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ss7ja [257]
4 years ago
7

The graph below shows the velocity of a car over time.

Physics
2 answers:
Vadim26 [7]4 years ago
6 0

Answer:

its calculate the area under the graph like the other person said :0

Explanation:

jekas [21]4 years ago
3 0

Answer:

D. Calculate the area under the graph.

Explanation:

The distance made during a particular period of time is calculated as (distance in m) = (velocity in m/s) * (time in s)

You can think of such a calculation as determining the area of a rectangle whose sides are velocity and time period. If you make the time period very very small, the rectangle will become a narrow "bar" - a bar with height determined by the average velocity during that corresponding short period of time. The area is, again, the distance made during that time. Now, you can cover the entire area under the curve using such narrow bars. Their areas adds up, approximately, to the total distance made over the entire span of motion. From this you can already see why the answer D is the correct one.

Going even further, one can make the rectangular bars arbitrarily narrow and cover the area under the curve with more and more of these. In fact, in the limit, this is something called a Riemann sum and leads to the definition of the Riemann integral. Using calculus, the area under a curve (hence the distance in this case) can be calculated precisely, under certain existence criteria.

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A puddle of water has a thin film of gasoline floating on it. A beam of light is shining perpendicular on the film. If the wavel
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Answer:

D.

Explanation:

To solve the problem it is necessary to apply the concepts of Destructive and constructive interference. The constructive interference in tin film is given by

2t = (m+\frac{1}{2})\frac{\lambda}{n}

Where,

t = thickness

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2nt = \frac{\lambda}{2}

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A.
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Answer:

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Determine
djyliett [7]

(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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