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3 years ago

The velocity of a car is 65 m/s and it’s mass is 2515 kg. What is it’s KE?

2 answers:

4 0

Example Problems. Kinetic Energy (KE = ½ m v2). 1) The velocity of a car is 65 m/s and its mass is 2515 kg. What is its KE? 2) If a 30 kg child were running at a rate of 9.9 m/s, what is his KE? Practice Problems. IN THIS ORDER…. Page 2: #s 6, 7, 8, 5. Potential Energy. An object can store energy as the result of its position.

spin [16.1K]3 years ago

3 0

Answer:

The KE will be 5312.9375 KJ

Explanation:

Mass of car = m = 2515 kg

Velocity of car = v = 65 m/s

Kinetic Energy = KE = ?

We know that,

KE = 1/2 mv^2 ...... (i)

By putting values in equation (i)

KE = 1/2 (2515)(65)^2

KE = 5312937.5 J

KE = 5312.9375 KJ

Hence, KE of the car will be 5312.9375 KJ.

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3 years ago

You take a couple of capacitors and connect them in series, to which you observe a total capacitance

Zepler [3.9K]

Answer:

Approximately $\rm 5.7\; \mu F$ and approximately $29\; \rm \mu F$ .

Explanation:

Let $C_1$ and $C_2$ denote the capacitance of these two capacitors.

When these two capacitors are connected in parallel, the combined capacitance will be the sum of $C_1$ and $C_2$ . (Think about how connecting these two capacitors in parallel is like adding to the total area of the capacitor plates. That would allow a greater amount of charge to be stored.)

$C(\text{parallel}) = C_1 + C_2$ .

On the other hand, when these two capacitors are connected in series, the combined capacitance should satisfy:

$\displaystyle \frac{1}{C(\text{series})} = \frac{1}{C_1} + \frac{1}{C_2}$ .

(Consider how connecting these two capacitors in series is similar to increasing the distance between the capacitor plates. The strength of the electric field ( $V$ ) between these plates will become smaller. That translates to a smaller capacitance if the amount of charge stored $(Q)$ stays the same.)

The question states that:

$C(\text{parallel}) = 35\; \rm \mu F$ , and
$C(\text{series}) = 4.8\; \rm \mu F$ .

Let the capacitance of these two capacitors be $x\; \rm \mu F$ and $y\; \rm \mu F$ . The two equations will become:

$\displaystyle \left\lbrace \begin{aligned}& x + y = 35 \\ & \frac{1}{x} + \frac{1}{y} = \frac{1}{4.8}\end{aligned}\right.$ .

From the first equation:

$y = 35 - x$ .

Hence, the $y$ in the second equation here can be replaced with $(35 - x)$ . That equation would then become:

$\displaystyle \frac{1}{x} + \frac{1}{35 - x} = \frac{1}{4.8}$ .

Solve for $x$ :

$\displaystyle \frac{x + (35 - x)}{x \, (35 - x)} = \frac{1}{4.8}$ .

$x\, (35 - x) = 4.8$ .

$x^2 - 35 \, x + 168 = 0$ .

Solve this quadratic equation for $x$ :

$x \approx 5.7$ or $x \approx 29.3$ .

Substitute back into the equation $y = 35 - x$ for $y$ :

$x \approx 5.7$ and $y \approx 29.3$ , or
$x \approx 29.3$ and $x \approx 5.7$ .

In other words, these two capacitors have only one possible set of capacitances (even though the previous quadratic equation gave two distinct real roots.) The capacitances of the two capacitors would be approximately $5.7\; \rm \mu F$ and approximately $29\; \rm \mu F$ (both values are rounded to two significant digits.)

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3 years ago