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Harlamova29_29 [7]
3 years ago
9

Shutting the fluid discharge of an air-operated reciprocating pump will cause the pump to ?

Physics
2 answers:
Misha Larkins [42]3 years ago
8 0
Had to look for the options and here is my answer. What happens when the fluid discharge of an air-operated reciprocating pump is shut, this will cause the pump to OVERSTROKE. Overstroke happens when the engine is switching in a normally-closed manner.  
KATRIN_1 [288]3 years ago
4 0

Explanation:

In pneumatic pumps, compressed air is utilized to generate a force that is used to channel fluids through the piping system. If the discharge line of the pneumatic pump is blocked the pump will stop and the fluid won't flow, the pump won't run and airflow will also stop. Pneumatic pumps have weak drivers. The compressed air which runs the pump is also at low pressure.

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What caused rutherford to propose a revised model of the atom
Wewaii [24]

Explanation:

Rutherford proposed a revised model for the atom, called the planetary model. The previous model of the atom was Thomson's Plum Pudding Model which consisted of freely moving positive and negative charges inside the atom.

Rutherford proposed his model after an experiment he conducted called the Gold Foil Experiment. This experiment consisted of a thin gold sheet into which alpha particles were shot upon and they were detected by a sensor. The image attached will give a better explanation of this. In this experiment he shot a beam of alpha particles(helium nucleus) at a thin sheet of gold. Rutherford hypothesised that there should be minimum deflection of the positively charged alpha particles occuring due to the repulsion of the alpha particle with the positive charges in the thin gold sheet. This was not the case.

However what he found was that most of alpha particles went straight through the thin sheet of gold but some were reflected back to him. This surprised him. Hence he proposed that most of the atom must be empty space as most of the alpha particles went straight through the sheet and there must be a heavy nucleus inside the atom causing the alpha particles to bounce back.

7 0
3 years ago
Calculate the period of a satellite orbiting the Moon, 98 kmkm above the Moon's surface. Ignore effects of the Earth. The radius
beks73 [17]

Answer:

3.6*10^18s

Explanation:

To find the period of the satellite

We need to apply kephler's third law

Which is

MP² = (4π²/G) d³

d=semi-major axis which is the distance from center of moon = 98km+1740km = 1838km

where M= mass of the moon = 7.3x10^22kg

P=period

G=newtonian gravatational constant= 6.67x10^-11

To find the Period solve for P

P = √[(4π²/G M)xd³]

P=√(4 π²/6.67x10^-22*7.3x10^22kg) x (1.838x10^6m)³]

= 3.6*10^18s

6 0
3 years ago
Explain why gases are more compressible than solid
Oduvanchick [21]

Answer:

If we put pressure on a solid or a liquid, there is essentially no change in volume. ... The kinetic-molecular theory explains why gases are more compressible than either liquids or solids. Gases are compressible because most of the volume of a gas is composed of the large amounts of empty space between the gas particles.

Explanation:

7 0
3 years ago
A 10kg box accelerates forward at a rate of 12 m/s^2. What is the force acting on the box?
Allushta [10]

Answer:

120N

Explanation:

Newton's second law formula: F= ma

given that m = 10 kg, a = 12 m/s^2

F = ma = 10 kg * 12 m/s^2 = 120 kgm/s^2 = 120 N

5 0
2 years ago
A 0.20-kg object is attached to the end of an ideal horizontal spring that has a spring constant of 120 N/m. The simple harmonic
Umnica [9.8K]

Answer:

<em>A = 6.9 cm</em>

Explanation:

<u>Simple Harmonic Motion</u>

A mass-spring system is a common example of a simple harmonic motion device since it keeps oscillating when the spring is stretched back and forth.

If a mass m is attached to a spring of constant k and they are set to oscillate, the angular frequency of the motion is

\displaystyle w=\sqrt{\frac{k}{m}}

The equation for the motion of the object is written as a sinusoid:

\displaystyle X=A\ cos\ w\ t

Where A is the amplitude.

The instantaneous speed is computed as the derivative of the distance

\displaystyle X'=V=-A\ w\ sin\ w\ t

And the maximum speed is

\displaystyle V_{max}= A\ w

Solving for the amplitude

\displaystyle A= \frac{V_{max}}{w}

Computing w

\displaystyle w =\sqrt{\frac{120}{0.2}}=24.5\   rad/ s

Calculating A

\displaystyle A=\frac{1.7}{24.5}=0.069\ m

\displaystyle \boxed{A=6.9\ cm}

7 0
3 years ago
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