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Slav-nsk [51]
3 years ago
14

Please answer quickly and correctly. if you do i will mark you brainliest

Physics
1 answer:
zhannawk [14.2K]3 years ago
5 0

Answer:

The answer is B

Explanation:

My answer to this quesstion is B because the black fur will absorb more light which then results for the black dog with black fur at a greater risk.

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The four living things seen here, dog, flower, algae, and bacteria, are all made of cells. A - C are multicellular, but D is uni
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B) They are all made up of at least one cell
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The diver is on a board3.00m above the water. She jumps straight up at 3.25m/s. How many seconds later does she hit the water.
tigry1 [53]
Speed =distance/time
3.25=3.00/time
3.25xt=3.00
t=3/3.25
s=0.9s
5 0
3 years ago
Plz Help me with this
AlexFokin [52]

Answer:

Cools ; size

Explanation:

The rate at which magma cools determines the size of the crystals in the new rock. Igneous rocks are formed from the cooling and solidification of molten magma which finds its way to the surface or depth of very low pressure beneath the surface. This place or depth of cooling of magma affects the cooling rate and hence the size of the crystals formed. Igneous rocks formed at depths below the surface have more time to cool and allows more time for Crystal growth and hence produce coarse grained crystal grains called Intrusive igneous rocks which have significantly larger crystals than those formed on the surface which cools rapidly and allowing very little time for crystal growth giving rise to the formation of fine grained crystals and are called extrusive igneous rocks.

6 0
3 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
3 years ago
characterizing vibration-assisted atomic force microscopy (afm)-based nanomachining via perception of acoustic emission phenomen
DaniilM [7]

The study 'characterizing vibration-assisted atomic force based nanomachining' aims to elucidate nanomachine properties for heterogeneous materials.

<h3>What is nanomachining?</h3>

The expression nanomachining makes reference to the study of nanometric machines (nanomachines) and related materials, which can be achieved by different approaches including sensor-based strategies related to acoustic auditive phenomena.

In conclusion, the study 'characterizing vibration-assisted atomic force based nanomachining' aims to elucidate nanomachine properties for heterogeneous materials.

Learn more about nanomachines here:

brainly.com/question/20875598

#SPJ1

8 0
1 year ago
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