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r-ruslan [8.4K]

2 years ago

11

A trough is filled with a liquid of density 810 kg/m3. The ends of the trough are equilateral triangles with sides 8 m long and

vertex at the bottom. Find the hydrostatic force on one end of the trough. (Use 9.8 m/s2 for the acceleration due to gravity.)

1 answer:

Ilya [14]2 years ago

6 0

Answer:

The hydrostatic force on one end of the trough is 54994.464 N

Explanation:

Given;

liquid density, ρ = 810 kg/m³

side of the equilateral triangle, L = 8m

acceleration due to gravity, g = 9.8 m/s²

Hydrostatic force is given as;

H = ρgh

where;

h is the vertical height of the equilateral triangle

Draw a line to bisect upper end of the trough, to the vertex at the bottom, this line is the height of the equilateral triangle.

let the half side of the triangle = x

x = ⁸/₂ = 4m

The half section of the triangle forms a right angled triangle

h² = 8² - 4²

h² = 48

h = √48

h = 6.928m

F = ρgh

F = 810 x 9.8 x 6.928

F = 54994.464 N

Therefore, the hydrostatic force on one end of the trough is 54994.464 N

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Answer:

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Explanation:

Average velocity of oxygen molecule at given temperature is

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now we have

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now for de Broglie wavelength we know that

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3 years ago

A rod extending between x = 0 and x = 13.0 cm has uniform cross-sectional area A = 8.00 cm2. Its density increases steadily betw

mezya [45]

Answer:

The mass is $m = 3.45408 kg$

Explanation:

From the question we are told that

The extension of the rod is $from , \ x_1 = 0 \to x_2 = 13.0$

The area is $A = 8.0 cm^2$

The density increase as follows $from \ \rho_1 =2.5 g/cm^2 \to \rho_2= 19.0 g/cm^3$

The equation $\rho = B + Cx$

at $x_1= 0$ $\rho_1 =2.5 g/cm^2$

So

$2.5 = B + 0$

=> $B =2.5$

So at $x_2 = 13.0$ , $\rho_2= 19.0 g/cm^3$

So

$19.0 = 2.5 + C(13)$

=> $C = 1.27$

Now

$m = 8 \int\limits^{13}_{0} {2.5 + 1.27x} \, dx$

$m = 8 [{2.5 +\frac{ 1.27x^2}{2} } ]\left | 13} \atop {0}} \right.$

$m = 8 [{2.5 +\frac{ 1.27(13)^2}{2} } ]$

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3 years ago

Will the 79 kg skier in the figure below slide down if f the coefficient of static friction is 0.25?

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Man will not slide down

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Coefficient of static friction = 0.25

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3 years ago

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2 years ago

Calculate the maximum deceleration (in m/s2) of a car that is heading down a 14° slope (one that makes an angle of 14° with the

lesya [120]

The question is incomplete. Here is the complete question.

Calculate the maximum deceleration of a car that is heading down a 14° slope (one that makes an anlge of 14° with the horizontal) under the following road conditions. You may assum that the weight of the car is evenlydistributed on all four tires and that the sttic coefficient of friction is involved - that is, the tires are not allowed to slip during the deceleration. (Ignore rolling) Calculate for a car: (a) On a dry concrete. (b) On a wet concrete. (c) On ice, assuming that μs = 0.100, the same as for shoes on ice.

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Explanation: The image in the attachment describe the forces acting on the car. Observing that, we know that:

$F_{net}$ = - $W_x$ - $f_s$

The $W_x$ is a x-component of force due to gravity (W) and, in this case, is given by: $W_x$ = W.sin(14)

W is described as: W = m.g

Force due to friction ( $f_s$ ) is given by: $f_s$ = μs.N

N is the normal force and, in the system, is equivalent of $W_y$ , so:

$W_y$ = m.g.cos(14)

Therefore, the formula will be:

$F_{net}$ = - $W_x$ - $f_s$

m.a = - (m.g.sin14) - (μs.mg.cos14)

a = - g (sin14 + μscos 14)

a) For dry concrete, μs = 1:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin14 + 1.cos14)

a = - 11.05 m/s²

b) For wet concrete, μs = 0.7:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin 14 + 0.7.cos14)

a = - 10.64 m/s²

c) For ice, μs = 0.1:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin14 + 0.1cos14)

a = - 9.84 m/s²

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3 years ago