years ago

An astronaut weighs 8.00 × 102 newtons on the sur- face of Earth. What is the weight of the astronaut 6.37 × 106 meters above th

kolbaska11 [484]

Answer:

$mg=200.4 N$ .

Explanation:

This problem can be solved using Newton's law of universal gravitation: $F=G\frac{m_{1}m_{2}}{r^{2}}$ ,

where F is the gravitational force between two masses $m_{1}$ and $m_{2}$ , $r$ is the distance between the masses (their center of mass), and $G=6.674*10^{-11}(m^{3}kg^{-1}s^{-2})$ is the gravitational constant.