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3 years ago

A___ remote control can be an advantage to an

Engineering

2 answers:

KATRIN_1 [288]3 years ago

8 0

Answer:

Wireless

Explanation:

Elina [12.6K]3 years ago

8 0

I think the correct choice would me wireless

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A gas tank is known to have a thickness of 0.5 inches and an internal pressure of 2.2 ksi. Assuming that the maximum allowable s

sergiy2304 [10]

Answer:

$D_o=11.9inch$

Explanation:

From the question we are told that:

Thickness $T=0.5$

Internal Pressure $P=2.2Ksi$

Shear stress $\sigma=12ksi$

Elastic modulus $\gamma= 35000$

Generally the equation for shear stress is mathematically given by

$\sigma=\frac{P*r_1}{2*t}$

Where

r_i=internal Radius

Therefore

$12=\frac{2.2*r_1}{2*0.5}$

$r_i=5.45$

Generally

$r_o=r_1+t$

$r_o=5.45+0.5$

$r_o=5.95$

Generally the equation for outer diameter is mathematically given by

$D_o=2r_o$

$D_o=11.9inch$

Therefore

Assuming that the thin cylinder is subjected to integral Pressure

Outer Diameter is

$D_o=11.9inch$

7 0

3 years ago

Problem 9.11 A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain f

BabaBlast [244]

Answer:

the critical flaw length is 10.06 mm

Explanation:

Given the data in the question;

plane strain fracture toughness $K_{tc$ = 92 Mpa√m

yield strength σ $_y$ = 900 Mpa

design stress is one-half of the yield strength ( 900 Mpa / 2 ) 450 Mpa

Y = 1.15

we know that;

Critical crack length $a_c$ = 1/π( $K_{tc$ / Yσ )²

we substitute

$a_c$ = 1/π( 92 Mpa√m / (1.15 × 450 Mpa )²

$a_c$ = 1/π( 92 Mpa√m / (517.5 Mpa )²

$a_c$ = 1/π( 0.177777 )²

$a_c$ = 1/π( 0.03160466 )

$a_c$ = 0.01006 m = 10.06 mm

Therefore, the critical flaw length is 10.06 mm

{ $a_c$ = ( 10.06 mm ) > 3 mm

The critical flow is subject to detection

5 0

3 years ago

An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. A

irina [24]

An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile at a constant acceleration of 1.96 m/s2 . The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s2

Find the total time required for the police car to over take the automobile.

Answer:

15.02 sec

Explanation:

The total time required for the police car to overtake the automobile is related to the distance covered by both cars which is equal from instant point of abreast.

So; we can say :

$D_{pursuit} =D_{police}$

By using the second equation of motion to find the distance S;

$S= ut + \dfrac{1}{2}at^2$

$D_{pursuit} = (15.65 *12 )+(15.65 (t)+ (\dfrac{1}{2}*(-3.05)t^2)$

$D_{pursuit} = (187.8)+(15.65 \ t)-0.5*(3.05)t^2)$

$D_{pursuit} = (187.8+15.65 \ t-1.525 t^2)$

$D_{police} = ut _P + \dfrac{1}{2}at_p^2$

where ;

u = 0

$D_{police} = \dfrac{1}{2}at_p^2$

$D_{police} = \dfrac{1}{2}*(1.96)*(t+12)^2$

$D_{police} = 0.98*(t+12)^2$

$D_{police} = 0.98*(t^2 + 144 + 24t)$

$D_{police} = 0.98t^2 + 141.12 + 23.52t$

Recall that:

$D_{pursuit} =D_{police}$

$(187.8+15.65 \ t-1.525 t^2)= 0.98t^2 + 141.12 + 23.52t$

$(187.8 - 141.12) + (15.65 \ t - 23.52t) -( 1.525 t^2 - 0.98t^2) = 0$

= 46.68 - 7.85 t -2.505 t² = 0

Solving by using quadratic equation;

t = -6.16 OR t = 3.02

Since we can only take consideration of the value with a positive integer only; then t = 3.02 secs

From the question; The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit;

Therefore ; the total time required for the police car to over take the automobile = 12 s + 3.02 s

Total time required for the police car to over take the automobile = 15.02 sec

8 0

2 years ago

Make Routh tables and tell how many roots of the following two polynomials are in the right half plane and in the left half plan

12345 [234]

Answer:

P(s) = s5 + 3s4 + 5s3 + 4s2 + s

For above Polynomial, there are two RHP Poles & Three LHP Poles.

& for the Second Polynomial,

P(s)=3s7 + 9s6 + 6s5 + 4s4 + 7s3 + 8s2 + 2s + 6

For this Polynomial, there are Four RHP Poles & Three LHP Poles.

Explanation:

As the equation is given by for the first polynomial,

P(s)=s5 + 3s4 + 5s3 + 4s2 + s

The Routh table is attached in the attachment, & from that table we can see that there are two sign changes for thuis first ploynomial. Moreover, there are two RHP poles & Three LHP Poles.

Similarly the equation for the second polynomial is given by as,

P(s)=3s7 + 9s6 + 6s5 + 4s4 + 7s3 + 8s2 + 2s + 6

The Routh Table is attached in the same attachment below in the 2nd slide & from this table we can see that there are four sign changes for this polynomial while it includes Four RHP Poles & three LHP Poles.

Download pptx

3 0

3 years ago

The water level in a tank z1, is 16 m above the ground. A hose is connected to the bottom of the tank, and the nozzle at the end

Volgvan

Answer:

40.7 m

Explanation:

Let point 1 represent the surface of the water, point 2 be the top of the water trajectory and the reference be the bottom of the tank. Hence:

$z_1=16\ m,P_1=2\ atm,V_1=0(velocity\ at\ the \ surface \ of\ water\ is\ low)\\V_2=0,P_2=P\\\\Using\ Bernoulli\ equation:\\\\\frac{P_1}{\rho g} +\frac{V_1^2}{2g}+z_1= \frac{P_2}{\rho g} +\frac{V_2^2}{2g}+z_2\\\\Sine\ V_1=0,V_2=0:\\\\\frac{P_1}{\rho g} +z_1=\frac{P}{\rho g} +z_2\\\\z_2=\frac{P_1}{\rho g} -\frac{P}{\rho g} +z_1\\\\z_2=\frac{P_1-P}{\rho g} +z_1\\\\z_2=\frac{P_{1.gage}}{\rho g} +z_1\\\\$

$z_2=\frac{2\ atm}{1000\ kg/m^3*9.81\ m/s^2}(\frac{101325\ N/m^2}{1\ atm} )(\frac{1\ kg.m/s^2}{1 \ N} ) +20\\\\z_2=20.7+20\\\\z_2=40.7\ m$

7 0

3 years ago