Answer:
![c=10\ J/kg^{\circ} C](https://tex.z-dn.net/?f=c%3D10%5C%20J%2Fkg%5E%7B%5Ccirc%7D%20C)
Explanation:
Given that,
Heat required, Q = 1200 J
Mass of the object, m = 20 kg
The increase in temperature, ![\Delta T=6^{\circ} C](https://tex.z-dn.net/?f=%5CDelta%20T%3D6%5E%7B%5Ccirc%7D%20C)
We need to find the specific heat of the object. The heat required to raise the temperature is given by :
![Q=mc\Delta T\\\\c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{1200}{20\times 6}\\\\c=10\ J/kg^{\circ} C](https://tex.z-dn.net/?f=Q%3Dmc%5CDelta%20T%5C%5C%5C%5Cc%3D%5Cdfrac%7BQ%7D%7Bm%5CDelta%20T%7D%5C%5C%5C%5Cc%3D%5Cdfrac%7B1200%7D%7B20%5Ctimes%206%7D%5C%5C%5C%5Cc%3D10%5C%20J%2Fkg%5E%7B%5Ccirc%7D%20C)
So, the specific heat of the object is
.
Answer:
Destructive interference
Explanation:
Destructive interference occurs when waves come together in such a way that they completely cancel each other out. When two waves interfere destructively, they must have the same amplitude in opposite directions.
Answer:
Answer is A) Fermi
Explanation:
Fermi is the expressive unit for nuclear sizes. Fermi = 10^-15 meter.
Answer:
a
The pressure will increase
b
![T_2 = 576^oC](https://tex.z-dn.net/?f=T_2%20%3D%20%20576%5EoC)
Explanation:
From the ideal gas law we have that
![PV = nRT](https://tex.z-dn.net/?f=PV%20%20%3D%20%20nRT)
We see that the temperature varies directly with the pressure so if there is an increase in temperature that pressure will increase
The initial temperature is ![T_i = 10^oC = 10 + 273 = 283 \ K](https://tex.z-dn.net/?f=T_i%20%20%3D%20%2010%5EoC%20%3D%2010%20%2B%20273%20%3D%20%20283%20%5C%20%20K%20)
The objective of this solution is to obtain the temperature of the gas where the pressure is tripled
Now from the above equation given that nR and V are constant we have that
![\frac{P}{T} = constant](https://tex.z-dn.net/?f=%5Cfrac%7BP%7D%7BT%7D%20%20%3D%20%20constant)
=> ![\frac{P_1}{T_1} =\frac{P_2}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%7BP_1%7D%7BT_1%7D%20%20%3D%5Cfrac%7BP_2%7D%7BT_2%7D)
Let assume the initial pressure is ![P_1 = 1 Pa](https://tex.z-dn.net/?f=P_1%20%3D%20%201%20Pa)
So tripling it will result to the pressure being ![P_2 = 3 Pa](https://tex.z-dn.net/?f=P_2%20%3D%20%203%20Pa)
So
=> ![T_2 = 3 * 283](https://tex.z-dn.net/?f=T_2%20%20%3D%20%203%20%2A%20%20283)
=> ![T_2 = 3 * 283](https://tex.z-dn.net/?f=T_2%20%20%3D%20%203%20%2A%20%20283)
=> ![T_2 = 849 \ K](https://tex.z-dn.net/?f=T_2%20%20%3D%20849%20%5C%20K%20)
Converting back to ![^oC](https://tex.z-dn.net/?f=%5EoC)
![T_2 = 849 - 273](https://tex.z-dn.net/?f=T_2%20%20%3D%20%20849%20-%20%20273)
=> ![T_2 = 576^oC](https://tex.z-dn.net/?f=T_2%20%3D%20%20576%5EoC)