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3 years ago

During normal conditions, we only see the Sun’s , but during a solar eclipse, we get to see the _____.

2 answers:

8 0

I would say A. because during the solar eclipse you cannot see the suns Photosphere at all, and the corona is the light that emits around the moon during the eclipse.

sattari [20]3 years ago

5 0

The corona of the sun is only visible during a solar eclipse, meaning you can't see it on a normal day. The photosphere is the deepest layer of the sun. The chromosphere is the layer outside of the photosphere. I would say the answer is B.

Hope this helps! I would greatly appreciate a brainliest! :)

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To practice Problem-Solving Strategy 30.1: Inductors in Circuits. A circuit has a 1 V battery connected in series with a switch.

ki77a [65]

Answer:

0.0133A

Explanation:

Since we have two sections, for the Inductor region there would be a current $i_1$ . In the case of resistance 2, it will cross a current $i_2$

Defined this we proceed to obtain our equations,

For $i_1$ ,

$\frac{di_1}{dt}+i_1R_1 = V$

$I_1 = \frac{V}{R_1} (1-e^{-\frac{R_1t}{L}})$

For $i_2$ ,

$I_2R_2 =V$

$I_2 = \frac{V}{R_2}$

The current in the entire battery is equivalent to,

$i_t = I_1+I_2$

$i_t = \frac{V}{R_2}+\frac{V}{R_1} (1-e^{-\frac{R_1t}{L}})$

Our values are,

$V=1V$

$R_1 = 95\Omega$

$L= 1.5*10^{-2}H$

$R_2 =360\Omega$

Replacing in the current for t= 0.4m/s

$i=\frac{1}{360}+\frac{1}{95}(1-e^{-\frac{95*0.4}{1.5*10^{-2}}})$

$i= 0.0133A$

$i_1 = 0.01052A$

3 0

3 years ago

A 90 kg man stands in a very strong wind moving at 17 m/s at torso height. As you know, he will need to lean in to the wind, and

victus00 [196]

Answer:

a) $t=195.948N.m$

b) $\phi=13.6 \textdegree$

Explanation:

From the question we are told that:

Density $\rho=1.225kg/m^2$

Velocity of wind $v=14m/s$

Dimension of rectangle:50 cm wide and 90 cm

Drag coefficient $\mu=2.05$

Generally the equation for Force is mathematically given by

$F=\frac{1}{2}\muA\rhov^2$

$F=\frac{1}{2}2.05(50*90*\frac{1}{10000})*1.225*17^2$

$F=163.29$

Therefore Torque

$t=F*r*sin\theta$

$t=163.29*1.2*sin90$

$t=195.948N.m$

Generally the equation for torque due to weight is mathematically given by

$t=d*Mg*sin90$

Where

$d=sin \phi$

Therefore

$t=sin \phi*Mg*sin90$

$195.948=833sin \phi$

$\phi=sin^{-1}\frac{195.948}{833}$

$\phi=13.6 \textdegree$

5 0

3 years ago

What is the momentum of a bird with a mass of 0.018kg flying at 15m/s?

Mazyrski [523]

Answer:

0.27 kg-m/s

Explanation:

i believe this is the correct answer

6 0

3 years ago

Suppose you are pushing a 3 kg box with a force of 25 N (directed parallel to the ground) over a distance of 15 m. Afterward, th

alex41 [277]

Answer: 321 J

Explanation:

Given

Mass of the box $m=3\ kg$

Force applied is $F=25\ N$

Displacement of the box is $s=15\ m$

Velocity acquired by the box is $v=6\ m/s$

acceleration associated with it is $a=\dfrac{F}{m}$

$\Rightarrow a=\dfrac{25}{3}\ m/s^2$

Work done by force is $W=F\cdot s$

$W=25\times 15\\W=375\ J$

change in kinetic energy is $\Delta K$

$\Rightarrow \Delta K=\dfrac{1}{2}m(v^2-0)\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 6^2\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 36\\\\\Rightarrow \Delta K=54\ J$

According to work-energy theorem, work done by all the forces is equal to the change in the kinetic energy

$\Rightarrow W+W_f=\Delta K\quad [W_f=\text{Work done by friction}]\\\\\Rightarrow 375+W_f=54\\\Rightarrow W_f=-321\ J$

Therefore, the magnitude of work done by friction is $321\ J$

3 0

3 years ago

Water is a compound.<br> O A. True<br> B. False

puteri [66]

True because , a compound forms when two or more atoms form a chemical bond and the chemical formula for water is H2O which means each molecule of water consists of one oxygen atom

7 0

3 years ago

Read 2 more answers

During normal conditions, we only see the Sun’s __________, but during a solar eclipse, we get to see the _______________.

During normal conditions, we only see the Sun’s , but during a solar eclipse, we get to see the _____.