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3 years ago

Why do some ladders have a white tip or reflective tape attached to the tip?

Engineering

1 answer:

MrMuchimi3 years ago

3 0

To increase visibility when you find yourself in dark conditions when it’s hard to see.

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Steel balls 12 mm in diameter are annealed by heating to 1150 K and then slowly cooling to 400 K in an air environment for which

luda_lava [24]

Answer:

It will take the steel 425.925sec to cool from 1150k to 400k

Explanation:

Detailed explanation and calculation is shown in the image

6 0

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

7 0

3 years ago

TOPO NÀO CÓ CẤU HÌNH ĐA ĐIỂM

bixtya [17]

Answer:

Tout à fait les gens sont nuls

3 0

4 years ago

Read 2 more answers

A small probe P is gently forced against the circular surface with a vertical force F as shown . Determine the n- and t-componen

Rina8888 [55]

This question is incomplete, its missing an image which will be uploaded along this Answer.

Answer:

the normal component of force F_n is F((√(r²-s²)) / r)

the tangential component of force F_t is F(s/r)

Explanation:

Given the data in the image;

from the free body diagram, we write the expression for ∅

sin∅ = s/r

cos∅ = (√(r²-s²)) / r

now expression for normal component of force is;

F_n = Fcos∅

we substitute

F_n = F((√(r²-s²)) / r)

Therefore, the normal component of force F_n is F((√(r²-s²)) / r)

Also for force F_t

F_t = Fsin∅

we substitute

F_t = F(s/r)

Therefore, the tangential component of force F_t is F(s/r)

6 0

3 years ago

If d=0.25m and D=0.40m. Assume headloss from the contraction to the end of the pipe can be found as hų = 0.9 (V is velocity in t

pychu [463]

Answer:

Height=14.25 m

Discharge= $0.8155 m^{3}/s$

Explanation:

Bernoulli’s equation for the two points, let’s say A and B is given by

$\frac {P_A}{\rho g}+\frac {(v_A)^{2}}{2g}+z_A=\frac {P_B}{\rho g}+\frac {(v_B)^{2}}{2g}+z_B+h_{L(A-B)}$

But since the elevations are the same then

$\frac {P_A}{\rho g}+\frac {(v_A)^{2}}{2g}=\frac {P_B}{\rho g}+\frac {(v_B)^{2}}{2g}+h_{L(A-B)}$

Since $h_{L(A-B)}=\frac {0.9v^{2}}{2g}$ then

$\frac {P_A}{\rho g}+\frac {(v_A)^{2}}{2g}=\frac {100}{\rho 9.81}+\frac {(v_B)^{2}}{2g}+\frac {0.9v^{2}}{2g}$

From continuity equation

$A_AV_A=A_BV_B=0.25\pi D^{2}V_A=0.25\pi d^{2}V_B hence [tex]D^{2}V_A= d^{2}V_B$ and substituting 0.4 for D and 0.25 for d

$V_B=\frac { D^{2}V_A}{d^{2}}=\frac { 0.4^{2}V_A}{0.25^{2}}=2.56V_A$

$\frac {P_A}{\rho g}=\frac {100}{\rho 9.81}+\frac {(v_B)^{2}}{2g}+\frac {0.9v^{2}}{2g}-\frac {(v_A)^{2}}{2g}$ and substituting $V_B$ with $2.56V_A$ we have