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denis23 [38]
2 years ago
8

Two pistons of a hydraulic lift have radii of 2. 67 cm and 20. 0 cm, respectively. The downward force on the 2. 67-cm piston tha

t is required to lift a mass of 2000 kg supported by the 20-cm piston is
Physics
1 answer:
Makovka662 [10]2 years ago
4 0

The downward force on the 2. 67-cm piston that is required to lift a mass of 2000 kg supported by the 20-cm piston is 350 N.

<h3>What is pascal law?</h3>

Pressure applied to a closed system of fluid will be transfer at each point of the fluid and the boundaries of the system.

Let suppose at the two point of such system the input force apllied is F₁ and output force we get is F₂. The area of this points is A₁ and A₂ respectively. As the pressure at both point is same. Then by the pascal law,

\dfrac{F_1}{A_1}=\dfrac{F_2}{A_2}\\

In case of radius,

\dfrac{F_1}{\pi r_1^2}=\dfrac{F_2}{\pi r_2^2}\\\\\dfrac{F_1}{r_1^2}=\dfrac{F_2}{ r_2^2}

Two pistons of a hydraulic lift have radii of 2. 67 cm and 20. 0 cm, respectively. Thus,

r_1=2.67\rm\; cm

r_2=20.0\rm\; cm

The downward force on the 2. 67-cm piston that is required to lift a mass of 2000 kg supported by the 20-cm piston is The force applied on second piston is,

F_2=mg\\F_2=2000\times9.81\\F_2=19620 \rm\; N

Put the values in the above formula,

\dfrac{F_1}{2.67^2}=\dfrac{19620}{20^2}\\F_1=\dfrac{19620\times2.67^2}{20^2}\\F_1\approx 350\rm\; N

Thus, the downward force on the 2. 67-cm piston that is required to lift a mass of 2000 kg supported by the 20-cm piston is 350 N.

Learn more about the pascal law here;

brainly.com/question/4262025

#SPJ4

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0.273 liters are needed to accomplish this task without boiling.

Explanation:

The minimum boiling point of kerosene is 150\,^{\circ}C. According to this question, we need to determine the minimum volume of liquid such that heat received is entirely sensible, that is, with no phase change.

If we consider a steady state process and that energy interactions with surrounding are negligible, then we get the following formula by the Principle of Energy Conservation:

\rho_{k}\cdot V_{k}\cdot c_{k}\cdot (T-T_{k,o}) = m_{t}\cdot c_{t}\cdot (T_{t,o}-T) (1)

Where:

\rho_{k} - Density of kerosene, measured in kilograms per cubic meter.

V_{k} - Volume of kerosene, measured in cubic meters.

c_{k}, c_{t} - Specific heats of the kerosene and tin, measured in joule per kilogram-Celsius.

T_{k,o}, T_{t,o} - Initial temperatures of kerosene and tin, measured in degrees Celsius.

T - Final temperatures of the kerosene-tin system, measured in degrees Celsius.

Please notice that the block of tin is cooled at the expense of the temperature of the kerosene until thermal equilibrium is reached.

From (1), we clear the volume of kerosene:

V_{k} = \frac{m_{t}\cdot c_{t}\cdot (T_{t,o}-T)}{\rho_{k}\cdot c_{k}\cdot (T-T_{k,o})}

If we know that m_{t} = 1.83\,kg, c_{t} = 218\,\frac{J}{kg\cdot ^{\circ}C}, T_{t,o} = 88\,^{\circ}C, T_{k,o} = 24.0\,^{\circ}C, T = 57\,^{\circ}C, c_{k} = 2010\,\frac{J}{kg\cdot ^{\circ}C} and \rho_{k} = 820\,\frac{kg}{m^{3}}, then the volume of the liquid needed to accomplish this task without boiling is:

V_{k} = \frac{(1.83\,kg)\cdot \left(218\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (88\,^{\circ}C-57\,^{\circ}C)}{\left(820\,\frac{kg}{m^{3}} \right)\cdot \left(2010\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (57\,^{\circ}C-24\,^{\circ}C)}

V_{k} = 2.273\times 10^{-4}\,m^{3}

V_{k} = 0.273\,L

0.273 liters are needed to accomplish this task without boiling.

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A monster truck enters a freeway with a speed of 0 m/s and accelerates at a rate of 25 m/s2 in 5 seconds. How far does the car t
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Answer:

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=312.5

plz mark as brainalist

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