Answer:
Explanation:
The corona is in the outer layer of the Sun's atmosphere—far from its surface. Yet the corona is hundreds of times hotter than the Sun's surface. ... In the corona, the heat bombs explode and release their energy as heat. But astronomers think that this is only one of many ways in which the corona is heated.
The chromosphere ("sphere of color") is the second of the three main layers in the Sun's atmosphere and is roughly 3,000 to 5,000 kilometers deep. Its rosy red color is only apparent during eclipses. The chromosphere sits just above the photosphere and below the solar transition region.
The photosphere is the visible "surface" of the Sun. The Sun is a giant ball of plasma (electrified gas), so it doesn't have a distinct, solid surface like Earth. ... The photosphere is much cooler than the Sun's core, which has a temperature well above 10 million degree
And thats all i know
Answer:
Velocity = v = 2.24m/s
Acceleration = a = 0.20m/s²
Explanation:
Please see attachment below.
Given
z=(−8 cosθ) and θ = 0.3t
z = -8Cos (0.3t)
V = dz/dt
a = v²/R.
Please see full solution below.
The change in temperature was [ (37) - (-5) ] = 42 Fahrenheit degrees.
Kelvins and Celsius degrees are 9/5 (or 1.8) the size of Fahrenheit degrees.
42 F-degrees / 1.8 K per F-degree =<em> 23 and 1/3</em> Kelvins.
<u>Check,</u> using °C = (5/9) (°F - 32) and K = 273.15 + °C :
-5° F = -20.566° C = 252.594 K.
+37° F = 2.778° C = 275.928 K
(275.928K - 252.594K) = <u>23.334 K</u> close enough yay !
Answer:
The minimum frequency is 702.22 Hz
Explanation:
The two speakers are adjusted as attached in the figure. From the given data we know that
=3m
=4m
By Pythagoras theorem
Now
The intensity at O when both speakers are on is given by
Here
- I is the intensity at O when both speakers are on which is given as 6
- I1 is the intensity of one speaker on which is 6
- δ is the Path difference which is given as
- λ is wavelength which is given as
Here
v is the speed of sound which is 320 m/s.
f is the frequency of the sound which is to be calculated.
where k=0,1,2
for minimum frequency , k=1
So the minimum frequency is 702.22 Hz