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3 years ago

What is the sun classified as on the hertzsprung-russell diagram?

Physics

2 answers:

Juliette [100K]3 years ago

8 0

Main sequence star!!!

solong [7]3 years ago

8 0

Answer:

Main Sequence star

Explanation:

Hertzsprung-Russell (H-R) diagram is a scatter plot showing the relation b/w a star's absolute magnitude and its temperature. Through this diagram one can predict the life cycle of a star and its classification. Our Star, the Sun, is a main sequence star on H-R diagram.

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If an engine can produce 670 J of energy in 1 min, how many watts can it output?

daser333 [38]

Answer:

11.2Watts

Explanation:

Given parameters:

Energy produced by the engine = 670J

Time taken = 1min = 60s

Unknown:

Number of watts it can output = ?

Solution:

This problem entails finding the power of the engine.

Power is the rate at which work is being done.

So;

Power = $\frac{Work done }{time}$

Power = $\frac{670}{60}$ = 11.2Watts

6 0

3 years ago

A pipe in a manufacturing plant is transporting superheated vapor at a mass flow rate of 0.3 kg/s. the pipe is 10 m long, has an

tatiyna

Draw a schematic and determine your thermal resistance values.

6 0

3 years ago

Explain how you can recognize the sound of a flute

lions [1.4K]

Answer:

It's airy, light, poetic, mellow, bright, wafting, ethereal, rich, soft, graceful, penetrating, brilliant, clear, shrill, silvery, wind-like, whistling, whispering, humming, filigree, sighing, aspirate.

5 0

3 years ago

Read 2 more answers

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:

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

8 0

3 years ago

Who knows how to do this

zlopas [31]

Answer:

1) 6

2) 2

3) 11

4) 38

5) 13

6) 14

7) 4

8) 41

Explanation:

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3 years ago