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anastassius [24]
3 years ago
11

The small piston of a hydraulic lift has a crosssectional area of 2.23 cm2 and the large piston 297 cm2 . What force must be app

lied to the small piston for the lift to raise a load of 2.4 kN? (In service stations, this force is usually exerted by compressed air.) Answer in units of N.
Physics
1 answer:
Lerok [7]3 years ago
8 0

Answer:

The force that must be applied to the small piston = 18.02 N.

Explanation:

Hydraulic Press: This is a device that produce a very large force to compress something e.g printing press.

From pascal's principle,

In an hydraulic press,

f/a = F/A......................... Equation 1

Where f = force applied to the small piston, a = area of the small piston, F = force required to lift the load/ force produced at the large piston. A = Area of the large piston.

Making f the subject of the equation above,

f = F×a/A.............................. Equation 2

Given: F = 2.4 kN = 2400 N, a = 2.23 cm², A = 297 cm².

Substituting into  equation 2

f = 2400(2.23)/297

f = 18.02 N.

Thus the force that must be applied to the small piston = 18.02 N.

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47.4 m

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Now we consider only the downward motion of the ball: it is a free fall motion, so we can find the vertical displacement by using the suvat equation

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Solving the  formula, we find

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3 years ago
Calculate the kinetic energy of a 750 kg compact car moving 50 m/s.
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Exposure to a sufficient quantity of ultraviolet will redden the skin, producing erythema - a sunburn. The amount of exposure ne
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Answer:

Energy = 7.83 x 10⁻¹⁹ J

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Explanation:

The energy of a photon in terms of wavelength can be calculated by the following formula:

Energy = \frac{hc}{\lambda}\\

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λ = wavelength of light

Now, for λ = 254 nm = 2.54 x 10⁻⁷ m:

Energy = \frac{(6.63\ x\ 10^{-34}\ Js)(3\ x\ 10^8\ m/s)}{2.54\ x\ 10^{-7}\ m}\\

<u>Energy = 7.83 x 10⁻¹⁹ J</u>

<u></u>

Now, for λ = 300 nm = 3 x 10⁻⁷ m:

Energy = \frac{(6.63\ x\ 10^{-34}\ Js)(3\ x\ 10^8\ m/s)}{3\ x\ 10^{-7}\ m}\\

<u>Energy = 6.63 x 10⁻¹⁹ J</u>

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