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3 years ago

I'm so confused to this question What is the difference between Science and Engineering?

Engineering

2 answers:

madreJ [45]3 years ago

5 0

Answer:

<u>Generally, Science is the study of the physical world, while Engineering applies scientific knowledge to design processes, structures or equipment.</u>

muminat3 years ago

4 0

Explanation:

Science is the body of knowledge that explores the physical and natural world. Engineering is the application of knowledge in order to design, build and maintain a product or a process that solves a problem and fulfills a need (i.e. a technology).

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Compared to 15 mph on a dry road, about how much longer will it take for

Marysya12 [62]

Answer:

8 to 10 times

Explanation:

For dry road

u= 15 mph ( 1 mph = 0.44 m/s)

u= 6.7 m/s

Let take coefficient of friction( μ) of dry road is 0.7

So the de acceleration a = μ g

a= 0.7 x 10 m/s ² ( g=10 m/s ²)

a= 7 m/s ²

We know that

v= u - a t

Final speed ,v=0

0 = 6.7 - 7 x t

t= 0.95 s

For snow road

μ = 0.4

de acceleration a = μ g

a = 0.4 x 10 = 4 m/s ²

u= 30 mph= 13.41 m/s

v= u - a t

Final speed ,v=0

0 = 30 - 4 x t'

t'=7.5 s

t'=7.8 t

We can say that it will take 8 to 10 times more time as compare to dry road for stopping the vehicle.

8 to 10 times

7 0

3 years ago

Read 2 more answers

Can you help me with this

Fiesta28 [93]

the function is to provide sealed combustion so that the loss of gas is minimized

6 0

2 years ago

Identify which sound type each line contains.

nydimaria [60]

Answer:i can not see it

Explanation:

4 0

3 years ago

Find a negative feedback controller with at least two tunable gains that (1) results in zero steady state error when the input i

IceJOKER [234]

Answer:

Gc(s) = $\frac{0.1s + 0.28727}{s}$

Explanation:

comparing the standard approximation with the plot attached we can tune the PI gains so that the desired response is obtained. this is because the time requirement of the setting is met while the %OS requirement is not achieved instead a 12% OS is seen from the plot.

attached is the detailed solution and the plot in Matlab

8 0

3 years ago

You are evaluating the lifetime of a turbine blade. The blade is 4 cm long and there is a gap of 0.16 cm between the tip of the

Tcecarenko [31]

Answer:

Explanation:

Given conditions

1)The stress on the blade is 100 MPa

2)The yield strength of the blade is 175 MPa

3)The Young’s modulus for the blade is 50 GPa

4)The strain contributed by the primary creep regime (not including the initial elastic strain) was 0.25 % or 0.0025 strain, and this strain was realized in the first 4 hours.

5)The temperature of the blade is 800°C.

6)The formula for the creep rate in the steady-state regime is dε /dt = 1 x 10-5 σ4 exp (-2 eV/kT)

where: dε /dt is in cm/cm-hr σ is in MPa T is in Kelvink = 8.62 x 10-5 eV/K

Young Modulus, E = Stress, $\sigma$ /Strain, ∈

initial Strain, $\epsilon_i = \frac{\sigma}{E}$

$\epsilon_i = \frac{100\times 10^{6} Pa}{50\times 10^{9} Pa}$

$\epsilon_i = 0.002$

creep rate in the steady state

$\frac{\delta \epsilon}{\delta t} = (1 \times {10}^{-5})\sigma^4 exp^(\frac{-2eV}{kT} )$

$\frac{\epsilon_{initial} - \epsilon _{primary}}{t_{initial}-t_{final}} = 1 \times 10^{-5}(100)^{4}exp(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )(800+273)K} )$

but Tinitial = 0

$\epsilon_{initial} - \epsilon _{primary}} = 0.002 - 0.003 = -0.001$

$\frac{-0.001}{-t_{final}} = 1 \times 10^{-5}(100)^{4}\times 10^{(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )1073K} )}$

solving the above equation,

we get

Tfinal = 2459.82 hr

3 0

3 years ago

I'm so confused to this question What is the difference between Science and Engineering? ​

I'm so confused to this question What is the difference between Science and Engineering?