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tresset_1 [31]
4 years ago
13

Among main sequence stars, those with the highest surface temperatures have ___________. Among main sequence stars, those with t

he highest surface temperatures have ___________. the
a) lowest masses and longest lifetimes
b) the highest masses and shortest lifetimes
c) the lowest masses and shortest lifetimes
d0 the highest masses and longest lifetimes
Physics
1 answer:
Helen [10]4 years ago
7 0

Answer:

Option b

Explanation:

  • Higher mass of the main sequence stars give rise to larger compression leading to higher surface temperatures which in turn results in faster rate of fusion leading to greater luminosity.
  • The luminosity depends on the reactions in the core. the higher the reaction sin the core the more is the luminosity of the main sequence star.
  • The lifetimes of the highest masses stars of the main sequence is the shortest  because of the reason that in spite of having a greater amount of hydrogen inside for nuclear reactions, the rate of consumption of this hydrogen is quite fast.
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The average standard rectangular building brick has a mass of 3.10 kg and dimensions of 225 m x 112 m x 75 m. The gravitational
zzz [600]

Answer:

P=3.61\times 10^{-3}\ Pa

Explanation:

Given that,

Mass of a brick, m = 3.1 kg

The dimensions of the brick 225 m x 112 m x 75 m

We need to find the maximum pressure created by the brick. We know that, the force acting per unit area is called pressure exerted. It is given by the formula as follows :

P=\dfrac{F}{A}

F = mg

A = area with minimum dimensions i.e. 112 m x 75 m

Pressure is maximum when the area is least.

So,

P=\dfrac{mg}{A}\\\\P=\dfrac{3.1\times 9.8}{112\times 75}\\\\P=3.61\times 10^{-3}\ Pa

So, the maximum pressure created by brick is 3.61\times 10^{-3}\ Pa.

5 0
3 years ago
When melting of a metamorphic rock occurs, it changes into what?
Thepotemich [5.8K]

Answer:

The upper limit of metamorphism occurs at the pressure and temperature of wet partial melting of the rock in question. Once melting begins, the process changes to an igneous process rather than a metamorphic process. During metamorphism the protolith undergoes changes in texture of the rock and the mineral make up of the rock.

Explanation:

i hope this helps

8 0
3 years ago
A 150 g sample of brass at 100 °C is placed in a Styrofoam cup of water containing 120 mL of water at 10 °C. No heat is lost to
fredd [130]

Answer:

≈19.144°C.

Explanation:

all the details are in the attachment.

Note, that c₁, m₁, t₁ are the parameters of the sample of brass; c₂, m₂ and t₂ are  the parameters of the sample of water.

P.S. change the provided design according Your requirements.

4 0
2 years ago
When steam condenses 1. All of these occur. 2. None of these occur. 3. molecules move closer together. 4. it changes from the ga
True [87]

2. None of these occur

5 0
4 years ago
Read 2 more answers
Sam, whose mass is 78 kg , stands at the top of a 11-m-high, 110-m-long snow-covered slope. His skis have a coefficient of kinet
Valentin [98]

Answer:

v = 8.09   m/s

Explanation:

For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.

Let's calculate the energy

       

starting point. Higher

         Em₀ = U = m gh

final point. To go down the slope

         Em_f = K = ½ m v²

The work of the friction force is

         W = fr L cos 180

to find the friction force let's use Newton's second law

Axis y

        N - W_y = 0

        N = W_y

X axis

        Wₓ - fr = ma

let's use trigonometry

        sin  θ = y / L

         sin θ = 11/110 = 0.1

         θ = sin⁻¹  0.1

          θ = 5.74º

         sin 5.74 = Wₓ / W

         cos 5.74 = W_y / W

         Wₓ = W sin 5.74

         W_y = W cos 5.74

the formula for the friction force is

         fr = μ N

         fr = μ W cos θ

Work is friction force is

         W_fr = - μ W L cos θ  

Let's use the relationship of work with energy

        W + ΔU = ΔK

         -μ mg L cos 5.74 + (mgh - 0) = 0  - ½ m v²

        v² = - 2 μ g L cos 5.74 +2 (gh)

        v² = 2gh - 2 μ gL cos 5.74

let's calculate

        v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74

        v² = 215.6 -150.16

        v = √65.44

        v = 8.09   m/s

6 0
3 years ago
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