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3 years ago

Dark areas on the surface of the Sun that exhibit intense magnetic activity are

2 answers:

7 0

A. sunspots
a spot or patch appearing from time to time on the suns surface, appearing dark by contrast with its surroundings, hope this helps

svp [43]3 years ago

6 0

I think it's A) Sunspots. I hope this helps:)

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Based on the following equation, answer the questions below. ρ = (2γϕ + ψ)/rg where ρ [=] moles per cubic foot [mol/ft3] γ [=] j

AlekseyPX

1) Fundamental units of $\Psi$ are $[\frac{mol}{m\cdot s^2}]$

2) Fundamental units of $\Phi$ are $[\frac{mol}{m^3}]$

Explanation:

The equation for the variable $\rho$ is

$\rho =\frac{2\gamma \Phi+\Psi}{rg}$

where we have:

$\rho$ measured in $[\frac{mol}{ft^3}]$

$\gamma$ measured in $[\frac{J}{kg}]$

$r$ measured in $[in]$

$g$ measured in $[\frac{m}{s^2}]$

We can re-write the equation as

$\rho rg = 2\gamma \Phi + \Psi$

And we notice that the units of the term on the left must be equal to the units of the term on the right.

This means that:

1) First of all, $\Psi$ must have the same units of $\rho r g$ . So,

$[\rho r g]=[\frac{mol}{ft^3}][in][\frac{m}{s^2}]$

However, both ft (feet) and in (inches) are not fundamental dimensions: this means that they can be expressed as meters. Therefore, the fundamental units of $\Psi$ are

$[\Psi]=[\frac{mol}{m^3}][m][\frac{m}{s^2}]=[\frac{mol}{m\cdot s^2}]$

The term $2\gamma \Phi$ must have the same units of $\Psi$ in order to be added to it. Therefore,

$[\gamma \Phi] = [\frac{mol}{m\cdot s^2}]$

We also know that the units of $\gamma$ are $[\frac{J}{kg}]$ , therefore

$[\frac{J}{kg}][\Phi]= [\frac{mol}{m\cdot s^2}]$

And so, the fundamental units of $\Phi$ are

$[\Phi]= [\frac{mol\cdot kg}{J\cdot m\cdot s^2}]$

However, the Joules can be written as

$[J]=[kg][\frac{m^2}{s^2}]$

Therefore

$[\Phi]= [\frac{mol\cdot kg}{(kg \frac{m^2}{s^2})\cdot m\cdot s^2}]=[\Phi]= [\frac{mol}{m^3}]$

#LearnwithBrainly

7 0

3 years ago

Whar are the types of energy associated with the microwave

Naily [24]

Type motion examples and subtypes
electromagnetic radiation disturbance propagating through electric and magnetic fields (classical physics) or the motion of photons (modern physics) radio waves, microwaves, infrared, light, ultraviolet, x-rays, gamma rays

Hope this helps and please mark me as brainlest and like:)

8 0

3 years ago

Which characteristics can be used to differentiate star systems? Select three options.

anyanavicka [17]

Answer:

1,3and4 just did and got correct

Explanation:

5 0

2 years ago

Four point charges of magnitudes +3q, -q, +2q, and -4q are arranged in the corners of a square of side length L. The charge -q c

mafiozo [28]

Answer:

d) 0 V

Explanation:

It can be showed that the potential due to a point charge q, to a distance d from the charge, can be expressed as follows:

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

where k = $\frac{1}{4*\pi*\epsilon0} = 9e9 N*m2/C2$

As the potential is an scalar, and is linear with the charge, we can apply the superposition principle, which means that we can find the potential due to one of the charges, as if the other were not present.

By symmetry, all four charges are at the same distance from the center, so we can write the total potential, as follows:

$V = \frac{k}{d} ( q1 + q2 + q3 + q4) (1)$

where d, is the semi-diagonal of the square, that we can find applying Pythagorean theorem, as follows:

$d = \sqrt{\frac{L^{2}}{4} + \frac{L^{2}}{4} } = L*\frac{\sqrt{2}}{2}$

Replacing by the values in (1) we have:

$V = \frac{9e9N*m2/C2}{\frac{L}{2}*\sqrt{2} }* ( +3q -q + 2q + -4q) = 0 V$

which is equal to the option d).

6 0

3 years ago

A small coin of mass m1 is undergoing a uniform circular motion at a velocity v. The radius of the circular path is r. A piece o

GarryVolchara [31]

Answer:

Answere is (D)

Explanation:

Given m1, v and m2

Total circular momentum is conserved: total circular momentum before putty coin collision is equal to total circular momentum after collision.

Let the final circular velocity of the system be V. Initial circular velocity of the putty is zero.

m1v + m2× 0 = (m1 + m2) V

V = m1v/(m1 + m2).

3 0

3 years ago