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3 years ago

12

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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Given that ethylene has a λmax of 175nm, butadiene has a λmax of 220nm, and 2-methyl-1,3-butadiene has a λmax or 215nm, what is

Vika [28.1K]

The λmax of 2,3,4-trimethylhexatriene is 280 nm.

Ethylene has a λmax of 175nm.

Butadiene has a λmax of 220nm.

2-methyl-1,3-butadiene has a λmax or 215nm.

1,3,5-hexatriene has a λmax of 258nm.

Woodward's rules, sometimes known as Woodward-Fieser rules (after Louis Fieser) and named after Robert Burns Woodward, are a number of sets of empirically developed principles that aim to forecast the wavelength of the absorption maximum (max) in an ultraviolet-visible spectrum of a certain molecule.

By using the Woodward Fieser rule,

R- (Alkyl Group) .... +5 nm = 5 × 2 = 10

RO- (Alkoxy Group) .. +6 = 6 × 2 = 12

Adding 22nm to the λmax of 1,3,5-hexatriene as it has 2 alkyl groups and 2 alkoxy groups to form 2,3,4-trimethylhexatriene.

The λmax of 2,3,4-trimethylhexatriene is 280 nm.

Learn more about Woodward-Fieser here:

brainly.com/question/16982345

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5 0

2 years ago

If the wave represents a sound wave, explain how increasing amplitude will affect the loudness of the sound? If we decrease the

Viktor [21]

Answer:

Explanation:

Think of a sound wave like a wave on the ocean, or lake... It's not really water moving, as much as it's energy moving through the water. Ever see something floating on the water, and notice that it doesn't come in with the wave, but rides over the top and back down into the trough between them? Sound waves are very similar to that. If you looked at a subwoofer speaker being driven at say... 50 cycles a second, you'd actually be able to see the speaker cone moving back and forth. The more power you feed into the speaker, the more it moves back and forth, not more quickly, as that would be a higher frequency, but further in and further out, still at 50 cycles per second. Every time it pushed out, it's compressing the air in front of it... the compressed air moves away from the speaker's cone, but not as a breeze or wind, but as a wave through the air, similar to a wave on the ocean

More power, more amplitude, bigger "wave", louder ( to the human ear) sound.

If you had a big speaker ( subwoofer ) and ran a low frequency signal with enough power in it, you could hold a piece of paper in front of it, and see the piece of paper move in and out at exactly the same frequency as the speaker cone. The farther away from the speaker you got, the less it'd move as the energy of the sound wave dispersed through the room.

Sound is a wave

We hear because our eardrums resonates with this wave I.e. our ear drums will vibrate with the same frequency and amplitude. which is converted to an electrical signal and processed by our brain.

By increasing the amplitude our eardrums also vibrate with a higher amplitude which we experience as a louder sound.

Of course when this amplitude is too high the resulting resonance tears our eardrums so that they can't resonate with the sound wave I.e. we become deaf

6 0

3 years ago

What is the de Broglie wavelength of an object with a mass of 2.50 kg moving at a speed of 2.70 m/s? (Useful constant: h = 6.63×

xxMikexx [17]

Answer:

9.82 × $10^{-35}$ Hz

Explanation:

De Broglie equation is used to determine the wavelength of a particle (e.g electron) in motion. It is given as:

λ = $\frac{h}{mv}$

where: λ is the required wavelength of the moving electron, h is the Planck's constant, m is the mass of the particle, v is its speed.

Given that: h = 6.63 × $10^{-34}$ Js, m = 2.50 kg, v = 2.70 m/s, the wavelength, λ, can be determined as follows;

λ = $\frac{h}{mv}$

= $\frac{6.63*10^{-34} }{2.5*2.7}$

= $\frac{6.63 * 10^{-34} }{6.75}$

= 9.8222 × $10^{-35}$

The wavelength of the object is 9.82 × $10^{-35}$ Hz.

4 0

3 years ago

What distance will a truck trvael in 3 hours at an average speed of 50 per hour

FromTheMoon [43]

The truck would of went 150 miles

5 0

3 years ago

As the concentration of a solute in a solution increases, the freezing point of the solution ________ and the vapor pressure of

kykrilka [37]

Answer:

As the concentration of a solute in a solution increases, the freezing point of the solution <u><em>decrease </em></u>and the vapor pressure of the solution <em><u>decrease </u></em>.

Explanation:

Depression in freezing point :

$\Delta T_f=K_f\times m$

where,

$\Delta T_f$ =depression in freezing point =

$K_f$ = freezing point constant

m = molality ( moles per kg of solvent) of the solution

As we can see that from the formula that higher the molality of the solution is directly proportionate to the depression in freezing point which means that:

If molality of the solution in high the depression in freezing point of the solution will be more.
If molality of the solution in low the depression in freezing point of teh solution will be lower .

Relative lowering in vapor pressure of the solution is given by :

$\frac{p_o-p_s}{p_o}=\chi_{solute}$

$p_o$ = Vapor pressure of pure solvent

$p_s$ = Vapor pressure of solution

$\chi_{solute}$ = Mole fraction of solute

$p_s\propto \frac{1}{\chi_{solute}}$

Vapor pressure of the solution is inversely proportional to the mole fraction of solute.

Higher the concentration of solute more will the be solute's mole fraction and decrease in vapor pressure of the solution will be observed.
lower the concentration of solute more will the be solute's mole fraction and increase in vapor pressure of the solution will be observed.

8 0

3 years ago