Which properties do all metals on the periodic table share?

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

Makovka662 [10]

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

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4 0

2 years ago

Why do people use the scientific method?

Len [333]

When conducting research, scientists use the scientific method to collect measurable, empirical evidence in an experiment related to a hypothesis (often in the form of an if/then statement), the results aiming to support or contradict a theory.
Hopefully this helped.

5 0

4 years ago

The charges Q1=Q and Q2=4Q that are a distance d apart, repel each other with a force of 1.60 N. What would be the force between

gladu [14]

Answer:

50.4 N

Explanation:

Q1 = Q

Q2 = 4 Q

Distance = d

The force is given by

$F = \frac{KQ_{1}Q_{2}}{d^{2}}$

$1.60 = \frac{4KQ^{2}}{d^{2}}$ .... (1)

Now,

Q3 = 2 Q

Q4 = 7 Q

distance = d/3

$F' = \frac{9KQ_{3}Q_{4}}{d^{2}}$

$F' = \frac{126KQ^{2}}{d^{2}}$ .... (2)

Divide equation (2) by equation (1), we get

F' / 1.60 = 126 / 4

F' = 50.4 N

Thus, the force is 50.4 N.

7 0

4 years ago

Determine the CM of the uniform thin L-shaped construction brace shown in (Figure 1) . Suppose that a = 2.20 m and b = 1.63 m De

aleksandrvk [35]

By weighted average method, the x-coordinates of the center of mass of the compound figure, the uniform thin L-shaped construction brace is $\bar x = \frac{1.10\cdot m_{A}+2.10\cdot m_{B}}{m_{A}+m_{B}}$ meters.

<h3>How to determine the coordinates of the center of mass of a compound figure </h3>

Let suppose that the entire construction has a <em>uniform</em> mass, then the coordinates of the center of mass can be determined by definition of <em>weighted</em> average:

$\bar x = \frac{\sum \limits_{i = 1}^{n}x_{i}\cdot m_{i}}{\sum \limits_{i=1}^{n}m_{i}}$ (1)

$\bar y = \frac{\sum \limits_{i = 1}^{n}y_{i}\cdot m_{i}}{\sum \limits_{i=1}^{n}m_{i}}$ (2)

If we know that a = 2.20 m and b = 1.63 m, then the x-coordinates of the center of mass of the compound figure is:

$\bar x = \frac{1.10\cdot m_{A}+2.10\cdot m_{B}}{m_{A}+m_{B}}$

By weighted average method, the x-coordinates of the center of mass of the compound figure, the uniform thin L-shaped construction brace is $\bar x = \frac{1.10\cdot m_{A}+2.10\cdot m_{B}}{m_{A}+m_{B}}$ meters.

To learn more on center of mass: brainly.com/question/8662931

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8 0

2 years ago

Find out the x-component and y-component of the following vectors:

Leno4ka [110]

Answer:

A = 15.0 cm; the vector is 30.00 north of west; = 15 cm N30W

Y - component = 7.5cm and X - component = -12.99 cm

B = 12.0 m; the vector is 55.00 north of east; = 12.0 m N55E

Y - component = 9.83 m and X - component = 6.88 m

C = 28.0 m/s; the vector is 40.00 south of west; = 28.0 m/s S40W

Y - component = - 17.998 m/s and X - component = - 21.448 m/s

D = 55.5 m/s2; the vector is 35.00 south of east; = 55.5 m/s² S35E

Y - component = - 31.835 m/s² and X - component = 45.466 m/s²

Explanation:

1. A = 15.0 cm; the vector is 30.00 north of west; = 15 cm N30W

Y - component = 15cm × sin30 = 7.5cm

X - component = -15cm × cos30 = -12.99 cm

2. B = 12.0 m; the vector is 55.00 north of east; = 12.0 m N55E

Y - component = 12.0 m × sin55 = 9.83 m

X - component =12.0 m × cos55 = 6.88 m

3. C = 28.0 m/s; the vector is 40.00 south of west; = 28.0 m/s S40W

Y - component = - 28.0 m/s × sin40 = - 17.998 m/s

X - component = - 28.0 m/s × cos40 = - 21.448 m/s

4. D = 55.5 m/s2; the vector is 35.00 south of east; = 55.5 m/s² S35E

Y - component = - 55.5 m/s² × sin35 = - 31.835 m/s²

X - component = 55.5 m/s² × cos35 = 45.466 m/s²

6 0

3 years ago

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