Answer: You could dissolve it by heating it back up, then just cooling it down again.
Hope that helps!
Answer: <u>In a divergent plate boundary</u>, seafloor spreading taking place. It leads to the formation of oceans as new materials are added here along the mid-oceanic ridge. There occur volcanism and shallow-focus earthquakes.
<u>In a convergent plate boundary</u>, two plates collide to form high mountain belts and also volcanic eruptions take place. There occur long chains of volcanic as well as island arcs, in association with deep-focus earthquakes.
<u>In a transform plate boundary</u>, two plates slide past each other, conserving the plates. Shallow-focus earthquakes are generated here.
The earth has experienced various geological processes, such as weathering and erosion of rocks, earthquakes, volcanic eruptions, mass extinction events, plate tectonic movements and many more. These continuous processes have configured the present shape of the earth's surface.
For example, the breaking up of the supercontinent Pangea divided into Laurasia and Gondwanaland and subsequently formed the present scenario. This separation of continents has taken place due to the convection current that generates in the mantle.
To solve this problem we will apply the principle of conservation of energy and the definition of kinematic energy as half the product between mass and squared velocity. So,
![KE_i = KE_f](https://tex.z-dn.net/?f=KE_i%20%3D%20KE_f)
![KE_f = \frac{1}{2} mv^2](https://tex.z-dn.net/?f=KE_f%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mv%5E2)
Here,
m = Mass
V = Velocity
Replacing,
![KE_f = \frac{1}{2} (12000)(11)^2](https://tex.z-dn.net/?f=KE_f%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2812000%29%2811%29%5E2)
![KE_f = 72600J](https://tex.z-dn.net/?f=KE_f%20%3D%2072600J)
Therefore the final kinetic energy of the two car system is 72.6kJ
Answer:
The temperature of the metal is ![T_m = 376.8 ^o C](https://tex.z-dn.net/?f=T_m%20%20%3D%20%20376.8%20%5Eo%20C)
Explanation:
From the question we are told that
The mass of the metal is ![M = 60 \ kg](https://tex.z-dn.net/?f=M%20%3D%20%2060%20%5C%20kg)
The specific heat of the metal is ![c_p = 0.1027 kcal/(kg \cdot ^oC)](https://tex.z-dn.net/?f=c_p%20%20%3D%20%200.1027%20kcal%2F%28kg%20%5Ccdot%20%5EoC%29)
The mass of the oil is ![M_o = 810 \ kg](https://tex.z-dn.net/?f=M_o%20%20%3D%20%20810%20%5C%20kg)
The temperature of the oil is ![T_o = 35^oC](https://tex.z-dn.net/?f=T_o%20%20%3D%20%2035%5EoC)
The specific heat of oil is ![c_o = 0.7167 kcal/(kg \cdot ^oC )](https://tex.z-dn.net/?f=c_o%20%20%3D%20%200.7167%20kcal%2F%28kg%20%5Ccdot%20%5EoC%20%29)
The equilibrium temperature is ![T_e = 39 ^oC](https://tex.z-dn.net/?f=T_e%20%20%3D%20%2039%20%5EoC)
According to the law of energy conservation
Heat lost by metal = heat gained by the oil
So
The quantity of heat lost by the metal is mathematically represented as
![Q = - Mc_p \Delta T](https://tex.z-dn.net/?f=Q%20%3D%20%20-%20Mc_p%20%5CDelta%20T)
=> ![Q = -Mc_p (T_m - T_c)](https://tex.z-dn.net/?f=Q%20%3D%20%20-Mc_p%20%28T_m%20%20-%20%20T_c%29)
Where
the temperature of metal before immersion
The negative sign show heat lost
The quantity of gained t by the metal is mathematically represented as
![Q = M_o c_o \Delta T](https://tex.z-dn.net/?f=Q%20%3D%20%20M_o%20c_o%20%5CDelta%20T)
=> ![Q = M_o c_o (T_c - T_o)](https://tex.z-dn.net/?f=Q%20%3D%20%20M_o%20c_o%20%28T_c%20-%20T_o%29)
So
![Mc_p (T_m - T_c) = M_o c_o (T_c - T_o)](https://tex.z-dn.net/?f=Mc_p%20%28T_m%20%20-%20%20T_c%29%20%20%20%3D%20%20%20M_o%20c_o%20%28T_c%20-%20T_o%29)
substituting values
![- 60 * 0.1027 (T_m - 39) = 810 * 0.7167 * (39 - 35)](https://tex.z-dn.net/?f=-%2060%20%2A%200.1027%20%28T_m%20%20-%2039%29%20%20%20%3D%20%20%20810%20%2A%200.7167%20%2A%20%20%2839%20-%2035%29)
=> ![T_m = 376.8 ^o C](https://tex.z-dn.net/?f=T_m%20%20%3D%20%20376.8%20%5Eo%20C)
Answer:
Please see below as the answer is self-explanatory.
Explanation:
The low band of the VHF TV Spectrum, spans channels 2-6, from 54 to 88 Mhz.
In the analog TV, in the Americas, the total bandwidth of any channel is 6 Mhz, with the visual carrier modulated in VSS (Vestigial Side Band) at 1.25 Mhz from the lowest frequency of the channel.
The aural carrier is located at 4.5 Mhz from the visual carrier, and is FM modulated.
For Channel 6, which spans between 82 and 88 Mhz, the visual carrier is at 83.25 Mhz, so the aural carrier is at 87.75 Mhz, which falls within the FM Band, so it is possible to listen the audio part of this channel in a FM radio receiver, even at a lower volume, due to the FM radio has a greater deviation than TV aural carrier.