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3 years ago

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

2 answers:

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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Which of the following best defines speed? (3 points) a It is an object's speed at a specific point in time. b It is the distanc

djverab [1.8K]

Answer:

Explanation:

they are all almost correct.

c is the only fully correct option

6 0

2 years ago

An AC voltage source and a resistor are connected in series to make up a simple AC circuit. If the source voltage is given by ΔV

ra1l [238]

Answer:

t = 5.48 × 10⁻³ s

Explanation:

Given:

ΔV = ΔVmax × sin(2πft)

frequency, f = 16.9Hz

thus,

ΔV = ΔVmax × sin(2π×16.9×ft)

Now,

Let 'R' be the resistance

Also according to the ohms law

i = V/R

where,

i = current

V = voltage

hence,

$i=\frac{\Delta V_{max}sin(2\pi \times 16.9\times t)}{R}$

also, given at time 't' the current in the circuit is 55.0% of the peak current

thus

$i=\frac{55}{100}\times \frac{\Delta V_{max}}{R}=0.55\times \frac{\Delta V_{max}}{R}$

thus,

$0.55\times \frac{\Delta V_{max}}{R}=\frac{\Delta V_{max}sin(2\pi \times 16.9\times t)}{R}$

$0.55=sin(2\pi \times 16.9\times t)}$

$0.5823=(2\pi \times 16.9\times t)}$

t = 5.48 × 10⁻³ s (Answer)

8 0

2 years ago

A regulation basketball has a 15 cm diameter and may be appropriated as a thin spherical shell.

REY [17]

Solution :

From the given data,

For the spherical shell is $\frac{k^2}{R^2}= \frac{2}{3}$

where x is the radius of gyration and the acceleration of a rolling body on an inclined plane = a

Therefore,

$a =\frac{g \sin \theta}{1+ \frac{k^2}{R^2}}$

$= \frac{g \sin \theta}{1 +\frac{2}{3}}$

$= \frac{3}{5} g \sin \theta$

= 0.6 x 9.81 x sin ( 29.7)

$= 2.913 \ m/s^2$

$h = \frac{1}{2} a(\Delta t)^2$

$\Delta t = \sqrt{\frac{2h}{a}}$

$\Delta t = \sqrt{\frac{2 \times 2.9}{2.913}}$

Δt = 1.411 s

6 0

2 years ago

An engine absorbs 1.69 kJ from a hot reservoir at 277°C and expels 1.25 kJ to a cold reservoir at 27°C in each cycle.

Anna35 [415]

Answer

Given,

Energy absorbed, $Q_H = 1.69\ kJ$

Energy expels, $Q_C = 1.25\ kJ$

Temperature of cold reservoir, T = 27°C

a) Efficiency of engine

$\eta = \dfrac{Q_H - Q_C}{Q_H}\times 100$

$\eta = \dfrac{1.69 - 1.25}{1.69}\times 100$

$\eta =26.03 %$

b) Work done by the engine

$W = Q_H- Q_C$

$W =1.69 - 1.25$

$W = 0.44\ kJ$

c) Power output

t = 0.296 s

$P = \dfrac{W}{t}$

$P = \dfrac{0.44}{0.296}$

$P = 1.486\ kW$

8 0

2 years ago

Visit this website and locate the element mercury, whose chemical symbol is Hg. Click on the square to read about the history, p

Irina-Kira [14]

We simply asked to name three uses for mercury.
The most common and well-known use of mercury is the production of thermometers. It's property to stay liquid at room temperature makes it ideal for a temperature indicator. However, the use of mercury is thermometers has been phased out due to health hazards.
It is also used to form an amalgam which is the result of its combination with silver or gold. Mercury has been used to mine gold and silver. This application has also been phased out.
Today's use of mercury includes mercury-vapor lamps which are the bright lamps used in high-ways.

6 0

2 years ago

Read 2 more answers