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r-ruslan [8.4K]
3 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.)
Physics
1 answer:
Ilya [14]3 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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3 years ago
List the largest-aperture single telescope currently in use in each of the following bands of the electromagnetic spectrum: radi
tankabanditka [31]

Explanation:

Radio Telescope:

1. FAST (Five Hundred Meter Aperture Spherical Telescope), China is the largest single dish radio telescope in the world with a dish of diameter 500 meters. Although its construction is over but its not yet functional.

2. Arecibo Observatory, Puerto Rico is the largest single dish <em>operational</em> radio telescope having a single dish of diameter 305 meters..

X - ray Telescope: Chandra X ray observatory also known as Advanced X-ray Astrophysics Facility (AXAF) is the largest x-ray telescope with aperture of 1.2 meters. It can detect X-ray sources which are 100 times fainter in comparison to any previous telescope.

Gamma ray telescope: Fermi Gamma ray telescope is the largest gamma ray telescope.

4 0
3 years ago
Which of the following tools can be used to determine humidity? Select all that apply
Y_Kistochka [10]
Thermometer there's others you can use but i know that's one of them
7 0
3 years ago
Read 2 more answers
Steam in a heating system flows through tubes whose outer diameter is 5 cm and whose walls are maintained at a temperature of 13
svet-max [94.6K]

Answer:

5945.27 W per meter of tube length.

Explanation:

Let's assume that:

  • Steady operations exist;
  • The heat transfer coefficient (h) is uniform over the entire fin surfaces;
  • Thermal conductivity (k) is constant;
  • Heat transfer by radiation is negligible.

First, let's calculate the heat transfer (Q) that occurs when there's no fin in the tubes. The heat will be transferred by convection, so let's use Newton's law of cooling:

Q = A*h*(Tb - T∞)

A is the area of the section of the tube,

A = π*D*L, where D is the diameter (5 cm = 0.05 m), and L is the length. The question wants the heat by length, thus, L= 1m.

A = π*0.05*1 = 0.1571 m²

Q = 0.1571*40*(130 - 25)

Q = 659.73 W

Now, when the fin is added, the heat will be transferred by the fin by convection, and between the fin and the tube by convection, thus:

Qfin = nf*Afin*h*(Tb - T∞)

Afin = 2π*(r2² - r1²) + 2π*r2*t

r2 is the outer radius of the fin (3 cm = 0.03 m), r1 is the radius difference of the fin and the tube ( 0.03 - 0.025 = 0.005 m), and t is the thickness ( 0.001 m).

Afin = 0.006 m²

Qfin = 0.97*0.006*40*(130 - 25)

Qfin = 24.44 W

The heat transferred at the space between the fin and the tube will be:

Qspace = Aspace*h*(Tb - T∞)

Aspace = π*D*S, where D is the tube diameter and S is the space between then,

Aspace = π*0.05*0.003 = 0.0005

Qspace = 0.0005*40*(130 - 25) = 1.98 W

The total heat is the sum of them multiplied by the total number of fins,

Qtotal = 250*(24.44 + 1.98) = 6605 W

So, the increase in heat is 6605 - 659.73 = 5945.27 W per meter of tube length.

5 0
3 years ago
if you run around a circle at 4.5 m/s and the circle has a radius of 7.7 m, what is your centripetal acceleration?
madreJ [45]

Answer:

Centripetal acceleration,

a_{c} =2.63\ m/s^{2} }

Explanation:

Centripetal acceleration:

Centripetal acceleration is the idea that any object moving in a circle, in something called circular motion, will have an acceleration vector pointed towards the center of that circle.

Centripetal means towards the center.

Examples of centripetal acceleration (acceleration pointing towards the center of rotation) include such situations as cars moving on the cicular part of the road.

An acceleration is a change in velocity.

Formula for Centripetal acceleration:

a_{c} =\frac{(velocity)^{2} }{radius}

Given here,

Velocity = 4.5 m/s

radius = 7.7 m

To Find :

a_{c} = ?

Solution:

We have,

a_{c} =\frac{(velocity)^{2} }{radius}

Substituting  given value in it we get

a_{c} =\frac{(4.5)^{2}}{7.7} \\\\a_{c} =\frac{20.25}{7.7}\\\\a_{c} =2.629\ m/s^{2} \\\\\therefore a_{c} =2.63\ m/s^{2

Centripetal acceleration,

a_{c} =2.63\ m/s^{2} }

7 0
3 years ago
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