1. What most commonly joins the structural panels of

Study the graphs below what type of trend and pattern in observed .

hichkok12 [17]

Answer:

negative and linear

Explanation:

its a straight line going down

8 0

3 years ago

Read 2 more answers

A Charpy V Notch test was conducted for an ASTM A572 Grade 50 bridge steel. The average values of the test results at four diffe

goldenfox [79]

Answer:

Explanation: see attachment below

5 0

4 years ago

A Pelton wheel is supplied with water from a lake at an elevation H above the turbine. The penstock that supplies the water to t

gayaneshka [121]

Answer:

Following are the proving to this question:

Explanation:

$\frac{D_1}{D} = \frac{1}{(2f(\frac{l}{D}))^{\frac{1}{4}}}$

using the energy equation for entry and exit value :

$\to \frac{p_o}{y} +\frac{V^{2}_{o}}{2g}+Z_0 = \frac{p_1}{y} +\frac{V^{2}_{1}}{2g}+Z_1+ f \frac{l}{D}\frac{V^{2}}{2g}$

where

$\to p_0=p_1=0\\\\\to Z_0=Z_1=H\\\\\to v_0=0\\\\AV =A_1V_1 \\\\\to V=(\frac{D_1}{D})^2 V_1\\\\\to V^2=(\frac{D_1}{D})^4 V^{2}_{1}$

$= (\frac{1}{(2f (\frac{l}{D} ))^{\frac{1}{4}}})^4\ V^{2}_{1}\\\\$

$= \frac{1}{(2f (\frac{l}{D}) )} \ V^{2}_{1}\\$

$\to \frac{p_o}{y} +\frac{V^{2}_{o}}{2g}+Z_0 =\frac{p_1}{y} +\frac{V^{2}_{1}}{2g}+Z_1+ f \frac{l}{D}\frac{V^{2}}{2g} \\\\$

$\to 0+0+Z_0 = 0 +\frac{V^{2}_{1} }{2g} +Z_1+ f \frac{l}{D} \frac{\frac{1}{(2f(\frac{l}{D}))}\ V^{2}_{1}}{2g} \\\\\to Z_0 -Z_1 = +\frac{V^{2}_{1}}{2g} \ (1+f\frac{l}{D}\frac{1}{(2f(\frac{l}{D}) )} ) \\\\\to H= \frac{V^{2}_{1}}{2g} (\frac{3}{2}) \\\\\to \frac{V^{2}_{1}}{2g} = H(\frac{3}{2})$

L.H.S = R.H.S

7 0

3 years ago

One good way to improve your gas Milage is to ___.

VashaNatasha [74]

Answer: B

Explanation:

One good way to improve your gas mileage is to accelerate smoothly and directly to a safe speed.

Hope this helps!

5 0

3 years ago

Read 2 more answers

The mean of hours that the average person watches television each day is 4.18 hours with a standard deviation of 1.19 hours. Fin

luda_lava [24]

Answer:

$z = \frac{3-4.18}{1.19}=-0.992$

$z = \frac{5-4.18}{1.19}=0.689$

And we can find this probability with this difference:

$P(-0.992$

And then we can conclude that the probability that someone watches between 3 and 5 hours a day is approximately 0.591 using a normal distribution

Explanation:

For this case we can define the random variable X as "hours that a person watches television". For this case we don't have the distribution for X but we have the following parameters:

$\mu = 4.18,\sigma =1.19$