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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:

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Explanation:

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3 years ago

Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is

Ronch [10]

Complete Question

A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.

If the acceleration of the projective is : a = c/s m/s2

Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?

Answer:

The value of the constant is $c = 28853.78 \ m^2 /s^2$

Explanation:

From the question we are told that

The acceleration is $a = \frac{c}{s}\ m/s^2$

The initial position of the projectile is s= 1.5m

The final position of the projectile is $s_f = 3 \ m$

The velocity is $v = 200 \ m/s$

Generally $time = \frac{ds}{dv}$

and acceleration is $a = \frac{v}{time }$

$a = v \frac{dv}{ds}$

=> $vdv = a ds$

$vdv = \frac{c}{s} ds$

integrating both sides

$\int\limits^a_b vdv = \int\limits^c_d \frac{c}{s} ds$

Now for the limit

a = 200 m/s

b = 0 m/s

c = s= 3 m

d = $s_f$ = 1.5 m

So we have

$\int\limits^{200}_{0} vdv = \int\limits^{3}_{1.5} \frac{c}{s} ds$

$[\frac{v^2}{2} ] \left | 200} \atop {0}} \right. = c [ln s]\left | 3} \atop {1.5}} \right.$

$\frac{200^2}{2} = c ln[\frac{3}{1.5} ]$

=> $c = \frac{20000}{0.69315}$

$c = 28853.78 \ m^2 /s^2$

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