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3 years ago

8

I’m NC the wind is blowing from the north. If the wind changes direction and begins to blow from the south, what will most likel

y result.

1 answer:

MrMuchimi3 years ago

8 0

NC means North Carolina.

Answer:

The temperatures will increase.

Explanation:

What will happen is that the temperature will increase. This is because, in wind direction and weather forecasting, a wind that shifts from the south means that warmer air/temperatures are approaching whereas a wind shift from the north implies that cooler air/temperatures are approaching.

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A researcher measures the thickness of a layer of benzene (n = 1.50) floating on water by shining monochromatic light onto the f

NNADVOKAT [17]

Answer:

Explanation:

This problem relates to interference of light in thin films .

The condition of bright fringe in thin films which is sandwitched by two layers of medium having lesser refractive index is as follows.

2nt = (2n+1) λ / 2 , n is refractive index of thin layer , t is its thickness , λ is wavelength of light .

2 x 1.5 t = λ / 2 , if n = 0 for minimum thickness.

2 x 1.5 t = 600 / 2 nm

t = 100 nm .

5 0

3 years ago

How much would you weigh on an imaginary planet that has no gravitational force?<br><br>

GarryVolchara [31]

0kg

If the gravitational pull is zero and I multiply by mass I get a zero

5 0

3 years ago

Mashutka [201]

Answer:

It cannot be constant because if it does not change and each time it increases its strength and speed.

Explanation:

4 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

A helium atom (mass 4.0 u) moving at 598 m/s to the right collides with an oxygen molecule (mass 32 u) moving in the same direct

labwork [276]

Answer:

Speed of the helium after collision = 246 m/s

Explanation:

Given that

Mass of helium ,m₁ = 4 u

u₁=598 m/s

Mass of oxygen ,m₂ = 32 u

u₂ = 401 m/s

v₂ =445 m/s

Given that initially both are moving in the same direction and lets take they are moving in the right direction.

Speed of the helium after collision = v₁

There is no any external force on the masses that is why the linear momentum will be conserve.

Initial linear momentum = Final linear momentum

P = m v

m₁u₁+m₂u₂ = m₁v₁+m₂v₂

598 x 4 + 32 x 401 = 4 x v₁+ 32 x 445

v₁ = 246 m/s

Speed of the helium after collision = 246 m/s

6 0

3 years ago