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Flura [38]
3 years ago
9

What is the surface temperature of Sirius B and A, in kelvins?​

Physics
1 answer:
Sindrei [870]3 years ago
3 0

Answer:

9,940 K

Image result for What is the surface temperature of Sirius B and A, in kelvins?

The bright component of the binary is a blue-white star 25.4 times as luminous as the Sun. It has a radius 1.71 times that of the Sun and a surface temperature of 9,940 kelvins (K), which is more than 4,000 K higher than that of the Sun

Explanation:

hope it help brainliest pls

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Gravity is an attractive force between two or more objects. The amount of force is directly proportional of the _______ of the o
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Susan's 12.0 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30∘ above the
Pavlova-9 [17]

Answer:1.71 m/s

Explanation:

Given

mass of Susan m=12 kg

Inclination \theta =30^{\circ}

Tension T=29 N

coefficient of Friction \mu =0.18

Resolving Forces Along x axis

F_x=T\cos \theta -f_r

where f_r=friction\ Force  

F_y=mg-N-T\sin \theta

since there is no movement in Y direction therefore

N=mg-T\sin \theta

and f_r=\mu N

Thus F_x=T\cos \theta -\mu N

F_x=29\cos (30)-\0.18\times (12\times 9.8-29\sin (30))                

F_x=25.114-18.558

F_x=6.556 N

Work done by applied Force is equal to change to kinetic Energy

F_x\cdot x=\frac{1}{2}\cdot mv_f^2-\frac{1}{2}\cdot mv_i^2

6.556\times 2.7=\frac{1}{2}\cdot 12\times v_f^2

v_f^2=\frac{6.556\times 2.7\times 2}{12}

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8 0
3 years ago
A 66-kg diver jumps off a 9.7-m tower. (a) Find the diver's velocity when he hits the water. (b) The diver comes to a stop 2.0 m
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Answer:

(a)  13.795 m/s.

(b) -3140.28 N.

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v² = u² + 2gs.......................... Equation 1

Where v = final velocity, u = initial velocity, s = height of the tower, g = acceleration due to gravity.

Given: s = 9.7 m, u = 0 m/s ( jump from a height), g = 9.81 m/s².

Substitute into equation 1

v² = 0² + 2×9.81×9.7

v² = 190.314

v = √(190.314)

v = 13.795 m/s.

Hence the velocity of the driver when he hits the water = 13.795 m/s.

(b)

F = ma.................... Equation 2

Where F = force exerted on the diver, m = mass of the diver, a = acceleration of the diver below the water surface.

Also using

v² = u² + 2as ............ Equation 3

Note: At the point when the diver enters the water, u = 13.795 m/s, and at the point when the diver comes to a complete stop, v = 0 m/s

Given: s = 2.0 m, u = 13.795 m/s, v = 0 m/s

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0 = 190.30203 + 4a

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a = 190.30203/-4

a = -47.58 m/s²

Also given: m = 66 kg,

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F = (-47.58)(66)

F = -3140.28

Note: The Force is negative because it act against the motion of the diver.

Hence the net force exerted on the diver while in the water = -3140.28 N.

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Here, the green part A is the chemical equation that includes various parts that are reactants B, methane, and oxygen, C is an arrow that indicates the formation of products.

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Answer A

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