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3 years ago

What two forces limit the height to which capillary action will take water up a tube?

1 answer:

3 0

Capillary action occurs when the adhesion to the walls is stronger than the cohesiveforces between the liquid molecules. The height to which capillary action will take water in a uniform circular tube (picture to left) is limited by surface tension and, of course, gravity.

Hope this helps. :)

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Bricks and insulation are used to construct the walls of a house. The

ira [324]

Answer:

$\dot Q=350.438\ W$

Explanation:

Given:

the thermal resistance in the form of

$R_1=\frac{x_1}{k_1} =0.095\ m^2.^{\circ}C.W^{-1}$

$R_2=\frac{x_2}{k_2} =0.704\ m^2.^{\circ}C.W^{-1}$

where:

$x_1\ \&\ x_2$ are the thickness of the respective bricks

$k_1\ \&\ k_2$ are the respective coefficient of conductivity

temperature inside the house, $T_h=24\ ^{\circ}C$

temperature outside the house, $\ T_c=10^{\circ}C$

area of the wall, $A=20\ m^2$

Since the bricks and insulation are used to construct a wall then they must be used in series for better shielding.

Using Fourier's law:

$\dot Q=k.A.\frac{dT}{x}$

$\dot Q={dT}\div {\frac{x}{k.A} }$

in series the resistances get add up

$\dot Q=dT\div (\frac{x_1}{k_1.A}+\frac{x_2}{k_2.A} )$

$\dot Q=(24-10)\div (\frac{0.095 }{20}+ \frac{0.704 }{20} )$

$\dot Q=350.438\ W$

7 0

4 years ago

A joule is equivalent to N/s N/m Ndm Nds

skelet666 [1.2K]

A joule is also called a newton metre so it is force in newtons multiplied by distance. Answer 3 seems to fit, assuming dm means “delta m” or a measure of the change of distance in metres.

7 0

4 years ago

How does friction help soccer players

Lena [83]

When soccer players run they are using friction to propell themselves

7 0

3 years ago

Read 2 more answers

Four charges are located at the corners of a square. Each side of the square is 37 m. The left two charges have a positive charg

nadezda [96]

Answer:

E = 781.12 N/C

Explanation:

Look at the attached graphic:

The 20µC charges are positive , then, the electric fields leave the charge.

The 22µC charges are negative, then, the electric fields enter the charge.

The electric field due to each of the charges is calculated by Coulomb's law:

E= k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

Problem development

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

E₁, E₂, E₃, E₄: Electric field at point P due to charge q₁, q₂, q₃, and q₄ respectively

The electric field in the direction of the y axis at the point P in the center of the square is equal to zero because the sum of the vertical components upwards is equal to the sum of the vertical components downwards.

Calculation of the electric field in the direction of the x-axis in the center of the square (point P)

Eₓ = E₁ₓ + E₂ₓ + E₃ₓ + E₄ₓ

$d = \sqrt{18.5^2+ 18.5^2} =26.16m$

E₁ₓ = E₂ₓ = (9*10⁹)*(20*10⁻⁶)*cos(45°)/(d²) = (9*10⁹)*(20*10⁻⁶)*cos(45°)/(26.16²) = 185.98 N/C

E₃ₓ = E₄ₓ = (9*10⁹)*(22*10⁻⁶)*cos(45°)/(d²) = (9*10⁹)*(22*10⁻⁶)*cos(45°)/(26.16²) = 204.58 N/C

Eₓ = E = 2*185.98 +2*204.58 = 781.12 N/C

5 0

3 years ago

The earth rotates about its pole once every 24 hrs. The distance from the pole to a location on the Earth is 35* north latitude

Oksana_A [137]

The angular velocity, ω=
2π/t; t = 24 hrs = 24 x 3600 seconds = 86400 s
ω = 7.27 x 10⁻⁵
v = ωr
= 7.27 x 10⁻⁵ x 3242.8 x 1.6 x 1000 (converting miles to meters)
= 377.2 m/s

4 0

3 years ago