Answer:
a) ![\mathbf{\sigma _ 1 = 4800 psi}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Csigma%20_%201%20%3D%204800%20psi%7D)
![\mathbf{ \sigma _2 = 0}](https://tex.z-dn.net/?f=%5Cmathbf%7B%20%5Csigma%20_2%20%3D%200%7D)
b)![\mathbf{\sigma _ 1 = 6000 psi}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Csigma%20_%201%20%3D%206000%20psi%7D)
![\mathbf{ \sigma _2 = 3000 psi}](https://tex.z-dn.net/?f=%5Cmathbf%7B%20%5Csigma%20_2%20%3D%203000%20psi%7D)
Explanation:
Given that:
diameter d = 12 in
thickness t = 0.25 in
the radius = d/2 = 12 / 2 = 6 in
r/t = 6/0.25 = 24
24 > 10
Using the thin wall cylinder formula;
The valve A is opened and the flowing water has a pressure P of 200 psi.
So;
![\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}](https://tex.z-dn.net/?f=%5Csigma_%7Bhoop%7D%20%3D%20%5Csigma%20_%201%20%3D%20%5Cfrac%7BPd%7D%7B2t%7D)
![\sigma_{long} = \sigma _2 = 0](https://tex.z-dn.net/?f=%5Csigma_%7Blong%7D%20%3D%20%5Csigma%20_2%20%3D%200)
![\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{200(12)}{2(0.25)}](https://tex.z-dn.net/?f=%5Csigma%20_%201%20%3D%20%5Cfrac%7BPd%7D%7B2t%7D%20%5C%5C%20%5C%5C%20%5Csigma%20_%201%20%3D%20%5Cfrac%7B200%2812%29%7D%7B2%280.25%29%7D)
![\mathbf{\sigma _ 1 = 4800 psi}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Csigma%20_%201%20%3D%204800%20psi%7D)
b)The valve A is closed and the water pressure P is 250 psi.
where P = 250 psi
![\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}](https://tex.z-dn.net/?f=%5Csigma_%7Bhoop%7D%20%3D%20%5Csigma%20_%201%20%3D%20%5Cfrac%7BPd%7D%7B2t%7D)
![\sigma_{long} = \sigma _2 = \frac{Pd}{4t}](https://tex.z-dn.net/?f=%5Csigma_%7Blong%7D%20%3D%20%5Csigma%20_2%20%3D%20%5Cfrac%7BPd%7D%7B4t%7D)
![\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{250*(12)}{2(0.25)}](https://tex.z-dn.net/?f=%5Csigma%20_%201%20%3D%20%5Cfrac%7BPd%7D%7B2t%7D%20%5C%5C%20%5C%5C%20%5Csigma%20_%201%20%3D%20%5Cfrac%7B250%2A%2812%29%7D%7B2%280.25%29%7D)
![\mathbf{\sigma _ 1 = 6000 psi}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Csigma%20_%201%20%3D%206000%20psi%7D)
![\sigma _2 = \frac{Pd}{4t} \\ \\ \sigma _2 = \frac{250(12)}{4(0.25)}](https://tex.z-dn.net/?f=%5Csigma%20_2%20%3D%20%5Cfrac%7BPd%7D%7B4t%7D%20%5C%5C%20%5C%5C%20%20%5Csigma%20_2%20%3D%20%5Cfrac%7B250%2812%29%7D%7B4%280.25%29%7D)
![\mathbf{ \sigma _2 = 3000 psi}](https://tex.z-dn.net/?f=%5Cmathbf%7B%20%5Csigma%20_2%20%3D%203000%20psi%7D)
The free flow body diagram showing the state of stress on a volume element located on the wall at point B is attached in the diagram below
Answer: Some activities that I do in my daily life that require energy are:
1. Doing ballet
2. Studying
3. Walking up and down stairs
4. Stretching
5. Running on the treadmill
Hope this helps! :)
Explanation:
Answer:
a) v = +/- 0.323 m/s
b) x = -0.080134 m
c) v = +/- 1.004 m/s
Explanation:
Given:
a = - (0.1 + sin(x/b))
b = 0.8
v = 1 m/s @ x = 0
Find:
(a) the velocity of the particle when x = -1 m
(b) the position where the velocity is maximum
(c) the maximum velocity.
Solution:
- We will compute the velocity by integrating a by dt.
a = v*dv / dx = - (0.1 + sin(x/0.8))
- Separate variables:
v*dv = - (0.1 + sin(x/0.8)) . dx
-Integrate from v = 1 m/s @ x = 0:
0.5(v^2) = - (0.1x - 0.8cos(x/0.8)) - 0.8 + 0.5
0.5v^2 = 0.8cos(x/0.8) - 0.1x - 0.3
- Evaluate @ x = -1
0.5v^2 = 0.8 cos(-1/0.8) + 0.1 -0.3
v = sqrt (0.104516)
v = +/- 0.323 m/s
- v = v_max when a = 0:
-0.1 = sin(x/0.8)
x = -0.8*0.1002
x = -0.080134 m
- Hence,
v^2 = 1.6 cos(-0.080134/0.8) -0.6 -0.2*-0.080134
v = sqrt (0.504)
v = +/- 1.004 m/s
Answer:
HIGH from the supply voltage
LOW from ground
Explanation:
The answer depends on the kind of system and the purpose of the signal. But for practical reasons, in a DIGITAL system where 5V is HIGH and 0 V is LOW, 5 volts can be taken from the supply voltage (usually the same as high, BUT must be verified), and the LOW signal from ground.
If the user has a multimeter, it must be set to continuous voltage on 0 to 20 V range. Then place the probe in the ground of the circuit (must be a big copper area). Finally leave one probe in the circuit ground and place the other probe in some test points to identify 5 v.
Answer:
peech synthesis is being used in programs where oral communication is the only means by which information can be received, while speech recognition is facilitating commu- nication between humans and computers, whereby the acoustic voice signals changes in the sequence of words making up a written text.
Explanation: