What causes the granular structure on the sun's photosphere?

Nvm i got the answer but <br> free points

faust18 [17]

Answer:

thank you 谢谢

Explanation:

8 0

3 years ago

Read 2 more answers

two charges are placed at corners A and B of a square of side length. How much work is needed to move a charge from point C to D

Burka [1]

A study occasionally the effect of anxiety (low vs. high) and stress (low vs. moderate vs. high) on test.

Everyone experiences anxiety occasionally, but persistent anxiety can reduce your quality of life. Though likely best known for altering behavior, worry can have negative effects on our physical health. Anxiety speeds up our heartbeat and breathing, concentrating blood flow to the parts of our brains that need it. You are getting ready for a challenging situation by having this extremely bodily reaction. Test performance may be impacted by anxiety. According to studies, pupils with low levels of test anxiety perform better on multiple-choice question (MCQ) exams than pupils with high levels of anxiety. Studies have occasionally that female students have greater levels of test anxiety than male students.

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6 0

1 year ago

A thin spherical shell of radius R has a total charge +Q uniformly distributed over its surface. Of the following distance r fro

grigory [225]

Answer:

The correct answer is B

Explanation:

Let's calculate the electric field using Gauss's law, which states that the electric field flow is equal to the charge faced by the dielectric permittivity

Φ $._{E}$ = ∫ E. dA = $q_{int}$ / ε₀

For this case we create a Gaussian surface that is a sphere. We can see that the two of the sphere and the field lines from the spherical shell grant in the direction whereby the scalar product is reduced to the ordinary product

∫ E dA = $q_{int}$ / ε₀

The area of a sphere is

A = 4π r²

E 4π r² = $q_{int}$ / ε₀

E = (1 /4πε₀ ) q / r²

Having the solution of the problem let's analyze the points:

A ) r = 3R / 4 = 0.75 R.

In this case there is no charge inside the Gaussian surface therefore the electric field is zero

E = 0

B) r = 5R / 4 = 1.25R

In this case the entire charge is inside the Gaussian surface, the field is

E = (1 /4πε₀ ) Q / (1.25R)²

E = (1 /4πε₀ ) Q / R2 1 / 1.56²

E₀ = (1 /4π ε₀ ) Q / R²

$E_{B}$ = Eo /1.56 ²

$E_{B}$ = 0.41 Eo

C) r = 2R

All charge inside is inside the Gaussian surface

$E_{B}$ =(1 /4π ε₀ ) Q 1/(2R)²

$E_{B}$ = (1 /4π ε₀ ) q/R² 1/4

$E_{B}$ = Eo 1/4

$E_{B}$ = 0.25 Eo

D) False the field changes with distance

The correct answer is B

4 0

2 years ago

What is the Coriolis Effect?

Crazy boy [7]

Answer:

An effect whereby a mass moving in a rotating system experienced a force acting perpendicular to the direction of motion and to the axis of rotation.

3 0

2 years ago

Brayden and Riku now use their skills to work a problem. Find the equivalent resistance, the current supplied by the battery and

Liono4ka [1.6K]

a) $5 \Omega$ , 1.6 A

b) $6 \Omega$ , 1.33 A

Explanation:

a)

In this situation, we have two resistors connected in series.

The equivalent resistance of resistors in series is equal to the sum of the individual resistances, so in this circuit:

$R=R_1+R_2$

where

$R_1=4\Omega$

$R_2=1 \Omega$

Therefore, the equivalent resistance is

$R=4+1=5 \Omega$

Now we can use Ohm's Law to find the current flowing through the circuit:

$I=\frac{V}{R}$

where

V = 8 V is the voltage supplied by the battery

$R=5\Omega$ is the equivalent resistance of the circuit

Substituting,

$I=\frac{8}{5}=1.6 A$

The two resistors are connected in series, therefore the current flowing through each resistor is the same, 1.6 A.

b)

In this part, a third resistor is added in series to the circuit; so the new equivalent resistance of the circuit is

$R=R_1+R_2+R_3$

where:

$R_1=4\Omega\\R_2=1\Omega\\R_3=1\Omega$

Substituting, we find the equivalent resistance:

$R=4+1+1=6 \Omega$

Now we can find the current through the circuit by using again Ohm's Law:

$I=\frac{V}{R}$

where

V = 8 V is the voltage supplied by the battery

$R=6\Omega$ is the equivalent resistance

Substituting,

$I=\frac{8}{6}=1.33 A$

And the three resistors are connected in series, therefore the current flowing through each resistor is the same, 1.33 A.

3 0

2 years ago