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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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Two rings of radius 5 cm are 15 cm apart and concentric with a common horizontal -axis. The ring on the left carries a uniformly
Lina20 [59]

Answer:

Explanation:

Radius of ring r = .05 m

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E = k Q x  / ( x² + r² )³/²   ;  Q is charge on the ring , x is distance of point  from the center of the ring and r is radius of the ring

electric field due to first ring at middle point or at x = 7.5 cm

E = 9 x 10⁹ x 33 x 10⁻⁹ x 7.5 x 10⁻² / ( 7.5² + 5² )³/² x 10⁻³

= 9 x 10⁹ x 33 x 10⁻⁹ x 7.5  / ( 7.5² + 5² )³/² x 10⁻¹

= 9 x 10⁹ x 33 x 10⁻⁹ x 7.5 / 73.23

=  30.41 N/C

The same field will be created by the other ring at the middle point because charge on the ring is same in magnitude . Due to negative charge on the second ring , field due to both the rings will align in the same direction.

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7 0
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4. A driver travels 135 km, east in 1.5 h, stops for 45 minutes for lunch,
stira [4]

Answer:

The driver's average velocity is 82.35 km/h.

Explanation:

Given:

The motion of the driver can be divided into 3 parts:

i. Displacement of the driver in 1.5 hours = 135 km

ii. Rest for 45 minutes.

iii. Displacement in next 2 hours = 215 km

The direction of motion remains same (east).

Now, total displacement of the driver is, D_{Total}=135+215=350 km.

Rest time is 45 minutes. Converting it to hours, we need to use the conversion factor 1\textrm{ min} = \frac{1}{60} hour.

So, 45 minutes in hours is equal to \frac{45}{60}=0.75 hours.

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Average velocity is given as:

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Therefore, the driver's average velocity is 82.35 km/h

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