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3 years ago

Ứng dụng của lưu chất trong máy nén thủy lực

Engineering

1 answer:

Mars2501 [29]3 years ago

7 0

Answer:

Hầu hết các công việc quan trọng sử dụng nguyên lý thủy lực. Trợ lực lái, giảm xóc, kính chắn gió và phanh là những ứng dụng phổ biến của hệ thống thủy lực trên xe. Thang máy hai trụ và thang nâng bốn trụ được sử dụng trong ngành công nghiệp ô tô để nâng các phương tiện phục vụ công tác bảo dưỡng và kiểm tra.

Explanation:

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sashaice [31]

Huh? Kffjkdkdodjenxkrkdkre

4 0

3 years ago

Read 2 more answers

A hollow, spherical shell with mass 2.00kg rolls without slipping down a slope angled at 38.0?.

mezya [45]

Answer:

$\mu = 0.31$

Explanation:

Given data:

mass = 2.00 kg

slope angle = 38.0

From figure

balancing force

$mgsin\theta - f = ma$ .....1

Balancing torque

$F_R = \frac{2}{3} mR^2 \alpha$ ......2

for pure rolling

$\alpha = \frac{a}{R}$

$F = \frac{2}{3} ma$

from 1 and 2nd equation

$mgsin\theta - \frac{2}{3}ma = ma$

$mgsin\theta = \frac{5}{3} ma$

$a = \frac{3}{5} g sin\theta$

$= \frac{3\theta 9.8 sin 38}{5} = 3.62 m/s^2$

$F =\mu N$

$= \frac{2}{3} ma$

$= \frac{2}{3} 2\times 3.62 = 4.83 N$

N =normal force $= mgsin\theta = 2 \times 9.8 sin 38 = 15.44 N$

$\mu \times 15.44 = 4.83$

solving for coefficent of friction we get

$\mu = 0.31$

4 0

3 years ago

Water "bubbles up" h2 = 9 in. above the exit of the vertical pipe attached to three horizontal pipe segments. The total length o

8_murik_8 [283]

The pressure needed at point (1) to produce the flow where water "bubbles up above the exit of the vertical pipe attached to three horizontal pipe segments is 0.750 psi.

<h3>What is Bernoulli's equation?</h3>

The Bernoulli's equation for incompressible fluid can be given as,

$P_1+\dfrac{1}{2}\rho v_1^2+\rho gh_1=P_2+\dfrac{1}{2}\rho v_2^2+\rho gh_2$

Here, ρ is density of fluid, g is acceleration due to gravity, (P) is the pressure v is velocity, h is height of elevation and subscript (1 and 2) is used for point 1 and 2.

Water "bubbles up" h2 = 9 in. Above the exit of the vertical pipe attached to three horizontal pipe segments.

The total length of the 1.75-in.- diameter galvanized iron pipe between point (1) and the exit is 21 inches. h3 = 18 in.

Reynolds number is,

$R_e=\dfrac{V_3D}{v}\\R_e=\dfrac{4.01\dfrac{0.75}{12}}{1.21\times10^{-15}}\\R_e=2.07\times10^4$

The formula for the ratio of pressure to radius, when pressure and velocity at point 2 is zero can be given as,

$\dfrac{p_1}{r}=z_2\left(f\dfrac{l}{d}+\sum k_L\right)\dfrac{v^2}{2g}-\dfrac{v_1^2}{2g}$

Here, f=0.039 and ∑K(L)=4.5. Put the values,

$\dfrac{p_1}{r}=\dfrac{7}{12}\left(0.03\dfrac{21}{0.75}+4.5-1\right)\dfrac{4.01^2}{2\times32.2}\\\dfrac{p_1}{r}=1.73\rm\; ft$

Put the value of r we get,

$p_1=62.4\times1.73\\p_1=108\rm\; ib/ft^2\\p_1=0.750\rm\; psi$

The pressure needed at point (1) to produce the flow where water "bubbles up above the exit of the vertical pipe attached to three horizontal pipe segments is 0.750 psi.

Learn more about the Bernoulli's equation here;

brainly.com/question/7463690

#SPJ1

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2 years ago

Select the correct answer.

gizmo_the_mogwai [7]

I think it’s C but I could be wrong I’m sorry

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3 years ago

Check engine light is illuminated on the instrument panel. what action should you take

Oksi-84 [34.3K]

Answer:

Connect your computer to the OBD(On-Board Diagnostics) port and see whats wrong

Explanation:

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mechanik

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meckanic

nvm, I fix cars

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2 years ago