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Sedbober [7]
3 years ago
8

While examining the color of an engine's coolant: Technician A says that because it is orange, the cooling system should be flus

hed and new coolant put into the system. Technician B says that if the coolant looks rusty, the cooling system should be flushed and new coolant put into the system. Who is correct?
Engineering
1 answer:
Stells [14]3 years ago
4 0

Answer:

Technician B is correct.

Explanation:

In every vehicles, coolants are very important as it keeps the engine cool and keep it running. The coolant serves the purpose of dissipating away the heat that is generated when the engine runs.

When the engine runs, the coolant passes all over the heating surfaces in the engine thus carrying away the heat along with it. With time the color of the coolant changes to rusty brown color when the coolant serves its purpose . It also washes away all the dirt particles from the engine part along with it. Thus the color changes and it is now can be flushed from the engine sump and fresh coolant can be put in to the engine.

Thus in the context, technician B is correct.

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Which regulations are related to guard rail height and dimensions and uniformity of stairs?
galina1969 [7]

Answer:

C.

structural safety

Explanation:

Guards protecting floor surfaces must be 36 inches in height, while guards for stairs must be 34 inches in height measured vertically from the tread nosing. A guard may also serve as the required handrail (34 to 38 inches high) provided the top rail meets the requirements for grip size.

4 0
3 years ago
Read 2 more answers
Which of the following are some of the problems found in city streets?
storchak [24]

Answer: drugs and rushing cars

Explanation: drug dealers are everywhere on city streets nowadays they have been killing young adults

rushing cars or reckless drivers cut curb fast and potentially someone can get hurt they are speeding and not worrying about other people lives at stake

4 0
3 years ago
Problem 34.3 The elevation of the end of the steel beam supported by a concrete floor is adjusted by means of the steel wedges E
Natasha2012 [34]

The image is missing, so i have attached it.

Answer:

A) P = 65.11 KN

B) Q = 30 KN

Explanation:

We are given;

The end reaction of the beam; F = 100KN

Coefficient of static friction between two steel surfaces;μ_ss = 0.3

Coefficient of static friction between steel and concrete;μ_sc = 0.6

So, F1 = μ_ss•F =0.3 x 100 = 30 KN

F2 = μ_ss•N_EF = 0.3N_EF

From the screen shot, we see that the angle is 12°

Sum of forces in the Y-direction gives;

F2•sin12 - N_EF•cos12 + 100 = 0

Rearranging gives;

N_EF•cos12 - F2•sin12 = 100

Let's put 0.3N_EF for F2 to give;

N_EF•cos12 - 0.3N_EF•sin12 = 100

Thus;

N_EF(0.9158) - 0.1247 = 100

N_EF(0.9781) = 100 + 0.1247

N_EF = 100.1247/0.9158

N_EF = 109.33 KN

Thus, F2 = 0.3N_EF = 0.3 x 109.33 = 32.8 KN

Wedge will move if;

P = (F1 + F2cos12 + N_EFsin12)

Thus;

P = 10 + (32.8 x 0.9781) + (109.33 x 0.2079)

P ≥ 65.11 KN

B) For static equilibrium, Q = F1

Thus, Q = 30 KN

3 0
4 years ago
Asolid rectangular rodhas a length of 90mm, made of steel material (E =207,000 MPa, Syield= 300 MPa), the cross section of the r
erastova [34]

Answer:

13.6mm

Explanation:

We consider diameter to be a chord that runs through the center point of the circle. It is considered as the longest possible chord of any circle. The center of a circle is the midpoint of its diameter. That is, it divides it into two equal parts, each of which is a radius of the circle. The radius is half the diameter.

See attachment for the step by step solution of the problem

3 0
3 years ago
A walrus loses heat by conduction through its blubber at the rate of 220 W when immersed in −1.00°C water. Its internal core tem
Llana [10]

Answer:

The average thickness of the blubber is<u> 0.077 m</u>

Explanation:

Here, we want to calculate the average thickness of the Walrus blubber.

We employ a mathematical formula to calculate this;

The rate of heat transfer(H) through the Walrus blubber = dQ/dT = KA(T2-T1)/L

Where dQ is the change in amount of heat transferred

dT is the temperature gradient(change in temperature) i.e T2-T1

dQ/dT = 220 W

K is the conductivity of fatty tissue without blood = 0.20 (J/s · m · °C)

A is the surface area which is 2.23 m^2

T2 = 37.0 °C

T1 = -1.0 °C

L is ?

We can rewrite the equation in terms of L as follows;

L × dQ/dT = KA(T2-T1)

L = KA(T2-T1) ÷ dQ/dT

Imputing the values listed above;

L = (0.2 * 2.23)(37-(-1))/220

L = (0.2 * 2.23 * 38)/220 = 16.948/220 = 0.077 m

7 0
4 years ago
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