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3 years ago

Need help ASAP!!!! here are the options: Amplitude, Compression, Rarefaction, and Wavelength

Physics

1 answer:

Strike441 [17]3 years ago

6 0

Amplitude is the pair of vertical buttons, so to speak. Compressions are the bunched up vertical lines with the purple arrows pointing left and right. Rarefactions are purple arrows pointing down. Wavelength is crest to crest purple buttons. Associated LH and RH pointing arrows.

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USPshnik [31]

Nabr is the answer bcoz in a chemical reaction reactants react to form new substances and according to your options NaBr should be the answer

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3 years ago

Do you wont to be friends

Angelina_Jolie [31]

Answer:

shore

Explanation:

8 0

4 years ago

The tape in a videotape cassette has a total

xxMikexx [17]

Answer:

$w=19.76 \ rad/s$

Explanation:

Circular Motion

Suppose there is an object describing a circle of radius r around a fixed point. If the relation between the angle of rotation by the time taken is constant, then the angular speed is also constant. If that relation increases or decreases at a constant rate, the angular speed is given by:

$w=w_o+\alpha \ t$

Where $\alpha$ is the angular acceleration and t is the time. If the object was instantly released from the circular path, it would have a tangent speed of:

$\displaystyle v_t=w.r$

We have two reels: one loaded with the tape to play and the other one empty and starting to fill with tape. They both rotate at different angular speeds, one is increasing and the other is decreasing as the tape goes from one to the other. We'll assume the tangent speed is constant for both (so the tape can play correctly). Let's call $w_1$ the angular speed of the loaded reel and $w_2$ that from the empty reel. We have

$w_1=w_{o1}+\alpha_1 \ t$

$w_2=w_{o2}+\alpha_2 \ t$

If $r_f=35\ mm=0.035\ m$ is the radius of the reel when it's full of tape, the angular speed for the loaded reel is computed by

$\displaystyle w_{01}=\frac{v_t}{r_f}$

The tangent speed is computed by knowing the length of the tape and the time needed to fully play it.

$t=1.8\ h=1.8*3600=6480\ sec$

$\displaystyle v_t=\frac{x}{t}=\frac{249\ m}{6480\ sec}=0.0384 \ m/s$

$\displaystyle w_{01}=\frac{0.0384}{0.035}=1,098 \ rad/s$

If $r_e=10\ mm=0.001\ m$ is the radius of the reel when it's empty, the angular speed for the empty reel is computed by

$\displaystyle w_{02}=\frac{0.0384}{0.001}=38.426 \ rad/s$

The full reel goes from $w_{01}$ to $w_{02}$ in 6480 seconds, so we can compute the angular acceleration:

$\displaystyle \alpha_1=\frac{w_{02}-w_{01}}{6480}$

$\displaystyle \alpha_1=\frac{38.426-1.098}{6480}=0.00576 \ rad/sec^2$

The empty reel goes from $w_{02}$ to $w_{01}$ in 6480 seconds, so we can compute the angular acceleration:

$\displaystyle \alpha_2=\frac{1.098-38.426}{6480}=-0.00576 \ rad/sec^2$

So the equations for both reels are

$w_1=1.098+0.00576 \ t$

$w_2=38.426-0.00576 \ t$

They will be the same when

$1.098+0.00576 \ t=38.426-0.00576 \ t$

Solving for t

$\displaystyle t=\frac{38.426-1.098}{0.0115}$

$t=3240 \ sec$

The common angular speed is

$w_1=1.098+0.00576 \ 3240=19.76 \ rad/s$

$w_2=38.426-0.00576 \ 3240=19.76 \ rad/s$

They both result in the same, as expected

$\boxed{w=19.76 \ rad/s}$

6 0

4 years ago

A retaining wall is 5m long (into the plane of the paper), and it is supported by an anchor rod. The soil it supports has a weig

vekshin1

Knowing that the greatest horizontal pressure is the pressure made by the material on the wall -considering for that the total height of the wall-, then P = 4.375 t/m

---------------------

Available data:

The wall is 5m long into the plane ⇒ H

The soil density is 1.4 metric tonnes per cubic metre ⇒ γ

K = 0.25 ⇒ Coefficient of earth pressure. Depends on the angle.

From this information, we need to calculate the greatest horizontal pressure.

Horizontal pressure refers to the force made by the supported soil on the retaining wall that tends to deflect the wall outward.

We can calculate horizontal pressure at different heights from the top. And since we need to calculate the greatest horizontal pressure, we need to consider the total height that equals the wall height, H.

To do it, we will use the following formula, and replace the terms.

P = [ k γ H² ] / 2

P = [ 0.25 x 1.4 t/m³ x 5m² ] / 2

P = 8.75/2

P = 4.375 t/m

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