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Archy [21]
3 years ago
15

You have a 81.9 81.9 mF capacitor initially charged to a potential difference of 11.1 11.1 V. You discharge the capacitor throug

h a 3.19 3.19 Ω resistor. What is the time constant?
Physics
2 answers:
-BARSIC- [3]3 years ago
8 0

Answer:

The time constant is 0.26 s

Explanation:

From ohm's law,

Current (I) = potential difference ÷ resistance = 11.1 ÷ 3.19 = 3.48 A

Quantity of electricity (Q) = capacitance × potential difference = 81.9×10×^-3 × 11.1 = 0.90909 C

time constant (t) = Q/I = 0.90909/3.48 = 0.26 s

lukranit [14]3 years ago
8 0
<h2>Answer:</h2>

0.261s

<h2>Explanation:</h2>

For an R-C discharging circuit, the time constant τ, is the product of the capacitance (C) of the capacitor and the resistance (R) of the resistor. i.e

τ = R x C               ------------------(i)

From the question;

R = 3.19Ω

C = 81.9mF =  81.9 x 10⁻³F

Substitute these values into equation (i) as follows;

τ = 81.9 x 10⁻³ x 3.19

τ = 261.261 x 10⁻³

τ = 0.261s

Therefore, the time constant is 0.261s

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Answer:

a)   x₁ = 290.50 feet ,  x₂ = 169.74 feet , b)  v_max= 41 mph

Explanation:

For this exercise we will work in two parts, the first with Newton's second law to find the acceleration of vehicles

X Axis          fr = m a

Y Axis          N-W = 0

                    N = W = mg

The force of friction has the expression

                  fr = μ N

We replace

                 μ mg = ma

                 a = μ g

                 g = 32 feet / s²

Let's calculate the acceleration for each coefficient and friction

μ              a (feet / s2)

0.599       19.168

0.536       17,152

0.480       15.360

0.350        11.200

These are the acceleration values, for the maximum distance we use the minimum acceleration (a₁ = 11,200 feet / s²) and for the minimum braking distance we use the maximum acceleration (x₂ = 19,168 feet / s²)

                 v² = v₀² - 2 a x

When the speed stops it is zero

                 x₁ = v₀² / 2 a₁

                         

Let's reduce speed

            v₀ = 55mph (5280 foot / 1 mile) (1h / 3600s) = 80,667 feet / s²

Let's calculate the maximum braking distance

            x₁ = 80.667² / (2 11.2)

            x₁ = 290.50 feet

The minimum braking distance

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            x₂ = 169.74 feet

b) maximum speed to stop at distance x = 155 feet

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            v₀ = √2 a x

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             v₀₁ = √ (2 11.2 155)

             v₀₁ = 58.92 feet / s

       

             v₀₂ = √ (2 19.168 155)

             v₀₂ = 77.08 feet / s

To stop at the distance limit in the worst case the maximum speed must be 58.92 feet / s = 40.85 mph = 41 mph

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