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3 years ago

Which bright solar feature is shown in the picture above?

Physics

2 answers:

Ghella [55]3 years ago

6 0

Answer : (B) Prominence

Explanation :

A large, glittering and gaseous characteristic which is extending outward from the surface of the sun is called Prominence.

Photosphere is one of the layer of sun where the prominence are anchored and then they move into the corona of the sun.

Corona is a region in the surface of the sun which is the constituent of hot ionized gases (plasma).

The prominence consists of colder plasma and this prominence plasma is much more shining and denser as compared to coronal plasma.

Hence, the correct option is (B) Prominence.

bearhunter [10]3 years ago

3 0

The hottest spot in the feature, is the core, But in this situation it might be Prominence, B.)

Hope this helps!

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In a discussion person A is talking 1.2 dB louder than person B, and person C is talking 3.2 dB louder than person A. What is th

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Answer: 3.84dB

Explanation:

Since person A is talking 1.2dB louder than B, we will have

A = 1.2B... (1)

Similarly, person C is talking 3.2 dB louder than person A, we have

C = 3.2A... (2)

From equation 1, B = A/1.2... (3)

To get the ratio of the sound intensity of person C to the sound intensity of person B, we will divide equation 2 by 3 to give

C/B = 3.2A/{A/1.2}

C/B = 3.2A×1.2/A

C/B = 3.2×1.2

C/B = 3.84dB

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Read 2 more answers

A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the

Softa [21]

Answer:

$Q=3.9825\times 10^{-9} C$

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S= $\frac{360}{10000}=0.036 m^2$

$\Delta d=0.8 cm=0.008 m$

$\Delta V=100 V$

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

$C=\frac{\epsilon_0 S}{d}$

Capacitance of capacitor after moving plates

$C_1=\frac{\epsilon_0 S}{(d+\Delta d)}$

$V=\frac{Q}{C}$

Potential difference between plates after moving

$V=\frac{Q}{C_1}$

$V+\Delta V=\frac{Q}{C_1}$

$\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}$

$\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100$

$\frac{Q\Delta d}{\epsilon_0 S}=100$

$\epsilon_0=8.85\times 10^{-12}$

$Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}$

$Q=3.9825\times 10^{-9} C$

Hence, the charge on positive plate of capacitor= $Q=3.9825\times 10^{-9} C$

6 0

3 years ago