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Dimas [21]
3 years ago
14

Three forces act on an object. a 3 n force acts due east and a 8 n force acts due north. if the net force on the object is zero,

what is the magnitude of the third force?"
Physics
1 answer:
Bingel [31]3 years ago
6 0

The third force C is in the opposite direction of the sum of the first two forces, A+B, but the magnitude is the same as A+B  

Since A & B are at right angles we can just use Pythagoras and trig:  

A+B magnitude = sqrt(6^2 + 3^2) = 6.71  

A+B angle = arctan(6/3) deg north of east = 63.4 deg (first quadrant)

So C magnitude = 6.71 (same as A+B)  

[C angle = 63.4 deg south of west (third quadrant) = -180+63.4 deg polar = -116.6 deg

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a. The ratio of their resistance is 2783:64

b. The ratio of their energy is 4:23

c. The charge on the first bulb is 5.75 C

The charge on the second bulb is 0.\overline {36} C

Explanation:

The voltage on one of the electric bulbs, V₁ = 40  volts

The power rating of the bulb, P₁ = 230 w

The voltage on the other electric bulbs, V₂ = 110 volts

The power rating of the bulb, P₂ = 40 w

a. The power is given by the formula, P = I·V = V²/R

Therefore, R = V²/P

For the first bulb, the resistance, R₁ = 40²/230 ≈ 6.96

The resistance of the second bulb, R₂ = 110²/40

The ratio of their resistance, R₂/R₁ = (110²/40)/(40²/230) = 2783/64

∴ The ratio of their resistance, R₂:R₁ = 2783:64

b. The energy of a bulb, E = t × P

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t = The time in which the bulb is powered on

∴ The energy of the first bulb, E₁ = 230 w × t

The energy of the second bulb, E₂ = 40 w × t

The ratio of their energy, E₂/E₁ = (40 w × t)/(230 w × t) = 4/23

∴ The ratio of their energy, E₂:E₁ = 4:23

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I = The current flowing through the bulb

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I = P/V

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The charge on the first bulb per second (t = 1) is therefore;

Q₁ = 5.75 A × 1 s = 5.75 C

The charge on the first bulb, Q₁ = 5.75 C

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The charge on the second bulb, Q₂ = 0.\overline {36} C.

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