Answer:
The volume is decreasing at 160 cm³/min
Explanation:
Given;
Boyle's law, PV = C
where;
P is pressure of the gas
V is volume of the gas
C is constant
Differentiate this equation using product rule:
![V\frac{dp}{dt} +P\frac{dv}{dt} = \frac{d(C)}{dt}](https://tex.z-dn.net/?f=V%5Cfrac%7Bdp%7D%7Bdt%7D%20%2BP%5Cfrac%7Bdv%7D%7Bdt%7D%20%3D%20%5Cfrac%7Bd%28C%29%7D%7Bdt%7D)
Given;
(increasing pressure rate of the gas) = 40 kPa/min
V (volume of the gas) = 600 cm³
P (pressure of the gas) = 150 kPa
Substitute in these values in the differential equation above and calculate the rate at which the volume is decreasing (
);
(600 x 40) + (150 x
) = 0
![\frac{dv}{dt} = -\frac{(600*40)}{150} = -160 \ cm^3/min](https://tex.z-dn.net/?f=%5Cfrac%7Bdv%7D%7Bdt%7D%20%3D%20-%5Cfrac%7B%28600%2A40%29%7D%7B150%7D%20%3D%20-160%20%5C%20cm%5E3%2Fmin)
Therefore, the volume is decreasing at 160 cm³/min
Answer:
Following are the solution to this question:
Explanation:
Law:
![\to \theta= 180^{\circ}- 50^{\circ}- 25^{\circ}](https://tex.z-dn.net/?f=%5Cto%20%5Ctheta%3D%20180%5E%7B%5Ccirc%7D-%2050%5E%7B%5Ccirc%7D-%2025%5E%7B%5Ccirc%7D)
![= 180^{\circ}- 75^{\circ}\\\\= 105^{\circ}](https://tex.z-dn.net/?f=%3D%20180%5E%7B%5Ccirc%7D-%2075%5E%7B%5Ccirc%7D%5C%5C%5C%5C%3D%20105%5E%7B%5Ccirc%7D)
![\to \frac{T_{L}}{\sin (90+50)}= \frac{T_{R}}{\sin (25+90)}=\frac{1400}{\sin (105)}](https://tex.z-dn.net/?f=%5Cto%20%5Cfrac%7BT_%7BL%7D%7D%7B%5Csin%20%2890%2B50%29%7D%3D%20%5Cfrac%7BT_%7BR%7D%7D%7B%5Csin%20%2825%2B90%29%7D%3D%5Cfrac%7B1400%7D%7B%5Csin%20%28105%29%7D)
![\to T_L=931.65 \ pounds \\\\ \to T_R=1313.59 \ pounds \\\\](https://tex.z-dn.net/?f=%5Cto%20T_L%3D931.65%20%5C%20pounds%20%5C%5C%5C%5C%20%5Cto%20T_R%3D1313.59%20%5C%20pounds%20%5C%5C%5C%5C)
The acceleration due to gravity
Brainliest please