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tatiyna
3 years ago
9

What are scientists who study space called?

Physics
2 answers:
ikadub [295]3 years ago
7 0

`•.,¸¸,.•´¯     '     ¯´•.,¸¸,.•`

Answer:

Brilliant_brown [7]3 years ago
4 0

Answer:

Astronomer

Explanation:

A scientist who studies the objects in the sky, including planets, galaxies, black holes, and stars, is called an astronomer. These days, the terms astronomer and astrophysicist are used interchangeably, to talk about any physicist who specializes in celestial bodies and the forces that affect them.

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Elaborate on the reason(s) that matter is said to move even as in a solid state.
Mariulka [41]

<u>Answer:</u>

The matter does not move in solid state but vibrates.

<u>Explanation:</u>

The atoms inside the matter cannot move or shift their positions without any external force but makes some small vibration movements. Generally in solids, the particles are bound by the attractive forces acting in between the atoms inside the matter.

The small vibrations that are happening inside the matter are because of the external factors like temperature. The increase in temperature raises the kinetic energy of the atoms inside and makes them move faster and this results in the vibration of the matter.

6 0
3 years ago
Read 2 more answers
A car is at velocity of 20 km/h. If the car traveled 120 km in 3 hours at constant acceleration, what is its final velocity?
dedylja [7]
The velocity is dueuueue. 100
8 0
3 years ago
The back wall of a home aquarium is a mirror that is a distance of 46.0 cm away from the front wall. The walls of the tank are n
monitta

Answer:

i know the questin but i got to try and find it

Explanation:

5 0
2 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
Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.00×104 Pa . Assum
cestrela7 [59]

Answer:

T_{2}=278.80 K

Explanation:

Let's use the equation that relate the temperatures and volumes of an adiabatic process in a ideal gas.

(\frac{V_{1}}{V_{2}})^{\gamma -1} = \frac{T_{2}}{T_{1}}.

Now, let's use the ideal gas equation to the initial and the final state:

\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}

Let's recall that the term nR is a constant. That is why we can match these equations.  

We can find a relation between the volumes of the initial and the final state.

\frac{V_{1}}{V_{2}}=\frac{T_{1}p_{2}}{T_{2}p_{1}}

Combining this equation with the first equation we have:

(\frac{T_{1}p_{2}}{T_{2}p_{1}})^{\gamma -1} = \frac{T_{2}}{T_{1}}

(\frac{p_{2}}{p_{1}})^{\gamma -1} = \frac{T_{2}^{\gamma}}{T_{1}^{\gamma}}

Now, we just need to solve this equation for T₂.

T_{1}\cdot (\frac{p_{2}}{p_{1}})^{\frac{\gamma - 1}{\gamma}} = T_{2}

Let's assume the initial temperature and pressure as 25 °C = 298 K and 1 atm = 1.01 * 10⁵ Pa, in a normal conditions.

Here,

p_{2}=8.00\cdot 10^{4} Pa \\p_{1}=1.01\cdot 10^{5} Pa\\ T_{1}=298 K\\ \gamma=1.40

Finally, T2 will be:

T_{2}=278.80 K

6 0
3 years ago
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