Answer:
did you ever get the answer
Explanation:
The given data is as follows.
= 57 kg,
= 79 kg
= 6.5 m,
= (6.5 - 1.9) m = 4.6 m
(a) The sum of torque ends about far end is as follows.
= 0
= 0
T = 828 N
Therefore, 828 N is the tension in the cable closer to the painter.
(b) Now, we will calculate the sum about close ends as follows.
= 0
T= 506 N
Therefore, 506 N is the tension in the cable further from the painter.
A wave can be described as the disturbance of particles in an area. Think about it this way: particles (matter) carry energy. For all the laws of physics to work, this energy must be "traded" somehow. This happens by miniscule vibrations in the particles, which are apparent disturbances. This creates a wave, and therefore a wave is, indeed, a disturbance.<span />
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
The answer is b/ cope a small section word-for-word