Answer:0.0704 kg
Explanation:
Given
initial Absolute pressure
=210+101.325=311.325
![T_1=25^{\circ}\approx 298 K](https://tex.z-dn.net/?f=T_1%3D25%5E%7B%5Ccirc%7D%5Capprox%20298%20K)
![V=0.025 m^3](https://tex.z-dn.net/?f=V%3D0.025%20m%5E3)
![T_2=50^{\circ}\approx 323 K](https://tex.z-dn.net/?f=T_2%3D50%5E%7B%5Ccirc%7D%5Capprox%20323%20K)
as the volume remains constant therefore
![\frac{P_1}{T_1}=\frac{P_2}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%7BP_1%7D%7BT_1%7D%3D%5Cfrac%7BP_2%7D%7BT_2%7D)
![\frac{311.325}{298}=\frac{P_2}{323}](https://tex.z-dn.net/?f=%5Cfrac%7B311.325%7D%7B298%7D%3D%5Cfrac%7BP_2%7D%7B323%7D)
![P_2=337.44 KPa](https://tex.z-dn.net/?f=P_2%3D337.44%20KPa)
therefore Gauge pressure is 337.44-101.325=236.117 KPa
Initial mass ![m_1=\frac{P_1V}{RT_1}=\frac{311.325\times 0.025}{0.0287\times 298}](https://tex.z-dn.net/?f=m_1%3D%5Cfrac%7BP_1V%7D%7BRT_1%7D%3D%5Cfrac%7B311.325%5Ctimes%200.025%7D%7B0.0287%5Ctimes%20298%7D)
![m_1=0.91 kg](https://tex.z-dn.net/?f=m_1%3D0.91%20kg)
Final mass ![m_2=\frac{P_2V}{RT_2}=\frac{311.325\times 0.025}{0.0287\times 323}](https://tex.z-dn.net/?f=m_2%3D%5Cfrac%7BP_2V%7D%7BRT_2%7D%3D%5Cfrac%7B311.325%5Ctimes%200.025%7D%7B0.0287%5Ctimes%20323%7D)
![m_2=0.839](https://tex.z-dn.net/?f=m_2%3D0.839)
Therefore
=0.91-0.839=0.0704 kg of air needs to be removed to get initial pressure back
Answer:E
Molecules at higher temperatures have more energy, thus they can vibrate faster. Since the molecules vibrate faster, sound waves can travel more quickly. Wavelength always increases with speed, as frequency is fixed at source and speed= freq x wavelength
The speed of sound in room temperature air is 346 meters per second.
Answer:
B. 4.29s
Explanation:
If the man appears to move 12.4m/s when the boat appears to move 11m/s, it means that the man moves 1.4m on the boat (12.4-11=1.4). So if the man moves 1.4m per second then it will take him 6/1.4=4.29s to move 6 m relative to the boat.
Hope this helped!
Answer:
final displacement lf = 0.39 m
Explanation:
from change in momentum equation:
![\delta p = m \sqrt(2g * y/x)* [\sqrt li + \sqrt lf]](https://tex.z-dn.net/?f=%5Cdelta%20p%20%3D%20m%20%5Csqrt%282g%20%2A%20y%2Fx%29%2A%20%5B%5Csqrt%20li%20%2B%20%5Csqrt%20lf%5D)
given: m = 0.4kg, y/x = 19/85, li = 1.9 m,
\delta p = 1.27 kg*m/s.
putting all value to get the final displacement value
![1.27 = 0.4\sqrt(2*9.81 *(19/85))* [\sqrt 1.9 + \sqrt lf]](https://tex.z-dn.net/?f=1.27%20%3D%200.4%5Csqrt%282%2A9.81%20%2A%2819%2F85%29%29%2A%20%5B%5Csqrt%201.9%20%2B%20%5Csqrt%20lf%5D)
final displacement lf = 0.39 m
I believe it's a great square