Answer:
b) It is impossible to tell without knowing the masses.
Explanation:
The temperature change of a substance when it receives/gives off a certain amount of heat Q is given by
![\Delta T= \frac{Q}{m C_s}](https://tex.z-dn.net/?f=%5CDelta%20T%3D%20%5Cfrac%7BQ%7D%7Bm%20C_s%7D)
where
Q is the amount of heat
m is the mass of the substance
Cs is the specific heat capacity of the substance
In this case, we have a hot piece of aluminum in contact with a cold piece of copper: the amount of heat given off by the aluminum is equal to the amount of heat absorbed by the copper, so Q is the same for the two substances. However, we see that the temperature change of the two substances depends on two other factors: the mass, m, and the specific heat, Cs. So, since we know only the specific heat of the two substances, but not their mass, we can't tell which object will experience the greater temperature change.
Answer: 1.96 m/s
Explanation:
Given
Mass of Professor ![m_1=113\ kg](https://tex.z-dn.net/?f=m_1%3D113%5C%20kg)
Velocity of professor ![u_1=1.56\ m/s](https://tex.z-dn.net/?f=u_1%3D1.56%5C%20m%2Fs)
mass of chair ![m_2=10\ kg](https://tex.z-dn.net/?f=m_2%3D10%5C%20kg)
velocity of chair ![u_2=6.5\ m/s](https://tex.z-dn.net/?f=u_2%3D6.5%5C%20m%2Fs)
Suppose after the collision, v is the common velocity
Conserving momentum
![\Rightarrow m_1u_1+m_2u_2=(m_1+m_2)v\\\\\Rightarrow v=\dfrac{113\times 1.56+10\times 6.5}{113+10}=\dfrac{241.28}{123}\\\\\Rightarrow v=1.96\ m/s](https://tex.z-dn.net/?f=%5CRightarrow%20m_1u_1%2Bm_2u_2%3D%28m_1%2Bm_2%29v%5C%5C%5C%5C%5CRightarrow%20v%3D%5Cdfrac%7B113%5Ctimes%201.56%2B10%5Ctimes%206.5%7D%7B113%2B10%7D%3D%5Cdfrac%7B241.28%7D%7B123%7D%5C%5C%5C%5C%5CRightarrow%20v%3D1.96%5C%20m%2Fs)
Answer:
Conduction is the transfer of heat energy by direct process.
Radiation is the transfer of heat energy with electromagnetic waves
Explanation:
Answer:
15.8 ft/s
Explanation:
= Velocity of car A = 9 ft/s
a = Distance car A travels = 21 ft
= Velocity of car B = 13 ft/s
b = Distance car B travels = ft
c = Distance between A and B after 4 seconds = √(a²+b²) = √(21²+28²) = √1225 ft
From Pythagoras theorem
a²+b² = c²
Now, differentiating with respect to time
![2a\frac{da}{dt}+2b\frac{db}{dt}=2c\frac{dc}{dt}\\\Rightarrow a\frac{da}{dt}+b\frac{db}{dt}=c\frac{dc}{dt}\\\Rightarrow \frac{dc}{dt}=\frac{a\frac{da}{dt}+b\frac{db}{dt}}{c}\\\Rightarrow \frac{dc}{dt}=\frac{21\times 9+28\times 13}{\sqrt{1225}}\\\Rightarrow \frac{dc}{dt}=15.8\ ft/s](https://tex.z-dn.net/?f=2a%5Cfrac%7Bda%7D%7Bdt%7D%2B2b%5Cfrac%7Bdb%7D%7Bdt%7D%3D2c%5Cfrac%7Bdc%7D%7Bdt%7D%5C%5C%5CRightarrow%20a%5Cfrac%7Bda%7D%7Bdt%7D%2Bb%5Cfrac%7Bdb%7D%7Bdt%7D%3Dc%5Cfrac%7Bdc%7D%7Bdt%7D%5C%5C%5CRightarrow%20%5Cfrac%7Bdc%7D%7Bdt%7D%3D%5Cfrac%7Ba%5Cfrac%7Bda%7D%7Bdt%7D%2Bb%5Cfrac%7Bdb%7D%7Bdt%7D%7D%7Bc%7D%5C%5C%5CRightarrow%20%5Cfrac%7Bdc%7D%7Bdt%7D%3D%5Cfrac%7B21%5Ctimes%209%2B28%5Ctimes%2013%7D%7B%5Csqrt%7B1225%7D%7D%5C%5C%5CRightarrow%20%5Cfrac%7Bdc%7D%7Bdt%7D%3D15.8%5C%20ft%2Fs)
∴ Rate at which distance between the cars is increasing three hours later is 15.8 ft/s