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d1i1m1o1n [39]
2 years ago
10

What are the BENEFITS and RISKS of using automobiles?

Engineering
1 answer:
Alona [7]2 years ago
5 0

Answer:

7 benefits of a car: What you should know

Health and Emergencies. The safest way to ensure your health in the pandemic is by owning a private car. ...

Independence and Freedom. Relying on others for travel or using public vehicles could be detrimental. ...

Privacy. ...

Safety. ...

Save Time. ...

Take Pleasure Trips. ...

Flexibility.

Explanation:

The modern negative consequences of heavy automotive use include the use of non-renewable fuels, a dramatic increase in the rate of accidental death, the disconnection of local community, the decrease of local economy, the rise in obesity and cardiovascular diseases, the emission of air and noise pollution, the ...

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Can you prove that the two bleu areas are the same without numbers please?
Svet_ta [14]

Answer:

\small{\boxed{\tt{\colorbox{green}{✓Verified\:answer}}}}\:

Just draw a line from point D join to point E

The triangle formed DME will be congruent to AMC

6 0
3 years ago
Calculate the reluctance of a 4-meter long toroidal coil made of low-carbon steel with an inner radius of 1.75 cm and an outer r
My name is Ann [436]

Answer:

R = 31.9 x 10^(6) At/Wb

So option A is correct

Explanation:

Reluctance is obtained by dividing the length of the magnetic path L by the permeability times the cross-sectional area A

Thus; R = L/μA,

Now from the question,

L = 4m

r_1 = 1.75cm = 0.0175m

r_2 = 2.2cm = 0.022m

So Area will be A_2 - A_1

Thus = π(r_2)² - π(r_1)²

A = π(0.0225)² - π(0.0175)²

A = π[0.0002]

A = 6.28 x 10^(-4) m²

We are given that;

L = 4m

μ_steel = 2 x 10^(-4) Wb/At - m

Thus, reluctance is calculated as;

R = 4/(2 x 10^(-4) x 6.28x 10^(-4))

R = 0.319 x 10^(8) At/Wb

R = 31.9 x 10^(6) At/Wb

8 0
4 years ago
If a student doesn't major in Engineering as an undergraduate, They could still find a successful
34kurt

Answer:

True

Explanation:

It could either be true or false because you dont really have to be great at something just to do it, you could try new things too.

5 0
2 years ago
Read 2 more answers
If we have silicon at 300K with 10 microns of p-type doping of 4.48*10^18/cc and 10 microns of n-type doping at 1000 times less
liq [111]

Answer:

The resistance is 24.9 Ω

Explanation:

The resistivity is equal to:

R=\frac{1}{N_{o}*u*V } =\frac{1}{4.48x10^{15}*1500*106x10^{-19}  } =0.93ohm*cm

The area is:

A = 60 * 60 = 3600 um² = 0.36x10⁻⁴cm²

w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{A} }+\frac{1}{N_{D} })

If NA is greater, then, the term 1/NA can be neglected, thus the equation:

w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{D} })

Where

V = 0.44 V

E = 11.68*8.85x10¹⁴ f/cm

V_{o} =\frac{KT}{p} ln(\frac{N_{A}*N_{D}}{n_{i}^{2}  } , if n_{i}=1.5x10^{10}cm^{-3}  \\V_{o}=0.02585ln(\frac{4.48x10^{18}*4.48x10^{15}  }{(1.5x10^{10})^{2}  } )=0.83V

w=\sqrt{\frac{2*11.68*8.85x10^{-14}*(0.83-0.44) }{1.6x10^{-19}*4.48x10^{15}  } } =3.35x10^{-5} cm=0.335um

The length is:

L = 10 - 0.335 = 9.665 um

The resistance is:

Re=\frac{pL}{A} =\frac{0.93*9.665x10^{-4} }{0.36x10^{-4} } =24.9ohm

7 0
3 years ago
how are the construction steps for bisecting an angle similar to the steps you would have taken to construct and bisect a line s
GREYUIT [131]

Answer:

does the question entail anything else?

Explanation:

5 0
3 years ago
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