Solution :
Given data :
p = 315612 Pa
![$V_1=7.07 \ m/sec$](https://tex.z-dn.net/?f=%24V_1%3D7.07%20%5C%20m%2Fsec%24)
At exit of B,
p = ![$P_{atm}$](https://tex.z-dn.net/?f=%24P_%7Batm%7D%24)
![$V_B = 26.1 \ m/sec$](https://tex.z-dn.net/?f=%24V_B%20%3D%2026.1%20%5C%20m%2Fsec%24)
At exit of A,
![p=P_{atm}](https://tex.z-dn.net/?f=p%3DP_%7Batm%7D)
![$V_{A} = 26.1 \ m/s$](https://tex.z-dn.net/?f=%24V_%7BA%7D%20%3D%2026.1%20%5C%20m%2Fs%24)
We need to determine X component of force (
) to hold in its place.
From figure,
![$\sum F_x = m_0'V_{0x} - m_iV_{ix} $](https://tex.z-dn.net/?f=%24%5Csum%20F_x%20%3D%20m_0%27V_%7B0x%7D%20-%20m_iV_%7Bix%7D%20%24)
![$=F_x+P_1A_1\sin 30=-mVA-mV_1 \sin 30$](https://tex.z-dn.net/?f=%24%3DF_x%2BP_1A_1%5Csin%2030%3D-mVA-mV_1%20%5Csin%2030%24)
![$=F_x=-pA_1\sin 30-m_AV_AA-m_B \sin30$](https://tex.z-dn.net/?f=%24%3DF_x%3D-pA_1%5Csin%2030-m_AV_AA-m_B%20%5Csin30%24)
Substitute all the values,
![$=F_x=[-315612 \times \frac{\pi}{4}(0.3)^2 \sin 30]-[26.1 \times 1000 \times 26.1 \frac{\pi}{4}(0.1)^2]-[7.07 \times 1000\times 0.5 \sin 30]$](https://tex.z-dn.net/?f=%24%3DF_x%3D%5B-315612%20%5Ctimes%20%5Cfrac%7B%5Cpi%7D%7B4%7D%280.3%29%5E2%20%5Csin%2030%5D-%5B26.1%20%5Ctimes%201000%20%5Ctimes%2026.1%20%5Cfrac%7B%5Cpi%7D%7B4%7D%280.1%29%5E2%5D-%5B7.07%20%5Ctimes%201000%5Ctimes%200.5%20%5Csin%2030%5D%24)
![$=F_x = -11154.64-5350.21-1767.28$](https://tex.z-dn.net/?f=%24%3DF_x%20%3D%20-11154.64-5350.21-1767.28%24)
![$F_x = -18.2733 \ kN$](https://tex.z-dn.net/?f=%24F_x%20%3D%20-18.2733%20%5C%20kN%24)
Therefore, the force required to hold the nozzle in its place along horizontal direction.
![$F_x = -18.2733 \ kN$](https://tex.z-dn.net/?f=%24F_x%20%3D%20-18.2733%20%5C%20kN%24)
Answer:
Explanation:
The capacitive reactance of the capacitor is determined by ![X_{C}=\frac{1}{\omega C}](https://tex.z-dn.net/?f=X_%7BC%7D%3D%5Cfrac%7B1%7D%7B%5Comega%20C%7D)
When ![f'=2f \rightarrow \omega' =2\omega](https://tex.z-dn.net/?f=f%27%3D2f%20%5Crightarrow%20%5Comega%27%20%3D2%5Comega)
and ![X_{C}'=\frac{1}{2}X_{C}](https://tex.z-dn.net/?f=X_%7BC%7D%27%3D%5Cfrac%7B1%7D%7B2%7DX_%7BC%7D)
(The capacitive reactance will be reduced into 1/2 times of the previous capacitive reactance)
Answer:
Please select the correct statements concerning use of a centrifuge. (Select all that apply.)
(a) When placing test tubes in the centrifuge, what are the best practices?
- Wait until all positions on the rotor contain a test tube before activating the centrifuge.
- <u>Place test tubes containing approximately equal volumes in opposite positions on the rotor.</u>
- Carefully stopper the test tubes to ensure the liquids do not spill out.
- <u>Never put stoppers or other items in the centrifuge with your tubes</u>.
(b) When operating the centrifuge, students should do which of the following?
- <u>Make sure that the lid is completely shut and the safety knob is turned to engage the safety switch.</u>
- <u>Centrifuge for approximately 30 minutes</u>.
- Centrifuge for approximately 30 seconds.
- Mid-way through the process, open the lid of the centrifuge and check to be sure it is operating.
(c) When the centrifuge process is complete, students should do which of the following?
- <u>Using the eye-hole on top, monitor the spinning rate and open the lid only after the rotor has slowed significantly</u>.
- Immediately open the lid of the centrifuge to remove the test tubes before they are damaged.
- <u>Wait for the rotor to come to a complete stop before attempting to remove your test tubes</u>.
- Reach out and carefully slow the rotor with your hand so you may finish the lab faster.
Explanation:
I UNDERLINED and BOLD the answers that apply
Answer:
28.6 kg
Explanation:
The final weight can be calculated from the mixing data and formulae which is given as follows:
cement content = ![\frac{water content}{water - cement ratio}](https://tex.z-dn.net/?f=%5Cfrac%7Bwater%20content%7D%7Bwater%20-%20cement%20ratio%7D)
Computing the parameters and checking the tables gives 28.6 kg.