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3 years ago

14

What are sun spots??????????

2 answers:

Blababa [14]3 years ago

7 0

The surface of the Sun has dark areas called sunspots. They are cooler than the hot, yellow areas. Sunspots were discovered by the astronomer Galileo Galilei. Some sunspots can last weeks or even months.

babymother [125]3 years ago

4 0

Sunspots<span> are temporary phenomena on the </span>Sun<span>'s photosphere that appear as </span>spots<span> darker than the surrounding areas. They are regions of reduced surface temperature caused by concentrations of magnetic field flux that inhibit convection. </span>Sunspots<span> usually appear in pairs of opposite magnetic polarity.

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which cell structure is responsible for taking in and breaking down foreign particles and damaged organelles?

Klio2033 [76]

Answer:

lysosomes

Explanation:

4 0

3 years ago

Anyone know how to do this?

Gala2k [10]

Answer:

I think, (remember think) it might be 2.0 m/s

Explanation:

If it's wrong I'm truly sorry.

6 0

3 years ago

Helium gas is compressed by an adiabatic compressor from an initial state of 14 psia and 50°F to a final temperature of 320°F in

iVinArrow [24]

Answer:

The value is $P_2 = 40.54 \ psla$

Explanation:

From the question we are told that

The initial pressure is $P_1 = 14\ psla$

The initial temperature is $T_1 = 50 \ F = (50 - 32) * [\frac{5}{9} ] + 273 = 283 \ K$

The final temperature is $T_2 = 320 \ F = (320 - 32) * [\frac{5}{9} ] + 273 =433 \ K$

Generally the equation for adiabatic process is mathematically represented as

$PT^{\frac{\gamma}{1- \gamma} } = Constant$

=> $P_1T_1^{\frac{\gamma}{1- \gamma} } = P_2T_2^{\frac{\gamma}{1- \gamma} }$

Generally for a monoatomic gas $\gamma = \frac{5}{3}$

So

$14 * 283^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} } =P_2 * 433^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} }$

=> $14 * 283^{-2.5} =P_2 * 433^{-2.5}$

=> $P_2 = 40.54 \ psla$

8 0

2 years ago

You push a 15N cat in a box horizontally across the floor for 5m. What is the work done on the cat? W=Fx

scZoUnD [109]

Answer:

w=fx

w=(15)(5)

w=75

the angle between the deplacement and N is 0

3 0

2 years ago

A wire loop of radius 0.37 m lies so that an external magnetic field of magnitude 0.35 T is perpendicular to the loop. The field

Galina-37 [17]

Answer:

168.57 mV

Explanation:

Initial magnetic flux = BA , B magnetic field and A is area of loop

= .35 x 3.14 x .37²

= .15 Weber

Final magnetic flux

= - .2 x 3.14 x .37²

= - .086 Weber

change in flux

.15 + .086

= .236 Weber

rate of change of flux

= .236 / 1.4

= .16857 V

= 168.57 mV

5 0

3 years ago