Why do engineers need to be particularly aware of the impact of hubris?

10. An engineer is designing a total hip implant. She intends to make the femoral stem out of titanium because it forms a good i

creativ13 [48]

Answer:

Yes. She should be worried about corrosion. The 18-8 stainless exhibits intergranular corrosion due to high (0.08%) carbon content and gross pitting due to low molybdenum content.

Explanation: lol

8 0

3 years ago

The human eye, as well as the light-sensitive chemicals on color photographic film, respond differently to light sources with di

jeka57 [31]

Answer:

a) at T = 5800 k

band emission = 0.2261

at T = 2900 k

band emission = 0.0442

b) daylight (d) = 0.50 μm

Incandescent ( i ) = 1 μm

Explanation:

To Calculate the band emission fractions we will apply the Wien's displacement Law

The ban emission fraction in spectral range λ1 to λ2 at a blackbody temperature T can be expressed as

F ( λ1 - λ2, T ) = F( 0 ----> λ2,T) - F( 0 ----> λ1,T )

Values are gotten from the table named: blackbody radiation functions

a) Calculate the band emission fractions for the visible region

at T = 5800 k

band emission = 0.2261

at T = 2900 k

band emission = 0.0442

attached below is a detailed solution to the problem

b)calculate wavelength corresponding to the maximum spectral intensity

For daylight ( d ) = 2898 μm *k / 5800 k = 0.50 μm

For Incandescent ( i ) = 2898 μm *k / 2900 k = 1 μm

3 0

2 years ago

If a heat engine has an efficiency of 30% and its power output is 600 W, what is the rate of heat input from the combustion phas

jarptica [38.1K]

Answer:

The heat input from the combustion phase is 2000 watts.

Explanation:

The energy efficiency of the heat engine ( $\eta$ ), no unit, is defined by this formula:

$\eta = \frac{\dot W}{\dot Q}$ (1)

Where:

$\dot Q$ - Heat input, in watts.

$\dot W$ - Power output, in watts.

If we know that $\dot W = 600\,W$ and $\eta = 0.3$ , then the heat input from the combustion phase is:

$\eta = \frac{\dot W}{\dot Q}$

$\dot Q = \frac{\dot W}{\eta}$

$\dot Q = \frac{600\,W}{0.3}$

$\dot Q = 2000\,W$

The heat input from the combustion phase is 2000 watts.

8 0

2 years ago

An automobile weighing 2500 lbf increases its gravitational potential energy by a magnitude of 2.25 × 104 Btu in going from an e

Mila [183]

Answer:

The elevation at the high point of the road is 12186.5 in ft.

Explanation:

The automobile weight is 2500 lbf.

The automobile increases its gravitational potential energy in $2.25 * 10^4 BTU$ . It means the mobile has increased its elevation.

The initial elevation is of 5183 ft.

The first step is to convert Btu of potential energy to adequate units to work with data previously presented.

British Thermal Unit - $1 BTU = 778.17 lbf*ft$

$2.25 * 10^4 BTU (\frac{778.17 lbf*ft}{1BTU} ) = 1.75 * 10^7 lbf * ft$

Now we have the gravitational potential energy in lbf*ft. Weight of the mobile is in lbf and the elevation is in ft. We can evaluate the expression for gravitational potential energy as follows:

$Ep = m*g*(h_2 - h_1)\\ W = m*g$

Where m is the mass of the automobile, g is the gravity, W is the weight of the automobile showed in the problem.

$h_2$ is the final elevation and $h_1$ is the initial elevation.

Replacing W in the Ep equation

$Ep = W*(h_2 -h_1)\\(h_2 -h_1) = \frac{Ep}{W} \\h_2 = h_1 + \frac{Ep}{W}\\\\$

Finally, the next step is to replace the variables of the problem.

$h_2 = 5183 ft + \frac{1.75 * 10^7 lbf*ft}{2500 lbf}\\h_2 = 5183 ft + 70003.5 ft\\h_2 = 12186.5 ft$

The elevation at the high point of the road is 12186.5 in ft.

3 0

2 years ago

Steep safety ramps are built beside mountain highway to enable vehichles with fedective brakes to stop safely. a truck enters a

Veronika [31]

Answer:

a. 6 seconds

b. 180 feet

Explanation:

Images attached to show working.

a. You have the position of the truck so you integrate twice. Use the formula and plug in the time t = 7 sec. Check out uniform acceleration. The time at which the truck's velocity is zero is when it stops.

b. Determine the initial speed. Plug in the time calculated in the previous step. From this we can observe that the truck comes to a stop before the end of the ramp.

7 0

2 years ago