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2 years ago

What is the surface temperature of Sirius B in Kelvins?

Physics

2 answers:

Sauron [17]2 years ago

7 0

Answer: The surface temperature of Sirius B is 25,200 Kelvins(K).

Explanation: You would think Sirius would have a surface temperature of 9,940 Fahrenheit. That is somewhat correct, but Sirius is a binary star consisting of a main-sequence star of spectral type A0 or A1, termed Sirius A, and a faint white dwarf companion of spectral type DA2, termed Sirius B. Sirius, Sirius A, and Sirius B, are all different stars. Sirius A has a temperature of 9,940 Kelvins, but Sirius B has a temperature of 25,200 Kelvins(K).

lions [1.4K]2 years ago

3 0

That would be 9,940 kelvins (K)

Hope it helps! :D

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Electrons go downward and positive ions move upward in a long, straight, vertical lightning stroke, creating a current of magnitude I = 20.0 kA.

A free electron travels through the air at a speed of v = 300 m/s at a place r = 50.0 m east of the stroke's center.

Let the magnetic field be B, and F be the magnetic force.

Counterclockwise horizontal arcs of field lines are produced by the upward lightning current.

We have, B = 8 × 10⁻⁵ T and;

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The time interval is Δt = 60 μs = 60 × 10⁻⁶

The angular frequency is given as:

ω = qB /m = 2πN / Δt

Where the number of revolutions is N.

So,

N = qBΔt /2πm

N = (l.60 × l0⁻¹⁹)(8 × l0⁻⁵)(60 × 10⁻⁶) / 2π(9.11 × 10⁻³¹ kg)

N = 134 revolutions

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2 years ago

Sam is pulling a box up to the second story of his apartment via a string. The box weighs 16.5 kg and starts from rest on the gr

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Answer:

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3 years ago

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Answer:

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3 years ago

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A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre

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Answer:

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

$y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}$

$v_1=v_{01}+a_1t$

We must consider that it's launched from the ground ( $y_{01}=0m$ ) and from rest ( $v_{01}=0m/s$ ), with an upwards acceleration $a_{1}=28m/s^2$ that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

$y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m$

And the velocity achieved in part 1:

$v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s$

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 ( $y_{02}=1317.26m$ ) and its initial velocity is the one achieved in part 1 ( $v_{02}=271.6m/s$ ), now in free fall, which means with a downwards acceleration $a_{2}=-9,8m/s^2$ . For the data we have it's faster to use the formula $v_f^2=v_0^2+2ad$ , where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have: