All categories

3 years ago

Types of telescopefor Spaceobservation

Physics

2 answers:

harina [27]3 years ago

8 0

Answer:

oii me manda mensagem fofo vc tem namorada fofo

VikaD [51]3 years ago

4 0

Answer:The three main types are reflecting telescopes, refracting telescopes, and catadioptric telescopes. Radio telescopes collect and focus radio waves from distant objects. Space telescopes orbit Earth, collecting wavelengths of light that are normally blocked by the atmosphere.

You might be interested in

Can atoms go bad?, not in the reversible way like ionization and isotopes but really malfunction our die.

Elza [17]

Answer:

atoms cannot go bad

Explanation:

Because they stay alive and get good nutriants

5 0

2 years ago

An ordinary egg can be approximated as a 5.5-cm diameter sphere. The egg is initially at a uniform temperature of 8°C and is dro

kupik [55]

Answer:

a) $Q_{in} = 13.742\,kW$ , b) $\Delta S = 370.15\,\frac{kJ}{K}$

Explanation:

a) The heat transfered to the egg is computed by the First Law of Thermodynamics:

$Q_{in} +U_{sys,1} - U_{sys,2} = 0$

$Q_{in} = U_{sys,2} - U_{sys,1}$

$Q_{in} = \rho_{egg}\cdot \left(\frac{4\pi}{3}\cdot r^{3}\right)\cdot c \cdot (T_{2}-T_{1})$

$Q_{in} = \left(1020\,\frac{kg}{m^{3}}\right)\cdot \left(\frac{4\pi}{3}\right)\cdot (0.025\,m)^{3}\cdot \left(3.32\,\frac{kJ}{kg\cdot ^{\textdegree}C} \right)\cdot (70\,^{\textdegree}C - 8\,^{\textdegree}C)$

$Q_{in} = 13.742\,kW$

b) The amount of entropy generation is determined by the Second Law of Thermodynamics:

$\Delta S = \frac{Q_{in}}{T_{in}}$

$\Delta S = \frac{13.742\,kJ}{370.15\,K}$

$\Delta S = 370.15\,\frac{kJ}{K}$

3 0

3 years ago

Oil having a density of 926 kg/m3 floats on water. A rectangular block of wood 3.69 cm high and with a density of 974 kg/m3 floa

zloy xaker [14]

Answer:

the position of the wood below the interface of the two liquids is 2.39 cm.

Explanation:

Given;

density of oil, $\rho _o$ = 926 kg/m³

density of the wood, $\rho _{wood}$ = 974 kg/m³

density of water, $\rho _w$ = 1000 kg/m³

height of the wood, h = 3.69 cm

Based on the density of the wood, it will position across the two liquids.

let the position of the wood below the interface of the two liquids = x

Let the wood be in equilibrium position;

$F_{wood} - F_{oil} - F_{water} = 0\\\\\rho _{wood} .gh - \rho _o .g(h-x) - \rho_w .gx = 0\\\\\rho _{wood} .h - \rho _o (h-x) - \rho_w .x = 0\\\\\rho _{wood} .h -\rho _o h + \rho _o x - \rho_w .x =0\\\\h (\rho _{wood} -\rho _o ) = x( \rho_w - \rho _o)\\\\x =h[\frac{ \rho _{wood} -\rho _o }{\rho_w - \rho _o} ]\\\\x = 3.69\ cm \times [\frac{974 - 926}{1000-926} ]\\\\x = 2.39 \ cm$