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3 years ago

these questions confuse me. i think I know the answer but I'm not sure. this is density/hydraulic devices. help?

Physics

2 answers:

Bad White [126]3 years ago

8 0

It is hydraulic. The key is to remember that the pressure will be constant through the whole liquid thus the force requuired on the left side will be much less and proportional to the difference in area of the right and the left. Do you need actual calculations or can you figure it out from this? Think Pascal's principle

Anna11 [10]3 years ago

5 0

Had a quick look at the q, here's some guesswork .. apologies in advance for the "humour" ...
pressure in liquid is elephant divided by area she's (After the song about an elephant ... ) standing on.
45000/250,000 N/cm^2=0.18
p=F/A. F=pA. p=0.18, A=500. F=0.18x500N
Elle La Fonte + Platform =45,000N

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Answer:

its true that Scientific endeavor is driven by both simple curiosity as well as societal demands.

Explanation:

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3 years ago

A uniformly charged rod (length = 2.0 m, charge per unit length = 3.0 nc/m) is bent to form a semicircle. What is the magnitude

Artist 52 [7]

Answer:

84.82N/C.

Explanation:

The x-components of the electric field cancel; therefore, we only care about the y-components.

The y-component of the differential electric field at the center is

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Now, let us call $\lambda$ the charge per unit length, then we know that

$dQ = \lambda Rd\theta$ ;

therefore,

$dE = \frac{k \lambda R d\theta }{R^2} sin(\theta )$

$dE = \frac{k \lambda d\theta }{R} sin(\theta )$

Integrating

$E = \frac{k \lambda }{R}\int_0^\pi sin(\theta )d\theta$

$E = \frac{k \lambda }{R}*[-cos(\pi )+cos(0) ]$

$E = \frac{2k \lambda }{R}.$

Now, we know that

$\lambda = 3.0*10^{-9}C/m,$

$k = 9*10^9kg\cdot m^3\cdot s^{-4}\cdot A^{-2},$

and the radius of the semicircle is

$\pi R = 2.0m,\\\\R = \dfrac{2.0m}{\pi };$

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$\boxed{E = 84.82N/C.}$

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Answer: The answer is A for sure

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That is, there will be no acceleration. If you are sitting at rest in a chair and the upward push of the chair is equal to the downward pull of gravity, you will stay at rest in the chair. ... You now have an unbalanced force acting on you and therefore, according to Newton's First Law, your motion is going to change.

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