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Gennadij [26K]
2 years ago
10

16. Driverless cars have already , and they look so cool.

Engineering
2 answers:
mash [69]2 years ago
8 0

Answer:C.exist

hope it helps

Gekata [30.6K]2 years ago
4 0
Answer:
C. Exist
Hope it helps!
You might be interested in
If you were driving the blue Prius in the situation pictured above, explain why the red Mustang should be given right-of-way at
dexar [7]

Answer:

The blue Prius, because the Mustang arrived almost in the same time. And when you arrive in an intersection at the same time of other vehicle you need yield for the car on your right if the car is on your left you have the right of way.

Explanation:

4 0
2 years ago
Initially when 1000.00 mL of water at 10oC are poured into a glass cylinder, the height of the water column is 1000.00 mm. The w
Dafna11 [192]

Answer:

\mathbf{h_2 =1021.9 \  mm}

Explanation:

Given that :

The initial volume of water V_1 = 1000.00 mL = 1000000 mm³

The initial temperature of the water  T_1 = 10° C

The height of the water column h = 1000.00 mm

The final temperature of the water T_2 = 70° C

The coefficient of thermal expansion for the glass is  ∝ = 3.8*10^{-6 } mm/mm  \ per ^oC

The objective is to determine the the depth of the water column

In order to do that we will need to determine the volume of the water.

We obtain the data for physical properties of water at standard sea level atmospheric from pressure tables; So:

At temperature T_1 = 10 ^ 0C  the density of the water is \rho = 999.7 \ kg/m^3

At temperature T_2 = 70^0 C  the density of the water is \rho = 977.8 \ kg/m^3

The mass of the water is  \rho V = \rho _1 V_1 = \rho _2 V_2

Thus; we can say \rho _1 V_1 = \rho _2 V_2;

⇒ 999.7 \ kg/m^3*1000 \ mL = 977.8 \ kg/m^3 *V_2

V_2 = \dfrac{999.7 \ kg/m^3*1000 \ mL}{977.8 \ kg/m^3 }

V_2 = 1022.40 \ mL

v_2 = 1022400 \ mm^3

Thus, the volume of the water after heating to a required temperature of  70^0C is 1022400 mm³

However; taking an integral look at this process; the volume of the water before heating can be deduced by the relation:

V_1 = A_1 *h_1

The area of the water before heating is:

A_1 = \dfrac{V_1}{h_1}

A_1 = \dfrac{1000000}{1000}

A_1 = 1000 \ mm^2

The area of the heated water is :

A_2 = A_1 (1  + \Delta t  \alpha )^2

A_2 = A_1 (1  + (T_2-T_1) \alpha )^2

A_2 = 1000 (1  + (70-10) 3.8*10^{-6} )^2

A_2 = 1000.5 \ mm^2

Finally, the depth of the heated hot water is:

h_2 = \dfrac{V_2}{A_2}

h_2 = \dfrac{1022400}{1000.5}

\mathbf{h_2 =1021.9 \  mm}

Hence the depth of the heated hot  water is \mathbf{h_2 =1021.9 \  mm}

4 0
3 years ago
Dean is buying a home for $170,000. The mortgage company he decided to use to finance the home requires a 10% down payment. What
ANTONII [103]
They answer is 3. $17,00
8 0
2 years ago
Read 2 more answers
Preferential stress will cause crystals to rotate into a preferred orientation unless the crystals are ________
velikii [3]

Answer:

Preferential stress will cause crystals to rotate into a preferred orientation unless the crystals are Equant

6 0
1 year ago
A 35 ft simply supported beam is loaded with concentrated loads 15 ft in from each support. On one end, the dead load is 8 kips
agasfer [191]

Answer: ASD = 306 kips-ft

               LRSD = 1387.5 k-ft

Explanation:

To begin, we will take a step by step process to solving this problem.

Attached below is a picture to guide us to solving this.

To begin, we have that to reaction of the support

ΣMд = 0

where;

RB * 35 - (8+18)15 - (4+9)20 = 0

RB = 18.57k

also Ey = 0;

RA + RB = 18 + 8 + 9 +4 = 20.43 k

taking the maximum moment at mid point;

Mc = RA * 35/2 - (8 +18) (35/2 -15)

Mc = 292.525

therefore, MD = RA * 15 = 20.43 * 15 = 306.45 k.ft

MD = 306.45 k.ft

ME = 279 k.ft i.e 18.57 * 15

considering the unsupported  length; 35 - (15*2 = 5ft

now we have that;

Lb = Lp = 5ft

where Lp = 1.76 ry(√e/fy)

Lp = 1.76 ry √29000/50 ......

ry = 1.4 inch

so we have that Mr = Mp for Lb = Lp where

Mp = 2 Fy ≤ 1.5 sx Fy

Recall from the expression,

RA + RB = (8+4) * 1.2 + (18+9) * 1.6 = 57.6

RA * 35 = 4 * 1.2 * 15 + 9 *1.6 * 15 + 8 * 1.2 * 20 + 18 * 1.6 * 20

RA = 30.17 k

the maximum moment at D = 30.17 * 15 = 452.55 k.ft

Zrequired = MD/Fy = 452.55 * 12 / 50 = 108.61 inch³

so we have Sx = 452.55 * 12 / 1.5 * 50 = 72.4 inch³

also r = 1.41 in

Taking LRFD solution:

where the design strength ∅Mn = 0.9 * Zx * Fy

given r = 2.97

Zx = 370 and Sx = 81.5, we have

∅Mn = 0.9 * 370 * 50 = 16650 k-inch = 1387.5 k-ft

this tells us it is safe.

ASD solution:

for Lb = Lp, and where Mn = Mp = Fcr Sx

we already have value for Sx as 81.5 so

Fcr = ZxFy/Sx

Fcr = 370 * 50 / 81.5 = 227 ksi

considering the strength;

Strength = Mn / Ωb = (0.6 * 81.5 * 50) * (1.5) / 12 = 306 kips-ft

This justifies that it is safe because is less than 306

cheers i hope this helps.

4 0
2 years ago
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