Answer:
26.466cm³/min
Explanation:
Given:
Volume 'V'= 320cm³
P= 95kPa
dP/dt = -11 kPa/minute
pressure P and volume V are related by the equation
P
=C
we need to find dV/dt, so we will differentiate the above equation
![V^{1.4} \frac{dP}{dt} + P\frac{d[V^{1.4} ]}{dt} = \frac{d[C]}{dt}](https://tex.z-dn.net/?f=V%5E%7B1.4%7D%20%5Cfrac%7BdP%7D%7Bdt%7D%20%2B%20P%5Cfrac%7Bd%5BV%5E%7B1.4%7D%20%5D%7D%7Bdt%7D%20%20%3D%20%5Cfrac%7Bd%5BC%5D%7D%7Bdt%7D)
lets solve for dV/dt, we will have
(plugged in all the values at the instant)
= 26.466
Therefore, the volume increasing at the rate of 26.466cm³/min at this instant
It i believe it would be 7.3 × 10∧3.
correct me if im wrong
Answer:
In my mind when I hear the word momentum it comes that it is the product of mass and velocity.
Answer:
230.53°
Explanation:
α = angle in rad / t^2
t =2mins = 2 × 60 = 120sec.
α = angle in rad / 2^2
α = angle in rad / 16
15+4(2) = angle in rad /4
23 = angle/4
23× 4 = angle in radian
92 = angle in radian
But π rad = 180°
1 rad = 180 /π
Therefore
92 rad = 92 ×180/π = 5270.53°
This value is far more so we subtract from 360° continuously till we get a value less than 360°
5270.53- (360×14) =5270.53- 5040
230.53°
Answer:
C. 5.35 × 10^-5
Explanation:
5.35 × 0.00001 = 0.0000535
Hope this helped!
