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3 years ago

Which type of engineer is needed in the following scenario?

Engineering

2 answers:

pickupchik [31]3 years ago

6 0

Answer:

Civil Engineer

Explanation:

Hope this helps!!! C:

Scorpion4ik [409]3 years ago

4 0

Answer:

civil engineer

Explanation:

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A water jet strikes normal to a fixed plate. If diameter of the outlet of the nozzle is 8 cm,and velocity of water at the outlet

zysi [14]

Answer:

The force required to keep the plate in place is 251.36 N

Explanation:

Force = pressure × area

velocity of water = 10 m/s

height of water = v^2/2g = 10^2/2×9.8 = 5.102 m

pressure = height × rho × g = 5.102 × 1000 × 9.8 = 49999.6 N/m^2

diameter = 8 cm = 8/100 = 0.08 m

area = πd^2/4 = 3.142×0.08^2/4 = 5.0272×10^-3 m^2

Force = 49999.6 N/m^2 × 5.0272×10^-3 m^2 = 251.36 N

8 0

3 years ago

The internal loadings at a critical section along the steel drive shaft of a ship are calculated to be a torque of 2300 lb⋅ft, a

Arturiano [62]

Answer:

Explanation:

Given that:

Torque T = 2300 lb - ft

Bending moment M = 1500 lb - ft

axial thrust P = 2500 lb

yield points for tension σY= 100 ksi

yield points for shear τY = 50 ksi

Using maximum-shear-stress theory

$\sigma_A = \dfrac{P}{A}+\dfrac{Mc}{I}$

where;

$A = \pi c^2$

$I = \dfrac{\pi}{4}c^4$

$\sigma_A = \dfrac{P}{\pi c^2}+\dfrac{Mc}{ \dfrac{\pi}{4}c^4}$

$\sigma_A = \dfrac{2500}{\pi c^2}+\dfrac{1500*12c}{ \dfrac{\pi}{4}c^4}$

$\sigma_A = \dfrac{2500}{\pi c^2}+\dfrac{72000c}{\pi c^3}}$

$\tau_A = \dfrac{T_c}{\tau}$

where;

$\tau = \dfrac{\pi c^4}{2}$

$\tau_A = \dfrac{T_c}{\dfrac{\pi c^4}{2}}$

$\tau_A = \dfrac{2300*12 c}{\dfrac{\pi c^4}{2}}$

$\tau_A = \dfrac{55200 }{\pi c^3}}$

$\sigma_{1,2} = \dfrac{\sigma_x+\sigma_y}{2} \pm \sqrt{\dfrac{(\sigma_x - \sigma_y)^2}{2}+ \tau_y^2}$

$\sigma_{1,2} = \dfrac{2500+72000}{2 \pi c ^3} \pm \sqrt{\dfrac{(2500 +72000)^2}{2 \pi c^3}+ \dfrac{55200}{\pi c^3}} \ \ \ \ \ ------(1)$

Let say :

$|\sigma_1 - \sigma_2| = \sigma_y$

Then :

$2\sqrt{( \dfrac{2500c + 72000}{2 \pi c^3})^2+ ( \dfrac{55200}{\pi c^3})^2 } = 100(10^3)$

$(2500 c + 72000)^2 +(110400)^2 = 10000*10^6 \pi^2 c^6$

$6.25c^2 + 360c+ 17372.16-10,000\ \pi^2 c^6 =0$

According to trial and error;

c = 0.75057 in

Replacing c into equation (1)

$\sigma_{1,2} = \dfrac{2500+72000}{2 \pi (0.75057) ^3} \pm \sqrt{\dfrac{(2500 +72000)^2}{2 \pi (0.75057)^3}+ \dfrac{55200}{\pi (0.75057)^3}}$

$\sigma_{1,2} = \dfrac{2500+72000}{2 \pi (0.75057) ^3} + \sqrt{\dfrac{(2500 +72000)^2}{2 \pi (0.75057)^3}+ \dfrac{55200}{\pi (0.75057)^3}} \ \ \ OR \\ \\ \\ \sigma_{1,2} = \dfrac{2500+72000}{2 \pi (0.75057) ^3} - \sqrt{\dfrac{(2500 +72000)^2}{2 \pi (0.75057)^3}+ \dfrac{55200}{\pi (0.75057)^3}}$

$\sigma _1 = 22193 \ Psi$

$\sigma_2 = -77807 \ Psi$

The required diameter d = 2c

d = 1.50 in or 0.125 ft

6 0

3 years ago

A commuter train traveling at 50 mi/h is 3 mi from a station. The train then decelerates so that its speed is 15 mi/h when it is

jonny [76]

Answer:

a) $t = 277.477\,s\,(4.625\min)$ , b) $v_{f} = 0\,\frac{mi}{h}$ , c) $a = -0.128\,\frac{ft}{s^{2}}$

Explanation:

a) The deceleration experimented by the commuter train in the first 2.5 miles is:

$a=\frac{[(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}-[(50\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}}{2\cdot (2.5\,mi)\cdot (\frac{5280\,ft}{1\,mi} )}$

$a = -0.185\,\frac{ft}{s^{2}}$

The time required to travel is:

$t = \frac{(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,fi} )\cdot(\frac{1\,h}{3600\,s} )-(50\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,fi} )\cdot(\frac{1\,h}{3600\,s} )}{-0.185\,\frac{ft}{s^{2}} }$

$t = 277.477\,s\,(4.625\min)$

b) The commuter train must stop when it reaches the station to receive passengers. Hence, speed of train must be $v_{f} = 0\,\frac{mi}{h}$ .

c) The final constant deceleration is:

$a = \frac{(0\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot(\frac{1\,h}{3600\,s} )-(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot(\frac{1\,h}{3600\,s} )}{(2.875\,min)\cdot (\frac{60\,s}{1\,min} )}$