Explanation:
Sorry I don't know the answer
Answer:
Negative intrapleural pressure is the correct answer
Explanation:
Intrapleural pressure is more subatmospheric in the uppermost part of the thorax than in the lowermost parts in the standing horse.
Air moves from a region of higher pressure to one of lower pressure. Therefore, for air to be moved into or out of the lungs, a pressure difference between the atmosphere and the alveoli must be established. If there is no pressure difference, no airflow will occur.
Under normal circumstances, inspiration is accomplished by causing alveolar pressure to fall below atmospheric pressure. When the mechanics of breathing are being discussed, atmospheric pressure is conventionally referred to as 0 cm H2O, so lowering alveolar pressure below atmospheric pressure is known as negative-pressure breathing.
Answer:(a) 4775.2Hz (b) 4.06m/s (c) 19382.15m/s²
Explanation: Given that the frequency of oscilation f, is 760Hz and the maximum displacement x, is 0.85mm= 0.00085m
(a) Angular frequency w= 2πf
w= 2π × 760 = 4775.2Hz
(b) Maximum speed v is given as the product of angular frequency and maximum displacement
V=wx
V= 4775.2 × 0.00085
V= 4.06m/s
(c) The maximum acceleration a
= w²x
= (4775.2)² × (0.00085)
a= 19382.15m/s².
Answer:
Explanation:
Given
Diameter of Pulley=10.4 cm
mass of Pulley(m)=2.3 kg
mass of book![(m_0)=1.7 kg](https://tex.z-dn.net/?f=%28m_0%29%3D1.7%20kg)
height(h)=1 m
time taken=0.64 s
![h=ut+frac{at^2}{2}](https://tex.z-dn.net/?f=h%3Dut%2Bfrac%7Bat%5E2%7D%7B2%7D)
![1=0+\frac{a(0.64)^2}{2}](https://tex.z-dn.net/?f=1%3D0%2B%5Cfrac%7Ba%280.64%29%5E2%7D%7B2%7D)
![a=4.88 m/s^2and [tex]a=\alpha r](https://tex.z-dn.net/?f=a%3D4.88%20m%2Fs%5E2%3C%2Fp%3E%3Cp%3Eand%20%5Btex%5Da%3D%5Calpha%20r)
where
is angular acceleration of pulley
![4.88=\alpha \times 5.2\times 10^{-2}](https://tex.z-dn.net/?f=4.88%3D%5Calpha%20%5Ctimes%205.2%5Ctimes%2010%5E%7B-2%7D)
![\alpha =93.84 rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D93.84%20rad%2Fs%5E2)
And Tension in Rope
![T=m(g-a)](https://tex.z-dn.net/?f=T%3Dm%28g-a%29)
![T=1.7\times (9.8-4.88)](https://tex.z-dn.net/?f=T%3D1.7%5Ctimes%20%289.8-4.88%29)
T=8.364 N
and Tension will provide Torque
![T\times r=I\cdot \alpha](https://tex.z-dn.net/?f=T%5Ctimes%20r%3DI%5Ccdot%20%5Calpha%20)
![8.364\times 5.2\times 10^{-2}=I\times 93.84](https://tex.z-dn.net/?f=8.364%5Ctimes%205.2%5Ctimes%2010%5E%7B-2%7D%3DI%5Ctimes%2093.84)
![I=0.463\times 10^{-2} kg-m^2](https://tex.z-dn.net/?f=I%3D0.463%5Ctimes%2010%5E%7B-2%7D%20kg-m%5E2)
![I_{original}=\frac{mr^2}{2}=0.31\times 10^{-2}kg-m^2](https://tex.z-dn.net/?f=I_%7Boriginal%7D%3D%5Cfrac%7Bmr%5E2%7D%7B2%7D%3D0.31%5Ctimes%2010%5E%7B-2%7Dkg-m%5E2)
Thus mass is uniformly distributed or some more towards periphery of Pulley