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3 years ago

6

Need help fast I been stuck in this for the longest

1 answer:

gavmur [86]3 years ago

6 0

Answer:

3rd and 4rth

Explanation:

a geologist studies the earth and both of these have something to do with the earth

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Hi all, could you solve this please?<br> What is the value of the resistance R

Kazeer [188]

Answer:

try to 36v power and take 1a and intersect to 3

5 0

3 years ago

The assembly consists of two 10 mm diameter red brass C83400 coper rods AB and CD, a 15 mm diameter 304 stainless steel of EF an

My name is Ann [436]

Answer:

Therefore, the horizontal displacement of end F of rod EF is 0.4797 mm

Explanation:

solution is mentioned in steps.

6 0

3 years ago

Find the resultant of two forces 130 N and 110 N respectively, acting at an angle whose tangent

monitta

Answer:

$F_r = 200N$

Explanation:

Given

Let the two forces be

$F_1 = 130N$

$F_2 = 110N$

and

$\tan(\theta) = \frac{12}{5}$

Required

Determine the resultant force

Resultant force (Fr) is calculated using:

$F_r^2 = F_1^2 + F_2^2 + 2F_1F_2\cos(\theta)$

This means that we need to first calculate $\cos(\theta)$

Given that:

$\tan(\theta) = \frac{12}{5}$

In trigonometry:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

By comparing the above formula to $\tan(\theta) = \frac{12}{5}$

$Opposite = 12$

$Adjacent = 5$

The hypotenuse is calculated as thus:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 12^2 + 5^2$

$Hypotenuse^2 = 144 + 25$

$Hypotenuse^2 = 169$

$Hypotenuse = \sqrt{169$

$Hypotenuse = 13$

$\cos(\theta)$ is then calculated using:

$\cos(\theta)= \frac{Adjacent}{Hypotenuse}$

$\cos(\theta)= \frac{5}{13}$

Substitute values for $F_1$ , $F_2$ and $cos(\theta)$ in

$F_r^2 = F_1^2 + F_2^2 + 2F_1F_2\cos(\theta)$

$F_r^2 = 130^2 + 110^2 + 2*130*110*\frac{5}{13}$

$F_r^2 = 16900 + 12100 + 11000$

$F_r^2 = 40000$

Take square roots of both sides

$F_r = \sqrt{40000$

$F_r = 200N$

<em>Hence, the resultant force is 200N</em>

4 0

3 years ago

If the feedforward path of a control system contains at least one integrating element, then the output continues to change as lo

Thepotemich [5.8K]

Answer:

The attached system shows that there’s an integrator between the point where disturbance enters the system and error measuring element. A any time when R(s)=0 then

$\frac {C(s)}{D(s)}=\frac {G(s)}{1+G_c(s)G(s)}$ and considering that $E(s)=D(s)-G_c(s)C(s)$ then

$\frac {E(s)}{D(s)}=1-(\frac {C(s)}{D(s)})G_c(s)$

$\frac {E(s)}{D(s)}=1-(\frac {G(s)}{1+G_c(s)D(s)})G_c(s)$

$\frac {E(s)}{D(s)}=\frac {1}{1+G_c(s)G(s)}$

$E(s)=\frac {D(s)}{1+G_c(s)G(s)}$

For ramp disturbance d(t)=at

$D(s)=\frac {a}{s^{2}}$ therefore, the steady state error is given by

$e(\infty)= \lim_{s \to 0} s E(s)$

$e(\infty)= \lim_{s \to 0} s [\frac {D(s)}{1+G_c(s)G(s)}]$

$e(\infty)= \lim_{s \to 0} s [\frac {a}{s^{2}+s^{2}G_c(s)G(s)}]$

$e(\infty)= \lim_{s \to 0} s [\frac {a}{s+sG_c(s)G(s)}]$

$e(\infty)= \lim_{s \to 0} s [\frac {a}{sG_c(s)G(s)}]$

Whenever $G_c(s)$ has a double intergrator, the error $e(\infty)$ becomes zero

3 0

3 years ago

On the position time curve, if the slope of a tangent at a point is positive, that means:

Gnoma [55]

Answer:

C. the object is moving forward

Explanation:

A positive slope means position is increasing when time is increasing. Generally, increasing position is "moving forward."

7 0

3 years ago

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