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hoa [83]
4 years ago
9

What website provides access to information used to diagnose vehicles?

Engineering
1 answer:
Sergeeva-Olga [200]4 years ago
5 0

Answer:

thedrive.com

Explanation:

You might be interested in
On a cold winter day, wind at 55 km/hr is blowing parallel to a 4-m high and 10-m long wall of a house. If the air outside is at
koban [17]

Answer:

16.21 kW

Explanation:

Solution

Given that,

The velocity of wind = 55 km/hr

The length of the wall L = 10m

The height of the wall w = 4m

The surface temperature at wall Ts = 12° C

Temperature of air T∞ = 5°C

Now,

The properties  of the air at atm and average film temperature =( 12 + 5)/2 = 8.5°C, which is taken from the air table properties.

k= 0.02428 W/m°C

v= 1.413 *10 ^⁻5

Pr =0.7340

Now,

Recall Reynolds number when air flow parallel to 10 m side

[ 55 * 1000/3600) m/s (10 m)/1.413 *10^⁻5 m²/s

Rel =1.081 * 10⁷

This value is greater than Reynolds number.

The nusselt number is computed as follows:

Nu =hL/k

(0.037Rel^0.08 - 871)Pr^1/3

Nu =1.336 * 10 ^4

The heat transfer coefficient is

h = k/L Nu

= 0.2428 W/m°C /10 m (1.336 * 10 ^4)

h = 32.43 W/m°C

The heat transfer area of surface,

As = 40 m²

= ( 4 m) (10 m)

As = 40 m²

The rate of heat transfer is determined as follows:

Q = hAs( Ts - T∞)

= (32.43 W/m²°C) (40 m) (12 - 5)°C

=9081 W

Q = 9.08 kW

When the velocity is doubled,

let say V = 110km/hr

The Reynolds number is

Rel = VL/v

= [110 * 100/3600) m/s] (10 m)/ 1.413 *10^⁻5 m²/s

Rel = 2.163 * 10 ^7

This value is greater for critical Reynolds number

The nusselt number is computed as follows:

Nu =hL/k

(0.037Rel^0.08 - 871)Pr^1/3

[0.037 ( 2.163 * 10 ^7)^0.08 - 871] (0.7340)^1/3

Nu =2.344 * 10^4

The heat transfer coefficient is

h = k/L Nu

= 0.2428 W/m°C /10 m (2.384 * 10 ^4)

h= 57.88 W/m²°C

The heat transfer area of surface,

As =  wL

= ( 10 m) (4 m)

As = 40 m²

he rate of heat transfer is determined as follows:

Q = hAs( Ts - T∞)

= (57.88 W/m²°C) (40 m²) (12 - 5)°C

= 16,207 W

= 16.21 kW

3 0
4 years ago
A large class with 1,000 students took a quiz consisting of ten questions. To get an A, students needed to get 9 or 10 questions
VMariaS [17]

Answer:

a. 0.11

b. 110 students

c. 50 students

d. 0.46

e. 460 students

f. 540 students

g. 0.96

Explanation:

(See attachment below)

a. Probability that a student got an A

To get an A, the student needs to get 9 or 10 questions right.

That means we want P(X≥9);

P(X>9) = P(9)+P(10)

= 0.06+0.05=0.11

b. How many students got an A on the quiz

Total students = 1000

Probability of getting A = 0.11 ---- Calculated from (a)

Number of students = 0.11 * 1000

Number of students = 110 students

So,the number of students that got A is 110

c. How many students did not miss a single question

For a student not to miss a single question, then that student scores a total of 10 out of possible 10

P(10) = 0.05

Total Students = 1000

Number of Students = 0.05 * 1000

Number of Students = 50 students

We see that 5

d. Probability that a student pass the quiz

To pass, a student needed to get at least 6 questions right.

So we want P(X>=6);

P(X>=) =P(6)+P(7)+P(8)+P(9)+P(10)

=0.08+0.12+0.15+0.06+0.05=0.46

So, the probability of a student passing the quiz is 0.46

e. Number of students that pass the quiz

Total students = 1000

Probability of passing the quiz = 0.46 ----- Calculated from (d)

Number of students = 0.46 * 1000

Number of students = 460 students

So,the number of students that passed the test is 460

f. Number of students that failed the quiz

Total students = 1000

Total students that passed = 460 ----- Calculated from (e)

Number of students that failed = 1000 - 460

Number of students that failed = 540

So,the number of students that failed is 540

g. Probability that a student got at least one question right

This means that we want to solve for P(X>=1)

Using the complement rule,

P(X>=1) = 1 - P(X<1)

P(X>=1) = 1 - P(X=0)

P(X>=1) = 1 - 0.04

P(X>=1) = 0.96

7 0
3 years ago
State five applications of thermochromic materials
rusak2 [61]

Explanation:

The end-use industries of thermochromic materials include packaging, printing & coating, medical, textile, industrial, and others. Printing & coating is the fastest-growing end-use industry of thermochromic materials owing to a significant increase in the demand for thermal paper for POS systems. The use of thermochromic materials is gaining momentum for interactive packaging that encourages consumers to take a product off the shelf and use it.

8 0
3 years ago
Out
olchik [2.2K]

ave you ever seen a Rube Goldberg machine in action? You probably have, even if you didn’t know what it was. A Rube Goldberg machine is a contraption that uses a chain reaction to carry out a simple task. It performs a very basic job in a complicated way.

7 0
3 years ago
Come and look on my attachment​
CaHeK987 [17]

Crazy Guy what do uh mean ?

4 0
3 years ago
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