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2 years ago

How an application of atmospheric device work?example siphon

Physics

1 answer:

Vika [28.1K]2 years ago

6 0

Answer:

A practical siphon, operating at typical atmospheric pressures and tube heights, works because gravity pulling down on the taller column of liquid leaves reduced pressure at the top of the siphon (formally, hydrostatic pressure when the liquid is not moving).

I hope it's helpful!

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What is one common electrostatic phenomenon

Mademuasel [1]

Answer:

There are many examples of electrostatic phenomena, from those as simple as the attraction of the plastic wrap to one's hand after it is removed from a package to the apparently spontaneous explosion of grain silos, the damage of electronic components during manufacturing, and photocopier & laser printer operation

6 0

3 years ago

A 0.12 kg body undergoes simple harmonic motion of amplitude 8.5 cm and period 0.20 s. (a) What is themagnitude of the maximum f

Neporo4naja [7]

Answer:

a)F=698.83 N

b)K=8221.56 N/m

Explanation:

Given that

mass ,m = 0.12 kg

Amplitude ,A= 8.5 cm

time period ,T = 0.2 s

We know that

$T=\dfrac{2\pi}{\omega}$

${\omega}=\dfrac{2\pi }{0.2}\ rad/s$

${\omega}=31.41\ rad/s$

We know that

${\omega}^2=m\ K$

K=Spring constant

$K=\dfrac{\omega^2}{m}$

$K=\dfrac{31.41^2}{0.12}\ N/m$

K=8221.56 N/m

The maximum force F

F= K A

F= 8221.56 x 0.085 N

F=698.83 N

a)F=698.83 N

b)K=8221.56 N/m

3 0

3 years ago

A tomato of mass 0.18 kg is dropped from a tall bridge. If the tomato has a speed of 11 m/s just before it hits the ground, what

kondor19780726 [428]

The kinetic energy of the tomato is :

K.E = 1/2 mv^2

K.E = 1/2 x 0.18 kg x 11 m/S^2

K.E = 0.99

Hope this helps

7 0

3 years ago

The six statements below represent Newton's three laws of motion and Kepler's three laws of planetary motion. Match each stateme

mote1985 [20]

Answer:

1. Force = mass x acceleration - Newton

2. A planet moves faster in the part of its orbit nearer the Sun and slower when farther from the Sun, sweeping out equal areas in equal times - Kepler

3. For any force, there is an equal and opposite reaction force - Newton .

4. An object moves at constant velocity if there is no net force acting upon it - Newton

5. The orbit of each planet about the Sun is an ellipse with the Sun at one focus - Kepler.

6. More distant planets orbit the Sun at slower average speeds, obeying the precise mathematical relationship p2-a3 - Kepler.

Explanation:

The three laws of planetary motion formulated by Johannes Kepler or Kepler's laws of planetary motion:

The first law states that the planets move around the Sun in an elliptical orbit with the Sun at one of the foci.
The second law states that the line segment joining a planet to the Sun sweeps out equal areas in equal time.
The third law states that the square of the orbital period (p) of a planet is directly proportional to the cube of the mean distance (a) from the Sun (or semi-major axis of its orbit) i.e., p² is proportional to a³.

The three laws of motion formulated by Sir Isaac Newton or Newton's laws of motion:

The first law, also known as the law of inertia states that an object at rest or moves at a constant velocity will remain at rest or keep moving at a constant velocity unless it is acted upon by a force.
The second law states that the total force (F) applied on an object is directly related to the acceleration (a) of that object produced by the applied force and the mass (m) of the object, i.e., F = ma (assuming the mass m is constant).
The third law, also known as the law of action and reaction states that when an object exerts a force on another object, then the latter exerts a force equal in magnitude and opposite in direction on the former object i.e., for every action, there is an equal and opposite reaction. The example includes the recoiling of a gun when it fires a bullet forward.

5 0

3 years ago

.Find the uncertainty in a calculated electrical potential difference from the measurements of current and resistance. Electric

maw [93]

Answer:

a) The uncertainty in calculated V, ΔV = 25.3

b) The uncertainty in calculated v, Δv = 0.41 m/s

c) The uncertainty in calculated V, ΔV = 22.2 V

Explanation:

We'll use Upper-Lower Bounds method of uncertainty to estimate the uncertainties.

a) I = 5.1 A, ΔI = 0.3 A

I = (5.1 ± 0.3) A

R = 77.5 ohms, ΔR = 0.4 ohms

R = (77.5 ± 0.4) ohms

V = IR = 5.1 × 77.5 = 395.25 V

The lower bound for the voltage will be calculated using the lower bounds for the current and resistance

Iₗ = 5.1 - 0.3 = 4.8 A

Rₗ = 77.5 - 0.4 = 77.1 ohms

Vₗ = 4.8 × 77.1 = 370.08 V

The upper bound for the voltage will be calculated using the upper bounds for the current and resistance

Iᵤ = 5.1 + 0.3 = 5.4 A

Rᵤ = 77.5 + 0.4 = 77.9 ohms

Vᵤ = 5.4 × 77.9 = 420.66 V

The average of the differences from the mean voltage/true value is 25.3 V

V = 395.25 V, Δ = 25.3V

V = (395.25 ± 25.3) V

b) x = 2.9 m, Δx = 0.3 m

x = (2.9 ± 0.3) m

t = 4.4 s, Δt = 1.8 s

t = (4.4 ± 1.8) ohms

v = x/t = 2.9/4.4 = 0.659 m/s

The lower bound for average speed will be calculated using the lower bounds for distance and upper bounds for time.

xₗ = 2.9 - 0.3 = 2.6 m

tᵤ = 4.4 + 1.8 = 6.2 s

vₗ = 2.6/6.2 = 0.419 m/s

The upper bound for the average speed will be calculated using the upper bound for the distance and lower bound for time

xᵤ = 2.9 + 0.3 = 3.2 m

tₗ = 4.4 - 1.8 = 2.6 s

vᵤ = 3.2/2.6 = 1.231 m/s

The average of the differences from the mean average speed/true value is 0.41 m/s

v = 0.659 m/s, Δv = 0.41 m/s

v = (0.659 ± 0.41) m/s

c) ) I = 9.8 A, ΔI = 0.5 A

I = (9.8 ± 0.5) A

R = 40.5 ohms, ΔR = 0.2 ohms

R = (40.5 ± 0.2) ohms

V = IR = 9.8 × 40.5 = 396.9 V

The lower bound for the voltage will be calculated using the lower bounds for the current and resistance

Iₗ = 9.8 - 0.5 = 9.3 A

Rₗ = 40.5 - 0.2 = 40.3 ohms

Vₗ = 9.3 × 40.3 = 374.79 V

The upper bound for the voltage will be calculated using the upper bounds for the current and resistance

Iᵤ = 9.8 + 0.5 = 10.3 A

Rᵤ = 40.5 + 0.2 = 40.7 ohms

Vᵤ = 10.3 × 40.7 = 419.21 V

The average of the differences from the mean voltage/true value is 22.2 V

V = 396.9 V, Δ = 22.2 V

V = (396.9 ± 22.2) V

7 0

3 years ago

How an application of atmospheric device work?example siphon​

How an application of atmospheric device work?example siphon