Which of the following devices would you expect to consume the most energy for each hour that it operates?

John has an exhaust leak in his Acura Integra GS-R, What steps would he take to fix the leak in time for his inspection?

SCORPION-xisa [38]

Answer:

Explanation:

Fist you need to identify where the leak is coming from. You can do this by either listening for the leak or spraying soapy water on the exhaust to look for air bubbles coming out of the exhaust. Depending on the spot of the leak there are many ways you can fix this leak.

1. Exhaust clamp

2. Exhaust putty

3. Exhaust tape

4. New exhaust

Exhaust clamp is best used for holes on straight pipes.

Putty is best used on welds or small holes like on exhaust manifolds or welds connecting various pieces like catalytic converters, mufflers, or resonators.

Tape will work best on straight pipes with holes.

New exhaust is for when the thig is beyond repair, like rust.

Now good luck because working on exhausts is a pain.

4 0

2 years ago

Hot water at an average temperature of 88°C and an average velocity of 0.4 m/s is flowing through a 5-m section of a thin-walled

SIZIF [17.4K]

Answer:

a) The rate of heat transfer will be 19.58 Watts.

b) The temperature drop of the hot water will be 0.024 Degree Celcius.

Explanation:

4 0

3 years ago

Before finishing and installing a shelved cabinet you just constructed, you need to check the

Greeley [361]

Answer:

Carpenter's square

Explanation:

The most common hand tool used to measure or set angles with its application extending to setting angles of roofs and rafters. Another name of a Carpenter's square is a framing square.

Other hand tools that are used to measure angles are;

The combination square that allows a user to set both 90° and 45° angles
A Bevel that allows users to set any angle they like.
A Protractor that resembles a bevel but its marks are marked in an arc.
An electromagnetic angle finder which gives a reading according to the measure of the arms adjusted by the user.

7 0

3 years ago

It is said that Archimedes discovered his principle during a bath while thinking about how he could determine if KingHiero‘s cro

Rudiy27

Answer:

the crown is false densty= 12556kg/m^3[/tex]

Explanation:

Hello! The first step to solve this problem is to find the mass of the crown, this is found using the weight of the crown in the air by means of the equation for the weight.

W=mg

W=weight(N)=31.4N

M=Mass

g=gravity=9.81m/S^2

solving for M

m=W/g

$m=\frac{31.4N}{9.81m/S^2}=3.2kg$

The second step is find the volume of crown remembering that when an object is weighed in the water the result is the subtraction between the weight of the object and the buoyant force of the water which is the product of the volume of the crown by gravity by density of water

$F=mg-\alpha V g$

Where

F=weight in water=28.9N

m=mass of crown=3.2kg

g=gravity=9.81m/S^2

α=density of water=1000kg/m^3

V= crown´s volume

solving for V

$V=\frac{mg-F }{g \alpha } =\frac{(3.2)(9.81)-28.9}{9.81(1000)} =0.000254m^3$

finally, we remember that the density is equal to the index between mass and volume

$\alpha =\frac{m}{v} =\frac{3.2}{0.000254} =12556kg/m^3$

To determine the density of the crown without using the weight in the water and with a bucket we can use the following steps.

1.weigh the crown in the air and find the mass

2. put water in a cylindrical bucket and measure its height with a ruler.

3. Put the crown in the bucket and measure the new water level with a ruler.

4. Subtract the heights, and find the volume of a cylinder knowing the difference in heights and the diameter of the bucket, in order to determine the volume of the crown.

5. find density by dividing mass by volume

7 0

3 years ago

A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.6 mm; the spe

ankoles [38]

Answer:

F = 8849 N

Explanation:

Given:

Load at a given point = F = 4250 N

Support span = L = 44 mm

Radius = R = 5.6 mm

length thickness of tested material = 12 mm

First compute the flexural strength for circular cross section using the formula below:

σ $_{fs} = F_{f} L / \pi R^{3}$

σ = FL / π R³

Putting the given values in the above formula:

σ = 4250 ( 44 x 10⁻³ ) / π ( 5.6 x 10⁻³ ) ³

= 4250 ( 44 x 10⁻³ ) / 3.141593 ( 5.6 x 10⁻³ ) ³

= 4250 (44 x 1 /1000 )) / 3.141593 ( 5.6 x 10⁻³ ) ³

= 4250 ( 11 / 250 ) / 3.141593 ( 5.6 x 10⁻³ ) ³

= 187 / 3.141593 ( 5.6 x 1 / 1000 ) ³

= 187 / 3.141593 (0.0056)³

= 338943767.745358

= 338.943768 x 10⁶

σ = 338 x 10⁶ N/m²

Now we compute the load i.e. F from the following formula:

$F_{f}$ = 2 σ $_{fs}$ d³/3 L

F = 2σd³/3L

= 2(338 x 10⁶)(12 x 10⁻³)³ / 3(44 x 10⁻³)

= 2 ( 338 x 1000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

= 2 ( 338000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

= 676000000 ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

= 676000000 ( 12 x 1/1000 )³ / 3 ( 44 x 10⁻³)

= 676000000 ( 3 / 250 )³ / 3 ( 44 x 10⁻³)

= 676000000 ( 27 / 15625000 ) / 3 ( 44 x 10⁻³)

= 146016 / 125 / 3 ( 44 x 1 / 1000 )

= ( 146016 / 125 ) / (3 ( 11 / 250 ))

= 97344 / 11

F = 8849 N

4 0

3 years ago