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xz_007 [3.2K]
3 years ago
14

Lewiston and Vernonville are 208 miles apart. A car leaves Lewiston traveling towards​ Vernonville, and another car leaves Verno

nville at the same​ time, traveling towards Lewiston. The car leaving Lewiston averages 10 miles per hour more than the​ other, and they meet after 1 hour and 36 minutes. What are the average speeds of the​ cars?

Physics
1 answer:
iragen [17]3 years ago
3 0

Answer:

Average speed of the car A = 70 miles per hour

Average speed of the car B = 60 miles per hour

Explanation:

Average speed of the car A is v_{A} =\frac{x_{A} }{t_{A} } (Equation A) and Average speed of the car B is v_{B} =\frac{x_{B} }{t_{B} } (Equation B), where x_{A} and x_{B} are the distances and t_{A} and t_{B} are the times at which are travelling the cars A and B respectively.

We have to convert the time to the correct units:

1 hour and 36 minutes = 96 minutes

96 minutes . \frac{1 hour}{60 minutes} = 1.6 h

From the diagram (Please see the attachment), we can see that at the time they meet, we have:

v_{A} = \frac{208-x}{1.6h} + 10\frac{miles}{h} (Equation C)

v_{B} = \frac{208-x}{1.6h} (Equation D)

From Equation A and C, we have:

\frac{208-x}{1.6}+10 = \frac{x}{1.6}

208-x+16 = x

208 + 16 = 2x

x = \frac{224}{2}

x = 112 miles

Replacing x in Equation A:

v_{A}  = \frac{112miles}{1.6h}

v_{A} = 70 miles per hour

Replacing x in Equation B:

v_{B}  = \frac{208miles-112miles}{1.6h}

v_{B}  = \frac{96miles}{1.6h}

v_{B}  = 60 miles per hour

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(i) The total capacitance for the circuit is 5 μF.

(ii) The total charge stored in the circuit is 1 x 10⁻⁴ C.

(iii) The charge stored in 3μF capacitor is  6 x 10⁻⁶ C.

<h3>Total capacitance of the circuit</h3>

The total capacitance of the circuit is determined by reolving the series capacitors separate and parallel capacitors separate as well.

<h3>C1 and C2 are in series </h3>

\frac{1}{C_{12}} = \frac{1}{C_1 } + \frac{1}{C_2} \\\\\frac{1}{C_{12}} = \frac{1}{4 } + \frac{1}{4} \\\\\frac{1}{C_{12}} = \frac{1}{2} \\\\C_{12} = 2 \ \mu F

<h3>C1 and C2 are parallel to C3</h3>

C_{123} = C_{12} + C_3\\\\C_{123} = 2\ \mu F + 2\ \mu F \\\\C_{123} = 4 \ \mu F

<h3>C(123) is series to C5 and C6</h3>

\frac{1}{C_{t} } = \frac{1}{C_{123}} + \frac{1}{C_5} + \frac{1}{C_6} \\\\\frac{1}{C_{t} } = \frac{1}{4} + \frac{1}{6} + \frac{1}{6} \\\\\frac{1}{C_{t} } = \frac{12}{24} \\\\\frac{1}{C_{t} } = \frac{1}{2} \\\\C_t = 2 \ \mu F

<h3>C7 and C8 are in series</h3>

\frac{1}{C_{78}} = \frac{1}{6} + \frac{1}{6} \\\\\frac{1}{C_{78}} = \frac{2}{6} \\\\\frac{1}{C_{78}} =\frac{1}{3} \\\\C_{78} = 3 \ \mu F

<h3>Total capaciatnce of the circuit</h3>

Ct + C(78) = 2 μF + 3 μF = 5 μF

<h3 /><h3>Total charge stored in the circuit</h3>

The total charge stored in the capacitor is calculated as follows;

Q = CV

Q = (5 x 10⁻⁶) x (20)

Q = 1 x 10⁻⁴ C

<h3>Charge stored in 3μF capacitor</h3>

Q =  (3 x 10⁻⁶) x (20)

Q = 6 x 10⁻⁶ C

Learn more about capacitance of capacitor here: brainly.com/question/13578522

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