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Margaret [11]
3 years ago
6

You observe a very large and very hot star in the constellation orion. on the same night, you observe another star in orion that

is much smaller but has the same temperature. which star is more luminous?
Physics
1 answer:
Neporo4naja [7]3 years ago
7 0
The bigger one as the luminosity has a direct exponential relationship with R of the star so the star which has more surface area will be more luminous than the smaller one
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Anyone know how to do this?
Gala2k [10]

Answer:

I think, (remember think) it might be 2.0 m/s

Explanation:

If it's wrong I'm truly sorry.

6 0
3 years ago
Which of two factors influence the weight of an object due to gravitational pull?
densk [106]
The weight of an object when calculated by multiplying with the pull of the gravity is dependent on the mass of the object and the value of g. The value of g is constant however is still dependent on the distance of the object from the center of the Earth. Thus, the answers are <em>mass and distance. </em>
5 0
3 years ago
The Mars Curiosity rover was required to land on the surface of Mars with a velocity of 1 m/s. Given the mass of the landing veh
Aliun [14]

Answer:

The value is      A   = 39315 \  m^2

Explanation:

From the question we are told that

    The velocity which the rover is suppose to land with is  v  =  1 \ m/s

    The  mass of the rover and the parachute is  m  =  2270 \ kg

     The  drag coefficient is  C__{D}}  =  0.5

      The atmospheric density of Earth  is  \rho =  1.2 \  kg/m^3

     The acceleration due to gravity in Mars is  g_m  =  3.689 \  m/s^2

     

Generally the Mars  atmosphere density is mathematically represented as

          \rho_m  =  0.71 *  \rho

=>        \rho_m  =  0.71 *  1.2

=>        \rho_m  = 0.852 \  kg/m^3

Generally the drag force on the rover and the parachute  is mathematically represented as

          F__{D}} =  m  *  g_{m}

=>       F__{D}} =  2270   *  3.689  

=>       F__{D}} =  8374 \ N  

Gnerally this drag force is mathematically represented as

         F__{D}} =   C__{D}} *  A *  \frac{\rho_m * v^2 }{2}

Here A is the frontal area

So  

         A   =  \frac{2 *  F__D }{ C__D}  *  \rho_m  * v^2   }

=>       A   =  \frac{2 * 8374 }{ 0.5 *  0.852    *  1 ^2   }

=>       A   = 39315 \  m^2

8 0
3 years ago
Measurements of two electric currents are shown in the chart. A 3-column table with 2 rows titled Electric Currents. The first c
Jobisdone [24]

The answer is B - Current Y has a greater potential difference, and the charges flow at a slower rate.

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I would say that this is the first law of thermodynamics.
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