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3 years ago

A rocket has a mass 280(103) slugs on earth. Specify its mass in SI units, and its weight in SI units.

1 answer:

8 0

To solve this problem we will apply the concepts related to the conversion of units for which we will have that 1 slug is equal to 14.59kg. At the same time we will use Newton's second law for which weight is defined as the product between mass and acceleration (Due to gravity). This is then

A: Using the conversion ratio of slug to kilogram we have to,

$1 slug = 14.59kg$

Then

$m = 280*10^3 slugs (\frac{14.59kg}{1slug})$

$m = 4.09*10^6kg$

B: Using Newton's second law we have to,

$W = mg$

$W = (4.09*10^6)(9.8)$

$W = 40034960N\approx 4*10^7N$

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3 years ago

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The answer is B. composition

Explanation:

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3 years ago

At a certain time a particle had a speed of 17 m/s in the positive x direction, and 3.0 s later its speed was 28 m/s in the oppo

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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