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irga5000 [103]
2 years ago
10

If a sunspot appears only 70 s bright as the surrounding photosphere, and the photosphere has a temperature of approximately 578

0 k. What is the temperature of the sunspot?
Physics
1 answer:
grin007 [14]2 years ago
8 0

The temperature of the sunspot is 5290K.

Flux is directly proportional to brightness (energy per second falling on a unit area of your detector) (energy per second being radiated away from a unit area of the source).

In turn, flux is proportional to the temperature to the fourth power (Stefan-Boltzmann Law).

Consequently, tsunspot/bphotosphere = T4sunspot/T4photosphere

(bsunspot/bphotosphere) = T4sunspot + T4photosphere

(bsunspot/bphotosphere) Tsunspot =Tphotosphere

By putting in the values, we get,

1/4 = 5780 K (0.7)

1/4 = 5780 x 0.915 = 5290 K

Therefore, temperature of the sunspot is 5290K.

Learn more about temperature of the sunspot here:

brainly.com/question/13687926

#SPJ4

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Answer:

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8 0
3 years ago
Students are completing a lab in which they let a lab cart roll down a ramp. The students record the mass of the cart, the heigh
Dennis_Churaev [7]

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second column

Explanation:

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3 years ago
Determine el valor del ángulo de tiro de un proyectil que tarda en impactar en 8s y que tuvo un alcance de 0,3km.
Gelneren [198K]

Answer:

46.3^{\circ}

Explanation:

The motion of a projectile consists of two independent motions:

- A uniform motion along the horizontal direction (constant velocity)

- A uniformly accelerated motion along the vertical direction (constant acceleration)

The time of flight of a projectile is given by

t=\frac{2u_y}{g}

where

u_y is the initial vertical velocity

g=9.8 m/s^2 is the acceleration due to gravity

Here the time of flight is

t = 8 s

So, the initial vertical velocity is:

u_y=\frac{gt}{2}=\frac{(9.8)(8)}{2}=39.2 m/s

At the same time, the horizontal distance covered by the projectile is given by

d=u_x t

where

u_x is the horizontal velocity, which is constant

t is the time of flight

Here we know that the range of the projectile is 0.3 km, so

d = 0.3 km = 300 m

So the horizontal velocity is

u_x=\frac{d}{t}=\frac{300}{8}=37.5 m/s

Therefore, we can find the angle of projection of the projectile by using:

\theta=tan^{-1}(\frac{u_y}{u_x})=tan^{-1}(\frac{39.2}{37.5})=46.3^{\circ}

5 0
3 years ago
A 1.50 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1
Tema [17]

Answer:

α = 1.114 × 10⁻³ (°C)⁻¹

Explanation:

Given that:

Length of rod (L) = 1.5 m,

Diameter (d) = 0.55 cm,

Area (A) = \pi r^2

Radius (r) = d / 2 = 0.275 cm,

Voltage across the rod (V) = 15.0 V.

At initial temperature (T₀) = 20°C, the current (I₀) = 18.8 A while at a temperature (T) = 92⁰C, the current (I) = 17.4 A

a) The resistance of the rod (R) is given as:

R=\frac{Voltage(V)}{I_0} \\R=\frac{15}{18.8}=0.798\Omega

Therefore the resistivity and for the material of the rod at 20 °C (ρ) is:

\rho=\frac{RA}{L}=\frac{0.798*\pi *0.275^2}{1.5}=0.126\Omega m  

b) The temperature coefficient of resistivity at 20°C for the material of the rod (α) can be gotten from the equation:

R_T=R_0[1-\alpha (T-T_0)]\\but,R_T=\frac{V}{I}=\frac{15}{17.4}=0.862\\

Rearranging to make α the subject of formula:

\frac{R_T}{R_0} =1+\alpha (T-T_0)\\\alpha (T-T_0)=\frac{R_T}{R_0}-1\\\alpha =\frac{\frac{R_T}{R_0}-1}{(T-T_0)} \\Substituting:\\\alpha =\frac{\frac{0.862}{0.798}-1 }{92-20} \\\alpha =\frac{0.0802}{72} =1.114*10^-3(^0C)^{-1

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3 years ago
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It is bracking down because of all the pollutoin 
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