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2 years ago

Why are large astronomical bodies such as planets and stars round

Physics

2 answers:

Tamiku [17]2 years ago

5 0

Answer:

Gravity pulled little bits of space rocks (or gases) with equivalent force in all directions when they formed .

Explanation:

Alborosie2 years ago

4 0

Gravity pulled little bits of space rocks (or gases) with equivalent force in all directions when they formed

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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:

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

3 0

2 years ago

A 0.500-kg block, starting at rest, slides down a 30.0° incline with static and kinetic friction coefficients of 0.350 and 0.250

Leviafan [203]

Answer:x=23.4 cm

Explanation:

Given

mass of block $m=0.5 kg$

inclination $\theta =30$

coefficient of static friction $\mu =0.35$

coefficient of kinetic friction $\mu _k=0.25$

distance traveled $d=77.3 cm$

spring constant $k=35 N/m$

work done by gravity+work done by friction=Energy stored in Spring

$mg\sin \theta d-\mu _kmg\cos \theta d=\frac{kx^2}{2}$

$mgd\left ( \sin \theta -\mu _k\cos \theta \right )=\frac{kx^2}{2}$

$0.5\times 9.8\times 0.773\left ( \sin 30-0.25\cos 30\right )=\frac{35\times x^2}{2}$

$x=\sqrt{\frac{2\times 0.5\times 9.8\times 0.773(\sin 30-0.25\times \cos 30)}{35}}$

$x=0.234 m$

$x=23.4 cm$

6 0

2 years ago

The unit light-year is a measure of

Virty [35]

One light-year is the distance that light travels in vacuum
in one year. It's a unit of distance.

3 0

2 years ago

Read 2 more answers

Why are bathroom tiles usually rough

Aleks04 [339]

-- Bathroom tiles are usually cool, so water condenses on them
when you take a hot bath or shower in the room.

-- The natural result is that a smooth tile would become slippery,
exactly when you're walking around with wet feet and nothing on them ...
a dangerous situation.

-- In order to circumvent this safety hazard, the tiles in the bathroom
should be rough, especially on the floor.

8 0

2 years ago

a 2000 kg truck is traveling at a velocity of 30 m/s. What velocity must a 1000 kg car have in order to have the same momentum a

Molodets [167]

The car should have a velocity of 60 m/s to attain the same momentum as that of the truck of 2000 kg.

Answer:

Explanation:

Momentum is measured as the product of mass of object with the velocity attained by that object.

Momentum of 2000 kg truck = Mass × Velocity

Momentum of 2000 kg truck = 2000×30 = 60000 N

Similarly, the momentum of 1000 kg car will be 1000× velocity of the 1000 kg car.

Since, it is stated that momentum of 2000 kg truck is equal to the momentum of 1000 kg of car, then the velocity of 1000 kg of car can be determined by equating the momentum of car and truck.

Momentum of 2000 kg truck = Momentum of 1000 kg car

60000=1000×velocity of 1000 kg car

Velocity of 1000 kg car = 60000/1000=60 m/s

So, the car should have a velocity of 60 m/s to attain the same momentum as that of the truck of 2000 kg.

4 0

2 years ago