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.

You might be interested in

What happens to the electricity in a strand of

rodikova [14]

Answer:

Either all the lights are very old and need new ones or the cable is wet.

Explanation:

5 0

3 years ago

Read 2 more answers

What is the amount of thermal energy needed to make 5 kg of ice at - 10 °C to

agasfer [191]

Answer:

The amount of thermal energy needed is 15167500 joules.

Explanation:

By First Law of Thermodynamics, we see that amount of thermal energy ( $Q$ ), in joules, is equal to the change in internal energy. From statement we understand that change in internal energy consisting in two latent components ( $U_{l,ice}$ , $U_{l,steam}$ ), in joules, and two sensible component ( $U_{s,w}$ ), in joules, that is:

$Q = U_{l,ice} + U_{s, w} + U_{s,ice} + U_{l,steam}$ (1)

By definitions of Sensible and Latent Heat, we expanded the formula:

$Q = m\cdot (h_{f,w}+h_{v,w}+c_{ice}\cdot \Delta T_{ice}+c_{w}\cdot \Delta T_{w})$ (2)

Where:

$m$ - Mass, in kilograms.

$h_{f,w}$ - Latent heat of fussion of water, in joules per kilogram.

$h_{v,w}$ - Latent heat of vaporization of water, in joules per kilogram.

$c_{ice}$ - Specific heat of ice, in joules per kilogram per degree Celsius.

$c_{w}$ - Specific heat of water, in joules per kilogram per degree Celsius.

$\Delta T_{ice}$ - Change in temperature of ice, measured in degrees Celsius.

$\Delta T_{w}$ - Change in temperature of water, measured in degrees Celsius.

If we know that $m = 5\,kg$ , $h_{f,w} = 3.34\times 10^{5}\,\frac{J}{kg}$ , $h_{v,w} = 2.26\times 10^{6}\,\frac{J}{kg}$ , $c_{ice} = 2.090\times 10^{3}\,\frac{J}{kg\cdot ^{\circ}C}$ , $c_{w} = 4.186\times 10^{3}\,\frac{J}{kg\cdot ^{\circ}C}$ , $\Delta T_{ice} = 10\,^{\circ}C$ and $\Delta T_{w} = 100\,^{\circ}C$ , then the amount of thermal energy is: