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2 years ago

13

A hydraulic press has one piston of diameter 4.0 cm and the other piston of diameter 8.0 cm. What force must be applied to the s

maller piston to obtain a force of 1600 N at the larger piston

1 answer:

yan [13]2 years ago

8 0

Answer:

The force is $F_1 = 400.8 \ N$

Explanation:

From the question we are told that

The first diameter is $d_1 = 4.0 \ cm = 0.04 \ m$

The second diameter is $d_2 = 8.0 \ cm = 0.08 \ m$

Generally the first area is

$A_1 = \pi * \frac{d^2_1 }{4}$

=> $A_1 = 3.142 * \frac{0.04^2}{4}$

=> $A_1 = 0.00126 \ m^2$

The second area is

$A_2 = \pi * \frac{d^2_2 }{4}$

$A_2 = 3.142 * \frac{0.08^2}{4}$

$A_2 = 0.00503 \ m^2$

For a hydraulic press the pressure at both end must be equal .

Generally pressure is mathematically represented as

$P = \frac{F}{A}$

=>

$\frac{F_1}{A_1 } = \frac{F_2}{A_2 }$

=> $F_1 = \frac{1600}{0.00503} * 0.00126$

=> $F_1 = 400.8 \ N$

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2 years ago

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3 years ago

Read 2 more answers

How many atoms are in a sample of 2.39 moles of neon (Ne) atoms?

Reil [10]

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2 years ago

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s

yawa3891 [41]

Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring ?

Data:

$E_{pe}\:(elastic\:potential\:energy) = 5184\:J$

$K\:(constant) = 16200\:N/m$

$x\:(displacement) =\:?$

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

$E_{pe} = \dfrac{k*x^2}{2}$

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

Solving:

$E_{pe} = \dfrac{k*x^2}{2}$

$5184 = \dfrac{16200*x^2}{2}$

$5184*2 = 16200*x^2$

$10368 = 16200\:x^2$

$16200\:x^2 = 10368$

$x^{2} = \dfrac{10368}{16200}$

$x^{2} = 0.64$

$x = \sqrt{0.64}$

$\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark$

Answer:

The displacement of the spring = 0.8 m

_______________________________

I Hope this helps, greetings ... Dexteright02! =)

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3 years ago

water in a cup and a kettle can have the same temperature even though the quantities are different . give reasons

jekas [21]

Answer:

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$\infty M_{(environ)} \times c_{(environ)} \times (T_2 - T_1) = m_{1} \times c_{(water)} \times (T_3 - T_2) + m_{2} \times c_{(water)} \times (T_4 - T_2)$

Due to the large heat sink, T₂ - T₁ ≈ 0 such that the temperature of the kettle and that of the cup will both cool to the temperature of the environment

Explanation:

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3 years ago