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adoni [48]
3 years ago
5

What are the disadvantages of having a liquid cooled engine?

Engineering
1 answer:
Feliz [49]3 years ago
5 0
One notable disadvantage of liquid cooling over air cooling is that it is considerably costly to set up. Cooling fans are prevalent in the market, and this overabundance of supply means they are cheap. The components of a liquid cooling system can be expensive.
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How long should the shafts remain in the furnace to achieve a desired centerline temperature of 800K? 2) Determine the temperatu
Phantasy [73]

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

6 0
3 years ago
g An analog voice signal, sampled at the rate of 8 kHz (8000 samples/second), is to be transmitted by using binary frequency shi
slamgirl [31]

Answer:

The module is why it’s goin to work

Explanation:

4 0
3 years ago
A bakery wants to determine how many trays of doughnuts it should prepare each day. Demand is normal with a mean of 15 trays and
shutvik [7]

The number of trays that should be prepared if the owner wants a service level of at least 95% is; 7 trays

<h3>How to utilize z-score statistics?</h3>

We are given;

Mean; μ = 15

Standard Deviation; σ = 5

We are told that the distribution of demand score is a bell shaped distribution that is a normal distribution.

Formula for z-score is;

z = (x' - μ)/σ

We want to find the value of x such that the probability is 0.95;

P(X > x) = P(z > (x - 15)/5) = 0.95

⇒ 1 -  P(z ≤ (x - 15)/5) = 0.95

Thus;

P(z ≤ (x - 15)/5) = 1 - 0.95

P(z ≤ (x - 15)/5) = 0.05

The value of z from the z-table of 0.05 is -1.645

Thus;

(x - 15)/5 = -1.645

x ≈ 7

Complete Question is;

A bakery wants to determine how many trays of doughnuts it should prepare each day. Demand is normal with a mean of 15 trays and standard deviation of 5 trays. If the owner wants a service level of at least 95%, how many trays should he prepare (rounded to the nearest whole tray)? Assume doughnuts have no salvage value after the day is complete. 6 5 4 7 unable to determine with the above information.

Read more about Z-score at; brainly.com/question/25638875

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4 0
2 years ago
How large a force is required to accelerate a 1300 kg car from rest to a speed of 20 m/s in a distance of 80 m?
topjm [15]

F=m*a

F=80*20

F =1600 ans"

7 0
3 years ago
Read 2 more answers
Primary Creep: slope (creep rate) decreases with time
Igoryamba

Answer:

true

Explanation:

Creep is known as the time dependent deformation of structure due to constant load acting on the body.

Creep is generally seen at high temperature.

Due to creep the length of the structure increases which is not fit for serviceability purpose.

When time passes structure gain strength as the structure strength increases with time so creep tends to decrease.

When we talk about Creep rate for new structure the creep will be more than the old structure i.e. the creep rate decreases with time.

5 0
3 years ago
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