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3 years ago

if a boat was moving at a velocity of 30 m/s to the south and it sailed for 30 minutes, what was the displacement of the boat?

Physics

1 answer:

kifflom [539]3 years ago

8 0

If you move south at 30 m/s for 30 minutes, your displacement is

D = (30 mtr/s) x (30 min) x (60 sec/min)

D = (30 x 30 x 60) (mtr-min-sec/sec-min)

D = 54,000 meters south

D = <em>54 km south</em>

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Which statement correctly describes the relationship between the volume of a gas and its temperature, in Kelvin, assuming pressu

nekit [7.7K]

The relationship is directly proportional; as temperature increases, volume increases in the same way.

Charles's law states that at a constant pressure, the volume of fixed a mass of a gas is directly proportional to its absolute temperature or kelvin temperature.

Mathematically, this law can be written as follows;

$V = kT\\\\where;\\k \ \ is \ a \ constant \\\\T \ is \ kelvin \ temperature\\\\V \ is \ the \ volume \ of \ the \ gas$

This law explains the direct relationship between Volume of the gas and its Kelvin temperature. That is, as Temperature increases, the volume of the gas increases.

Thus, the correct statement is "The relationship is directly proportional; as temperature increases, volume increases in the same way".

Learn more here: brainly.com/question/16927784

5 0

3 years ago

Learning Goal: To be able to calculate work done by a constant force directed at different angles relative to displacement

svlad2 [7]

Answer:

Woke done, W = 4156.92 Joules

Explanation:

The work done by the force can be calculated as :

$W=F\times s$

$W=Fs\ cos\theta$

$\theta$ is the angle between force and the displacement

It is assumed to find the work done for the given parameters i.e.

Force, F = 30 N

Distance travelled, s = 160 m

Angle between force and displacement, $\theta=30$

Work done is given by :

$W=Fs\ cos\theta$

$W=30\times 160\ cos(30)$

W = 4156.92 Joules

So, the work done by the object is 4156.92 Joules. Hence, this is the required solution.

5 0

3 years ago

A charge is divided q1 and (q-q1)what will be the ratio of q/q1 so that force between the two parts placed at a given distance i

Arturiano [62]

Answer:

$q / q_{1} = 2$ , assuming that $q_{1}$ and $(q - q_{1})$ are point charges.

Explanation:

Let $k$ denote the coulomb constant. Let $r$ denote the distance between the two point charges. In this question, neither $k$ and $r$ depend on the value of $q_{1}$ .

By Coulomb's Law, the magnitude of electrostatic force between $q_{1}$ and $(q - q_{1})$ would be:

$\begin{aligned}F &= \frac{k\, q_{1}\, (q - q_{1})}{r^{2}} \\ &= \frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\end{aligned}$ .

Find the first and second derivative of $F$ with respect to $q_{1}$ . (Note that $0 < q_{1} < q$ .)

First derivative:

$\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}$ .

Second derivative:

$\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}$ .

The value of the coulomb constant $k$ is greater than $0$ . Thus, the value of the second derivative of $F$ with respect to $q_{1}$ would be negative for all real $r$ . $F\!$ would be convex over all $q_{1}$ .

By the convexity of $\! F$ with respect to $\! q_{1} \!$ , there would be a unique $q_{1}$ that globally maximizes $F$ . The first derivative of $F\!$ with respect to $q_{1}\!$ should be $0$ for that particular $\! q_{1}$ . In other words: