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wolverine [178]

3 years ago

12

Suppose that you travel to a planet that has 4 times the Earth's mass and 4 times the Earth's radius. Calculate how much more or

less you would weigh on this planet compared to your weight on Earth.

1 answer:

sweet [91]3 years ago

4 0

Mass on the planet would be 4 times the mass on Earth.

<u>Explanation: </u>

<u />

Given -

Let the mass of Earth = Me

Let the radius of Earth = Re

Mass of the planet, Mp = 4Me

Radius of the planet, Rp = 4Re

Weight on other planet, W = ?

Weight on planets is determined by the gravitational force

Gravitational force, F = GMm/r²

where, G = gravitational constant

M = Mass of the planet

m = our mass

r = radius of the planet

Gravitational force on Earth:

F = G X Me X ma/ Re² (ma = mass on earth)

Gravitational force on planet :

F = G X 4Me X mₓ/ (4Re)² (mx = mass on other planet)

Dividing both the equations, we get

mx = 4ma

Therefore, mass on the planet would be 4 times the mass on Earth.

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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:

$\displaystyle \frac{k}{r^{2}}\, (q - 2\, q_{1}) = 0$ <em>.</em>

$2\, q_{1} = q$ .

$q_{1} = q / 2$ .

In other words, the force between the two point charges would be maximized when the charge is evenly split:

$\begin{aligned} \frac{q}{q_{1}} &= \frac{q}{q / 2} = 2\end{aligned}$ .

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2 years ago

A plane flying at a steady speed of 100 m/s accelerates to 150 m/s in 10 seconds. What is the plane’s acceleration?

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A plane flying initially at 100 m/s uses an acceleration of 5 m/s² to reach a velocity of 150 m/s in 10 seconds.

<h3>What is acceleration?</h3>

Acceleration is the change in velocity over time.

A plane is flying initially at 100 m/s (u) and it accelerates to 150 m/s (v) in 10 s (t). We can calculate its acceleration using the following expression.

a = v - u / t = (150 m/s - 100 m/s) / 10 s = 5 m/s²

A plane flying initially at 100 m/s uses an acceleration of 5 m/s² to reach a velocity of 150 m/s in 10 seconds.

Learn more about acceleration here: brainly.com/question/14344386

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1 year ago

In-s [12.5K]

Answer:

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Explanation:

To be able to solve this problem we must first find the work done. Work is defined as the product of force by distance.

$W = F*d$

where:

W = work [J] (units of Joules)

F = force [N] (units of Newton)

d = distance [m]

We need to bear in mind that the force can be calculated by multiplying the mass by the gravity acceleration.

Now replacing:

$W = (80*10)*3\\W = 2400 [J]$

Power is defined as the work done over a certain time. In this way by means of the following formula, we can calculate the required power.

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W = work [J]

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$P = 2400/40\\P = 60 [W]$

The calculated power is the required power. Now as we have the efficiency of the machine, we can calculate the power that is introduced, to be able to do that work.

$Effic=0.7\\Effic=P_{required}/P_{introduced}\\P_{introduced}=60/0.7\\P_{introduced}=85.71[W]$

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