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3 years ago

Find the value of P(-1.5≤Z≤2)

Engineering

1 answer:

Vladimir [108]3 years ago

7 0

Answer:

0.9104

Explanation:

Suitable technology can tell you the probability.

P(-1.5≤Z≤2) ≈ 0.9104

A phone app gives the probability as 0.9104426667829628.

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2 years ago

The advantages of solar cells include all of the following, except a.moderate net energy yield b.little or no direct emissions o

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3 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

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2 years ago

weight of 1000 pounds is suspended from two cables. The allowable stress in the cables is 1500 psi. Find the minimum diameter fo

kari74 [83]

Answer:

The minimum diameter for each cable should be 0.65 inches.

Explanation:

Since, the load is supported by two ropes and the allowable stress in each rope is 1500 psi. Therefore,

(1/2)(Weight/Cross Sectional Area) = Allowable Stress

Here,

Weight = 1000 lb

Cross-sectional area = πr²

where, r = minimum radius for each cable

(1/2)(1000 lb/πr²) = 1500 psi

500 lb/1500π psi = r²

r = √1.061 in²

r = 0.325 in

Now, for diameter:

Diameter = 2(radius) = 2r

Diameter = 2(0.325 in)

<u>Diameter = 0.65 in</u>

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3 years ago