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3 years ago

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A hydraulic lift has pistons with diameter 28cm and 70cm, respectively. If a force of 500 N is exerted at the input piston. What

is the maximum load that can be lifted the output piston?

1 answer:

Crazy boy [7]3 years ago

4 0

Answer:

3125 N

Explanation:

diameter /2 =radius

so r1 =14cm , r2 =35cm

f1/A1 =f2/A2.

f2 = f1 × A2 / A1

=500×1225 pi cm² / 96 pi cm²

f2 =3125N

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Piano tuners tune pianos by listening to the beats between the harmonics of two different strings. When properly tuned, the note

Sophie [7]

(a) 2 Hz

The frequency of the nth-harmonic is given by

$f_n = n f_1$

where

$f_1$ is the fundamental frequency

Therefore, the frequency of the third harmonic of the A ( $f_1 = 440 Hz$ ) is

$f_3 = 3 \cdot f_1 = 3 \cdot 440 Hz =1320 Hz$

while the frequency of the second harmonic of the E ( $f_1 = 659 Hz$ ) is

$f_2 = 2 \cdot f_1 = 2 \cdot 659 Hz =1318 Hz$

So the frequency difference is

$\Delta f = 1320 Hz - 1318 Hz = 2 Hz$

(b) 2 Hz

The beat frequency between two harmonics of different frequencies f, f' is given by

$f_B = |f'-f|$

In this case, when the strings are properly tuned, we have

- Frequency of the 3rd harmonic of A-note: 1320 Hz

- Frequency of the 2nd harmonic of E-note: 1318 Hz

So, the beat frequency should be (if the strings are properly tuned)

$f_B = |1320 Hz - 1318 Hz|=2 Hz$

(c) 1324 Hz

The fundamental frequency on a string is proportional to the square root of the tension in the string:

$f_1 \propto \sqrt{T}$

this means that by tightening the string (increasing the tension), will increase the fundamental frequency also*, and therefore will increase also the frequency of the 2nd harmonic.

In this situation, the beat frequency is 4 Hz (four beats per second):

$f_B = 4 Hz$

And since the beat frequency is equal to the absolute value of the difference between the 3rd harmonic of the A-note and the 2nd harmonic of the E-note,

$f_B = |f_3-f_2|$

and $f_3 = 1320 Hz$ , we have two possible solutions for $f_2$ :

$f_2 = f_3 - f_B = 1320 Hz - 4 Hz = 1316 Hz\\f_2 = f_3 + f_B = 1320 Hz + 4 Hz = 1324 Hz$

However, we said that increasing the tension will increase also the frequency of the harmonics (*), therefore the correct frequency in this case will be

1324 Hz

8 0

3 years ago

26. The equation E=hf describes the energy of each photon in a beam of light. If Planck’s constant, h, were larger, would photon

Elenna [48]

The equation of the energy of a photon is E=h*f.

If we increase the Planck's constant h, the energy would increase.

For example, lets double the value of Planck's constant and name it H:

H=2*h. Now lets put that into the equation for energy that we will call E₂:

E₂=H*f=2*h*f=2*E.

So we can clearly see that E₂=2*E or that if we double Planck's constant, the energy also doubles.

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3 years ago

Describe one common critique or criticism of Utilitarianism. To what extent do you agree or disagree with this criticism?

lisabon 2012 [21]

The second most common criticism of utilitarianism is that it is impossible to apply - that happiness (etc) cannot be quantified or measured, that there is no way of calculating a trade-off between intensity and extent, or intensity and probability (etc), or comparing happiness to suffering.

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2 years ago

Help with these three

Elenna [48]

The first: alright, first: you draw the person in the elevator, then draw a red arrow, pointing downwards, beginning from his center of mass. This arrow is representing the gravitational force, Fg.
You can always calculate this right away, if you know his mass, by multiplying his weight in kg by the gravitational constant
$g = 9.81 \frac{m}{s {}^{2} }$
let's do it for this case:
$f_{g} = m \times g \\ f _{g} = 65kg \times 9.81 \frac{m}{s {}^{2} } = 637.65$
the unit of your fg will be in Newton [N]
so, first step solved, Fg is 637.65N
Fg is a field force by the way, and at the same time, the elevator is pushing up on him with 637.65N, so you draw another arrow pointing upwards, ending at the tip of the downwards arrow.
now let's calculate the force of the elevator
$f = m \times a \\ f = 65 \times 5 \frac{m}{s {}^{2} } \\ f = 325n$
so you draw another arrow which is pointing downwards on him, because the elevator is accelating him upwards, making him heavier
the elevator force in this case is a contact force, because it only comes to existence while the two are touching, while Fg is the same everywhere

8 0

3 years ago

In a mail-sorting facility, a 2.50-kg package slides down an inclined plane that makes an angle of 20.0° with the horizontal. Th

lawyer [7]

Answer:

The coefficient of kinetic friction is 0.382.

Explanation:

Given:

Angle of inclination is, $\theta=20.0$ °

Mass of package is, $m=2.50\ kg$

Initial speed of package is, $u=2.00\ m/s$

Final speed of the package at the bottom is, $v=0\ m/s$

Distance of travel along the incline is, $d=12.0\ m$

Acceleration due to gravity is, $g=9.8\ m/s^2$

Let the coefficient of kinetic friction be $\mu$ .

Now, the frictional force will be acting along the incline but in the direction opposite to the direction of motion.

So, the net acceleration acting on the package will be up the incline and is equal to:

$a=\mu g\cos\theta-g\sin\theta$ ----------------- 1

Now, using equation of motion, we have:

$v^2-u^2=2ad\\\\0-(2.00)^2=2a(12.0)$

Solving for 'a', we get:

$-4.00=24.0a\\\\a=-\frac{4}{24}=-\frac{1}{6}\ m/s^2$

Now, plug in the value of 'a' in equation (1). This gives,

$\mu g\cos\theta-g\sin\theta=\frac{1}{6}$ ( Neglecting negative sign)

Plug in all the given values and solve for $\mu$ . This gives,

$9.8(-sin(20)+\mu cos(20))=\frac{1}{6}\\\\-0.342+\mu\times 0.94=0.017\\\\0.94\mu=0.342+0.017\\\\0.94\mu=0.359\\\\\mu=\frac{0.359}{0.94}=0.382$

Therefore, the coefficient of kinetic friction is 0.382.

5 0

3 years ago