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1 year ago

Which of the following courses might a mechanical engineer take in college?

Physics

1 answer:

timofeeve [1]1 year ago

6 0

The course to be likely taken by a mechanical engineer in college is agriculture operations (option A).

<h3>What is engineering?</h3>

Engineering is application of mathematics and the physical sciences to the needs of humanity and the development of technology.

Mechanical engineering is a subfield of engineering concerned with designing and building machines and mechanical systems.

A mechanical engineer is an individual who specialises in the field of mechanical engineering. This means that a mechanical engineer is expected to be vast in the knowledge of designing and building machines for operations in other fields.

One of those fields that a mechanical engineer can build or design machine for is agriculture. Therefore, agriculture operations is the most likely course to be taken by a mechanical engineer in college.

Learn more about mechanical engineering at: brainly.com/question/20434227

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Answer:

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3 years ago

A freely suspended magnet does not point exactly at the geographical N-S direction

elena-s [515]

Because their is nothing at the geographical poles that attracts the magnet

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4 years ago

55. (III) A uniform circular plate of radius 2R has a circular hole of radius R cut out of it. The center of the smaller circle

Aleksandr [31]

Answer:

P = 0.27R from the center

Explanation:

Given,

The radius of the uniform circular plate, R = 2R

The radius of the hole, r = R

The center of the smaller circle from the center is, d = 0.8R

The center of mass of a circular disc with a hole in it given by the formula

P = dr²/R² - r²

Where P is the distance from the center of mass located in the line joining the two centers opposite to the hole.

Substituting the given values in the above equation,

P = 0.8R x R² / 4R² - R²

= 0.27R³/R²

= 0.27R

Hence the center of mass of plate is at a distant P = 0.27R from the center

6 0

3 years ago

A flat roof is very susceptible to wind damage during a thunderstorm and/or tornado. If a flat roof has an area of 500 m2 and wi

Evgesh-ka [11]

Answer: The magnitude of the force exerted on the roof is 490522.5 N.

Explanation:

The given data is as follows.

Below the roof, $v_{1}$ = 0 m/s

At top of the roof, $v_{2}$ = 39 m/s

We assume that $P_{1}$ is the pressure at lower surface of the roof and $P_{2}$ be the pressure at upper surface of the roof.

Now, according to Bernoulli's theorem,

$P_{1} + 0.5 \times \rho \times v^{2}_{1} = P_{2} \times 0.5 \rho \times v^{2}_{2}$

$P_{1} - P_{2} = 0.5 \times \rho \times (v^{2}_{2} - v^{2}_{1})$

= $0.5 \times 1.29 \times [(39)^{2} - (0)^{2}]$

= $0.645 \times 1521$

= 981.045 Pa

Formula for net upward force of air exerted on the roof is as follows.

F = $(P_{1} - P_{2})A$

= $981.045 \times 500$

= 490522.5 N

Therefore, we can conclude that the magnitude of the force exerted on the roof is 490522.5 N.

5 0

3 years ago

How long must a simple pendulum be if it is to make exactly six swings per second? (that is, one complete vibration takes exactl

MA_775_DIABLO [31]

The period of a simple pendulum is given by:
$T=2 \pi \sqrt{ \frac{L}{g} }$
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The pendulum in our problem makes one complete vibration in 0.333 s, so its period is T=0.333 s. Using this information, we can re-arrange the previous formula to find the length of the pendulum, L:
$L= g \frac{T^2}{(2 \pi)^2}=(9.81 m/s^2) \frac{(0.333 s)^2}{(2 \pi)^2}=0.028 m$

6 0

3 years ago