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Pavlova-9 [17]
3 years ago
6

PLEASE HELP!!!! :D When you look at the Sun through a filtered telescope, the visible portion of the Sun appears blotchy.

Physics
2 answers:
olya-2409 [2.1K]3 years ago
5 0
<span>When an individual looks through a filtered telescope in which he or she observes the sun, the portion where it appears blotchy is likely to be called the sunspots while the layer of the sun where it shows where it occurs is called the photosphere.</span>
olga2289 [7]3 years ago
4 0

The answer is; sunspots, photosphere.

Sunspots are cooler than the rest of the surface of the sun  hence their appearance of being darker than the rest of the photosphere. The photosphere is the outer sun’s shell that emits radiation to space. These sunspots are associated with increased magnetic influx that is accompanied by coronal mass ejection. It is thought that these magnetic fluxes hinder efficient convection currents in the regions.


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What is the function of the muscles around the lens in the human eye
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Its to capture light or to focus. don't forget to like. :D

6 0
3 years ago
an ice skater starts with a velocity of 2.25 m/s in a 50.0 direction. after 8.33 m/s in a 120 direction. what is the y-component
Nata [24]

The y-component of the acceleration is 0.28 m/s^2

Explanation:

The y-component of the ice skater acceleration can be calculated with the equation

a_y = \frac{v_y-u_y}{t}

where

v_y is the y-component of the final velocity

u_y is the y-component of the initial velocity

t is the time elapsed

Here we have:

  • Initial velocity is u=2.25 m/s at \theta_1=50.0^{\circ}, so its y-component is u_y = u sin \theta_1 = (2.25)(sin 50.0^{\circ})=1.72 m/s
  • Final velocity is v=4.65 m/s at \theta_2=120.0^{\circ}, so its y-component is v_y = v sin \theta_2 = (4.65)(sin 120.0^{\circ})=4.03 m/s

The time elapsed is

t = 8.33 s

Therefore, the y-component of the acceleration is

a_y = \frac{4.03-1.72}{8.33}=0.28 m/s^2

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7 0
3 years ago
Please help me on this l​
eduard

Answer:

16

Explanation:

because

7 0
3 years ago
Consider a physical situation in which a particle moves from point A to point B. This process is described from two coordinate s
Fudgin [204]

Coordinates, displacement

8 0
3 years ago
Read 2 more answers
a 15-nC point charge is at the center of a thin spherical shell of radius 10cm, carrying -22nC of charge distributed uniformly o
Aleks [24]

Answer:

A) E = 278925.62 N/C with direction; radially out.

B) E = 43048.47 N/C with direction radially out.

C) E = -3214.29 N/C with direction radially in.

Explanation:

From Gauss' Law, the Electric field for any spherically symmetric charge or charge distribution is the same as the point charge formula. Thus;

E = kQ/r²

where;

Q is the net charge within the distance r.

We are given the charge Q = 15-nC and

spherical shell of radius 10cm

A) The distance r = 2.2 cm = 0.022 m is between the surface and the point charge, so only the point charge lies within this distance and Q = 15 nC = 15 x 10^(-9) C

While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²

E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.022)²

E = 278925.62 N/C

This will be radially out ,since the net charge is positive.

B) The distance r = 5.6 cm = 0.056 m is between the surface and the point charge, so only the point charge lies within this distance and Q = 15 nC = 15 x 10^(-9) C

While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²

E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.056)²

E = 43048.47 N/C

This will be radially out ,since the net charge is positive.

C) The distance r = 14 cm = 0.14 m is outside the sphere so the "net" charge within this distance is due to both given charges. Thus;

Q = 15 nC - 22 nC

Q = -7 nC = -7 x 10^(-9) C

and;

E = (9 x 10^(9)*(-7 x 10^(-9))/(0.14)²

E = -3214.29 N/C

This will be radially in, since the net charge is negative. You can indicate this with a negative answer.

8 0
3 years ago
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