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3 years ago

A task is something that you can physically do.

Engineering

2 answers:

Jet001 [13]3 years ago

5 0

The answer is T because a task is getting something done

mr Goodwill [35]3 years ago

5 0

Answer:

false

some task can we done mentally

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What is the basic concept of the finite element method.

KIM [24]

Answer:

The FEM is a general numerical method for solving partial differential equations in two or three space variables (i.e., some boundary value problems). To solve a problem, the FEM subdivides a large system into smaller, simpler parts that are called finite elements.

5 0

2 years ago

In successive an object moves from start position, then moves 1ft, 4ft and 8ft. This is an example of a. non-uniform motion b. u

Nikolay [14]

Answer:Non-uniform motion

Explanation:

This is an example of Non-uniform motion because unequal distance traveled by an object in equal interval of time is termed as Non-Uniform motion and here also the distance traveled by object is 1 ft ,4 ft and 8 ft which is different from each other.

In Uniform motion distance traveled is equal in equal interval of time .

6 0

3 years ago

: A freeway exit ramp has a single lane and consist of entirely of a horizontal curve with a central angle of 90 degree and a le

Mariana [72]

The design speed was used for the freeway exit ramp is 11 mph.

<h3>Design speed used in the exit ramp</h3>

The design speed used in the exit ramp is calculated as follows;

f = v²/15R - 0.01e

where;

v is designated speed

v = ωr

v = (θ/t) r

θ = 90⁰ = 1.57 rad

v = (1.57 x 19.4)/2.5 s

v = 12.18 ft/s = 8.3 mph

<h3>Design speed</h3>

f = v²/15R - 0.01e

let the maximum superelevation, e = 1%

f = (8.3)²/(15 x 19.4) - 0.01

f = 0.22

0.22 is less than value of f which is 0.4

<h3>next iteration, try 10 mph</h3>

f = (10)²/(15 x 19.4) - 0.01

f = 0.33

0.33 is less than 0.4

<h3>next iteration, try 11 mph</h3>

f = (11)²/(15 x 19.4) - 0.01

f = 0.4

Thus, the design speed was used for the freeway exit ramp is 11 mph.

Learn more about design speed here: brainly.com/question/22279858

#SPJ1

4 0

2 years ago

The volume of 1.5 kg of helium in a frictionless piston-cylinder device is initially 6 m3. Now, helium is compressed to 2 m3 whi

coldgirl [10]

Answer:

The initial temperature will be "385.1°K" as well as final will be "128.3°K".

Explanation:

The given values are:

Helium's initial volume, v₁ = 6 m³

Mass, m = 1.5 kg

Final volume, v₂ = 2 m³

Pressure, P = 200 kPa

As we know,

Work, $W=p(v_{2}-v_{1})$

On putting the estimated values, we get

⇒ $=200000(2-6)$

⇒ $=200000\times (-4)$

⇒ $=800,000 \ N.m$

Now,

Gas ideal equation will be:

⇒ $pv_{1}=mRT_{1}$

On putting the values. we get

⇒ $200000\times 6=1.5\times 2077\times T_{1}$

⇒ $T_{1}=\frac{1200000}{3115.5}$

⇒ $=385.1^{\circ}K$ (Initial temperature of helium)

and,

⇒ $pv_{2}=mRT_{2}$

On putting the values, we get

⇒ $200000\times 2=1.5\times 2077\times T_{2}$

⇒ $T_{2}=\frac{400000}{3115.5}$

⇒ $=128.3^{\circ}K$ (Final temperature of helium)

3 0

3 years ago

The phasor technique makes it pretty easy to combine several sinusoidal functions into a single sinusoidal expression without us

devlian [24]

Answer:

The phasor technique can't be applied directly in the following cases:

a) 45 sin(2500t – 50°) + 20 cos(1500t +20°)

b) 100 cos(500t +40°) + 50 sin(500t – 120°) – 120 cos(500t + 60°)

c) -100 sin(10,000t +90°) + 40 sin(10, 100t – 80°) + 80 cos(10,000t)

d) 75 cos(8t+40°) + 75 sin(8t+10°) – 75 cos(8t + 160°)

Explanation:

For a) and c), it is not possible to use the phasor technique, due this technique is only possible when the sinusoidal signals to be combined are all of the same frequency.

This is due to the vector representing a signal is showed as a fixed vector in the graph( which magnitude is equal to the amplitude of the sinusoid and his angle is the phase angle with respect to cos (ωt)), which is rotating at an angular speed equal to the angular frequency of the sinusoidal signal that represents, like a radius that shows a point rotating in a circular uniform movement.

This rotating vector represents a sinusoidal signal, in the form of a cosine (as the real part of the complex function $e^{j(wt+\alpha)}$ ), so it is not possible to combine with functions expressed as a sine, even though both have the same frequency.

If we look at the graphs of cos (ωt) and sin (ωt) we can say that the sin lags the cos in 90º, so we can say the following:

sin (ωt) = cos (ωt-90º)

This means that in order to be able to represent a sine function as a cosine, we need to rotate it 90º in the plane clockwise.

This is the reason why before doing this transformation, it is not possible to use the phasor technique for b) and d).

8 0

3 years ago