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2 years ago

Compare the current in the 8-ohm resistors to the current in the 4-ohm resistors.

Physics

1 answer:

Gemiola [76]2 years ago

7 0

Answer:

a) i₈ = 0.5 i₄, b) i₁₀ = 0.3 i₃, i₁₀ = 0.8 i₈

Explanation:

For this exercise we use ohm's law

V = i R

i = V / R

we assume that the applied voltage is the same in all cases

let's find the current for each resistance

R = 4 Ω

i₄ = V / 4

R = 8 Ω

i₈ = V / 8

we look for the relationship between these two currents

i₈ /i₄ = 4/8 = ½

i₈ = 0.5 i₄

R = 3 Ω

i₃ = V3

R = 10 Ω

i₁₀ = V / 10

we look for relationships

i₁₀ / 1₃ = 3/10

i₁₀ = 0.3 i₃

i₁₀ / 1₈ = 8/10

i₁₀ = 0.8 i₈

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How do I go about this?

Anna71 [15]

Hi there!

(a)

Recall that:
$W = F \cdot d = Fdcos\theta$

W = Work (J)
F = Force (N)
d = Displacement (m)

Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.

$W =248(56)cos(30) = 12027.36 J$

To the nearest multiple of ten:
$W_A = \boxed{12030 J}$

(b)
The object is not being displaced vertically. Since the displacement (horizontal) is perpendicular to the force of gravity (vertical), cos(90°) = 0, and there is NO work done by gravity.

Thus:
$\boxed{W_g = 0 J}$

(d)

Recall that the force of kinetic friction is given by:
$F_{f} =\mu_k mg$

Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:
$W_f = -\mu_k mgd\\W_f = - (0.1)(56)(9.8)(56) = -3073.28 J$

In multiples of ten:
$\boxed{W_f = -3070 J}$

(e)
Simply add up the above values of work to find the net work.

$W_{net} = W_A + W_f \\\\W_{net} = 12027.36 + (-3073.28) = 8954.08 J$

Nearest multiple of ten:
$\boxed{W_{net} = 8950 J}}$

(f)
Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)
$F_{net} = F_{Ax} - F_f$

$W = F_{net} \cdot d = (F_{Ax} - F_f)$

$W = (F_Acos(30) - \mu_k mg)d\\W = (248cos(30) - 0.1(56)(9.8)) * 56 \\\\W = 8954.08 J$

Nearest multiple of ten:
$\boxed{W_{net} = 8950 J}$

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