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2 years ago

Why do you think that some asteroids tumble end over end through space while other asteroids rotate around their axis?

1 answer:

7 0

Imagine a lonely asteroid there in space, untouched, not rotating, just still. Then another asteroid passes by and comes into contact with that first asteroid. Upon collision, based on the conservation of elastic momentum, the asteroid that was once still moves; it may even spin if the incoming asteroid hit it at its side. Now asteroids have three angles or rotation (three dimensions): θ (theta - to the x-axis), φ (phi - to the y-axis), and let's say ψ (psi - to the z-axis). So these asteroids wobble through space, spinning like crazy.

I may have gone too in depth. Sorry, lol.
I hope this helped!

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A curved line going up indicates the object is

Vesna [10]

Explanation:

Both graphs show plotted points forming a curved line. Curved lines have changing slope; they may start with a very small slope and begin curving sharply (either upwards or downwards) towards a large slope. In either case, the curved line of changing slope is a sign of accelerated motion (i.e., changing velocity).

8 0

2 years ago

Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of +6q. Sphere B caries a charge of-2q. Sphere C

miskamm [114]

<h2>20. How much charge is on sphere B after A and B touch and are separated?</h2><h3>Answer:</h3>

$\boxed{q_{B}=+2q}$

<h3>Explanation:</h3>

We'll solve this problem by using the concept of electric potential or simply called potential $V$ , which is <em>the energy per unit charge, </em>so the potential $V$ at any point in an electric field with a test charge $q_{0}$ at that point is:

$V=\frac{U}{q_{0}}$

The potential $V$ due to a single point charge q is:

$V=k\frac{q}{r}$

Where $k$ is an electric constant, $q$ is value of point charge and $r$ is the distance from point charge to where potential is measured. Since, the three spheres A, B and C are identical, they have the same radius $r$ . Before the sphere A and B touches we have:

$V_{A}=k\frac{q_{A}}{r_{A}} \\ \\ V_{B}=k\frac{q_{B}}{r_{A}} \\ \\ But: \\ \\ \ r_{A}=r_{B}=r$

When they touches each other the potential is the same, so:

$V_{A}= V_{B} \\ \\ k\frac{q_{A}}{r}=k\frac{q_{B}}{r} \\ \\ \boxed{q_{A}=q_{B}}$

From the principle of conservation of charge <em>the algebraic sum of all the electric charges in any closed system is constant. </em>So:

$q_{A}+q_{B}=q \\ \\ q_{A}=+6q \ and \ q_{B}=-2q \\ \\ So: \\ \\ \boxed{q_{A}+q_{B}=+4q}$

Therefore:

$(1) \ q_{A}=q_{B} \\ \\ (2) \ q_{A}+q_{B}=+4q \\ \\ (1) \ into \ (2): \\ \\ q_{A}+q_{A}=+4q \therefore 2q_{A}=+4q \therefore \boxed{q_{A}=q_{B}=+2q}$

So after A and B touch and are separated the charge on sphere B is:

$\boxed{q_{B}=+2q}$

<h2>21. How much charge ends up on sphere C?</h2><h3>Answer:</h3>

$\boxed{q_{C}=+1.5q}$

<h3>Explanation:</h3>

First: A and B touches and are separated, so the charges are:

$q_{A}=q_{B}=+2q$

Second: C is then touched to sphere A and separated from it.

Third: C is to sphere B and separated from it

So we need to calculate the charge that ends up on sphere C at the third step, so we also need to calculate step second. Therefore, from the second step:

Here $q_{A}=+2q$ and C carries no net charge or $q_{C}=0$ . Also, $r_{A}=r_{C}=r$

$V_{A}=k\frac{q_{A}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}$

Applying the same concept as the previous problem when sphere touches we have:

$k\frac{q_{A}}{r} =k\frac{q_{C}}{r} \\ \\ q_{A}=q_{C}$

For the principle of conservation of charge:

$q_{A}+q_{C}=+2q \\ \\ q_{A}=q_{C}=+q$

Finally, from the third step:

Here $q_{B}=+2q \ and \ q_{C}=+q$ . Also, $r_{B}=r_{C}=r$

$V_{B}=k\frac{q_{B}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}$

When sphere touches we have:

$k\frac{q_{B}}{r} =k\frac{q_{C}}{r} \\ \\ q_{B}=q_{C}$

For the principle of conservation of charge:

$q_{B}+q_{C}=+3q \\ \\ q_{A}=q_{C}=+1.5q$

So the charge that ends up on sphere C is:

$q_{C}=+1.5q$

<h2>22. What is the total charge on the three spheres before they are allowed to touch each other.</h2><h3>Answer:</h3>

$+4q$

<h3>Explanation:</h3>

Before they are allowed to touch each other we have that:

$q_{A}=+6q \\ \\ q_{B}=-2q \\ \\ q_{C}=0$

Therefore, for the principle of conservation of charge <em>the algebraic sum of all the electric charges in any closed system is constant, </em>then this can be expressed as:

$q_{A}+q_{B}+q_{C}=+6q -2q +0 \\ \\ \therefore q_{A}+q_{B}+q_{C}=+4q$

Lastly, the total charge on the three spheres before they are allowed to touch each other is:

$+4q$

8 0

2 years ago

planet a takes one year to go around the sun at a distance of one au. .planet b is three a u. from the sun. how many years does

shtirl [24]

Answer:

3 years

Explanation:

Find the circumference of each orbit in AU.

2xπx1=6.283185307

2xπx3=18.84955592

Divide them.

18.84955592/6.283185307=3

3 years

4 0

2 years ago

A spherical tank with radius 4 m is half full of a liquid that has a density of 900 kg/m3. The tank has a 1 m spout at the top.

kompoz [17]

The answer & explanation for this question is given in the attachment below.

7 0

2 years ago

Which has the larger kinetic energy, a 10 g bullet fired at 400 m/s or a 80 kg bowling ball rolled at 6.5 m/s ?

mixer [17]

Formulae for Kinetic energy is:
Kinetic Energy= 1/2xmassx(velocity)^2

For comparison we need to have same units,thus we convert 10g into Kg.
10g/1000=0.01Kg

Input the value of bullet in the formulae;
Kinetic Energy= 1/2x0.01kgx(400)^2
K.E=800J

Input value of the ball:
Kinetic Energy=1/2x80kgx(6.5)^2
K.E=1690J

Which means that th Energy of the ball is more than the bullet.

7 0

2 years ago