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3 years ago

Part 1: What are the two groups of planets based on composition?

1 answer:

3 0

The two groups of planets based on their composition are the terrestrial planets, and the jovian planets. The terrestrial planets are small, and their composition is different than that of the jovian planets. The terrestrial planets have a solid iron inner core, liquid iron outer core, silicate mantle, and silicate crust. The jovian planets are also called gas giants because they are much larger and are gassy, not hard as the terrestrial planets. These planets have a silicate inner core, ice outer core, liquid metallic hydrogen layer, liquid hydrogen layer, and on the top gaseous hydrogen.

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What is analogy a powerful rhetorical device?

wolverine [178]

The other person who answered this is wrong btw

5 0

2 years ago

A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.

MakcuM [25]

Answer:

b. 9.5°C

Explanation:

$m_i$ = Mass of ice = 50 g

$T_i$ = Initial temperature of water and Aluminum = 30°C

$L_f$ = Latent heat of fusion = $3.33\times 10^5\ J/kg^{\circ}C$

$m_w$ = Mass of water = 200 g

$c_w$ = Specific heat of water = 4186 J/kg⋅°C

$m_{Al}$ = Mass of Aluminum = 80 g

$c_{Al}$ = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

$m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C$

The final equilibrium temperature is 9.50022°C

4 0

3 years ago

Which way do the players on the court rotate during a game? What happens if the players serve out of order?

tigry1 [53]

Answer:

Explanation:

on volleyballit's clockwise and What Would Happen If Someone Served Out Of Order? If the wrong player on your team serves because it's not their turn, the point and the possession of the serve will go to your opponents.

5 0

2 years ago

A child has an ear canal that is 1.3 cm long. Assume the speed of sound is v = 344 m/s.

kap26 [50]

Answer:

The frequencies are $(f, f_1) = (6615.4 \ Hz , 19846.2\ Hz)$

Explanation:

From the question we are told that

The length of the ear canal is $l = 1.3 \ cm =\frac{1.3}{100} = 0.013 \ m$

The speed of sound is assumed to be $v_s = 344 \ m/s$

Now taking look at a typical ear canal we see that we assume it is a closed pipe

Now the fundamental harmonics for the pipe(ear canal) is mathematically represented as

$f = \frac{v_s}{4 * l }$

substituting values

$f = \frac{344}{4 * 0.013 }$

$f = 6615.4 \ Hz$

Also the the second harmonic for the pipe (ear canal) is mathematically represented as

$f_1 = \frac{3v_s}{4 * l}$

substituting values

$f_1 = \frac{3 * 344}{4 * 0.013}$

$f_1 = 19846.2 \ Hz$

Given that sound would be loudest in the pipe at the frequency, it implies that the child will have an increased audible sensitivity at this frequencies

6 0

2 years ago

A flat coil having 160 turns, each with an area of 0.20 m 2, is placed with the plane of its area perpendicular to a magnetic fi

S_A_V [24]

Answer:

10.2 Watt

Explanation:

$N$ = number of turns in flat coil = 160