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Bond [772]
3 years ago
13

En una barra de 6m que se utiliza como palanca se coloca el fulcro a 2 m de distancia del extremo derecho, como se muestra en la

figura. En ese mismo extremo se requiere soportar una carga de 90N. Despreciando el peso de la palanca, determinada Fp, VMI, VMR y la eficiencia del sistema.
Physics
1 answer:
Anna007 [38]3 years ago
4 0

Answer:

Fp = 45 N

VMR = 2

VMI = 2

Efficiency = 100%

Explanation:

English Translation

In a 6m bar that is used as a lever, the fulcrum is placed 2m away from the right end, as shown in the figure. At that same end it is required to support a load of 90N. Neglecting lever weight, determined Fp, VMI, VMR and system efficiency

Solution

With the diagram not massively needed to solve this question.

Although, the load of 90 N is placed at the right end of the bar, the fulcrum is placed 2 m from that right end and the balancing force or effort is placed at the extreme most left end of the 6 m bar.

Taking moment about the fulcrum,

The sum of clockwise moments must balance the sum of anti-clockwise moments.

(Load) × (Distance of load from the fulcrum) = (Balancing force) × (Distance of balancing force from fulcrum)

Load = L = 90 N

Distance of load from the fulcrum = 2 m

Balancing force or Effort = Fp = ?

Distance of balancing force from fulcrum = 6 - 2 = 4 m

90 × 2 = Fp × 4

Fp = (180/4) = 45 N

The mechanical advantages are them given as

VMR = (Load)/(Effort) = (90/45) = 2

VMI = (Distance of Effort from the fulcrum)/(Distance of Load from fulcrum) = (4/2) = 2

Efficiency is then given as the VMR divided by VMI in percentage terms

Efficiency = 100% × (VMR/VMI)

Efficiency = 100% × (2/2) = 100%

Hope this Helps!!!

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brainly an andean condor with a wingspan and a mass soars along a horizontal path. model its wings as a rectangle with a width.
Anna35 [415]

The difference between the pressure at the top surfaces of the condor's wings and the pressure at the bottom surfaces is 101,204 Pa.

<h3>Difference in pressure between the top and bottom of the wingspan</h3>

The difference in pressure between the top and bottom of the wingspan is calculated as follows;

ΔP = P(top) - P(bottom)

<h3>Area of the wingspan</h3>

A = bh

A = 2.7 m x 0.27 m

A = 0.729 m²

<h3>Weight of the Andean condor</h3>

W = mg

W = 9 x 9.8

W = 88.2 N

<h3>Pressure at the top surface of condor's wings</h3>

The pressure at the top surface of condor's wings is due to atmospheric pressure

P(top) = 14.7 Psi = 101,325 Pa

<h3>Pressure at the bottom surface of condor's wings</h3>

The pressure at the bottom surface is due to weight of andean condor.

P = W/A

P(bottom) = 88.2/0.729

P(bottom) = 120.99 Pa

The difference between the pressure at the top surfaces of the condor's wings and the pressure at the bottom surfaces is calculated as;

ΔP = P(top) - P(bottom)

ΔP = 101,325 Pa - 120.99 Pa

ΔP = 101,204 Pa

The complete question is below;

An Andean condor with a wingspan of 270 cm and a mass of 9.00 kg soars along a horizontal path. Model its wings as a rectangle with a width of 27.0 cm.

Learn more about pressure here: brainly.com/question/25736513

5 0
3 years ago
Three students have been studying relative motion and decide to do an experiment to demonstrate their knowledge. The experiment
miskamm [114]

Answer with Explanation:

We are given that

Constant speed of Jane=12.6 m/s

a.When Fred can throw the ball 30  m/s

We have to find the angle relative to the horizontal when he throw the ball in order for Sue to see the ball travel vertically upward.

Let \theta be the angle .

Therefore,

30 cos\theta=12.6

cos\theta=\frac{12.6}{30}=0.42

\theta=cos^{-1}(0.42)=65.165^{\circ}

b.We have to find the height to which ball reach.

v^2-v^2_0=2aS

S=\frac{v^2-v^2_0}{2a}=\frac{0-(30 sin65.165)^2}{2(-9.81)}

S=37.78 m

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3 years ago
Describe what the effect of increasing the power of a camera would have on the battery life
patriot [66]

Answer:

. Cut Down on the LCD

The biggest battery drain in a camera is the LCD – both the rear screen and the electronic viewfinder. This is the big reason why DSLRs almost always have longer battery life specifications than mirrorless cameras – the optical viewfinder lets you skip LCDs altogether.

However, if you use your DSLR in live view, you’ll notice that its battery life slides dramatically. Side by side against a mirrorless camera, there’s actually a good chance it will die first. LCDs just take a lot of power to run.

What does this imply? Quite simply, you should always do what you can to cut down on LCD usage when your battery is running low.

For DSLR users, that means switching to the optical viewfinder. For mirrorless photographers, it means turning off the camera frequently, or setting it so the viewfinder only activates when you hold it to your eye.

And regardless of the camera you use, drastically cut down on the amount of time you spend reviewing photos. Chimping has its place, but not while your battery warning is blinking red.Optimize Your Battery Saver Settings

Most cameras have menu options designed to improve battery life and maximize your shooting time. For example, the “metering timeout” setting lets you select how long you want the camera to wait during inactivity before shutting off its metering system.

Beyond that, a number of cameras today have an “Eco mode” that minimizes power consumption from the camera’s LCD. On the Canon EOS R, for example, Eco mode dims and then turns off the LCD when not in use, improving your battery life significantly – from 370 to 540 shots per charge, according to Canon’s official specifications.

It’s also important to note that mirrorless cameras are generally more efficient using the rear LCD than the electronic viewfinder. In terms of the EOS R again, Canon only rates 350 shots using the EVF, with no Eco mode to improve it. On the Sony side of things, the new A7R IV is rated for 530 shots via the viewfinder and 670 via the rear LCD.

If none of that applies to you, one option at your disposal is always to lower the brightness of your rear LCD. It might make photography a bit trickier in bright conditions, but the payoff is getting the shot rather than missing it completely due to a dead battery.

Other camera settings and extras that harm battery life include:

Image stabilization (both in-body and in-lens)

Popup flash

Bluetooth and WiFi

Most external accessories: GPS dongles, lightning triggers, wireless remote releases, shotgun mics, etc.

Sometimes, these capabilities are essential for your photo, so it’s worth the battery life sacrifice. But if you’re down to your last bar, double check to ensure that you’re not using any of the above settings or accessories without good reason.

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3 years ago
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vladimir2022 [97]

Answer:

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4 0
3 years ago
Two tuning forks of frequency 480 hz and 484 hz are struck simultaneously. what is the beat frequency resulting from the two sou
umka21 [38]
Ans: Beat frequency = f_b = 4Hz

Explanation: 
The beat frequency is equal to the absolute value of the difference in frequency of the two waves. In other words, the number of beats per second is equal to the difference in frequency. It is due to the destructive and constructive interference. <span>According to this interference, sound will be soft or loud.

Hence. the formula is:
</span>Beat frequency = f_b = |f_2 - f_1|
<span>
Since,
</span>f_1 = 480Hz
f_2 = 484Hz

Therefore,
Beat frequency = f_b = |484 - 480|

=> Beat frequency = f_b = 4Hz
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3 years ago
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