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scoray [572]
1 year ago
12

oil is flowing through a tube that has two different cross-sectional areas. at position a where the radius of the tube is 7.0 cm

, the mass flow rate of the oil is 0.025 kg/s. what is the mass flow rate at position b where the radius of the tube is 3.5 cm?
Physics
1 answer:
AleksAgata [21]1 year ago
5 0

According to the continuity equation, the rate at which mass enters the system equals the rate at which mass exits in any steady state process.

An equation that explains the movement of a particular quantity is a continuity equation, also known as a transport equation. Although it can be applied generally to any significant quantity, it is extremely simple and useful when used with preserved quantities.

The radius is seven centimeters, and the mass flow rate is 0 to 5 kg/s. Find the mass flow rate at a point with a 3.5 cm radius. We can consequently deduce that based on the equation. As we all know, the mass flow rate is constant.

If the rate of mass entering and leaving the system is equal, the rate of mass leaving the system should be processed.

The mass flow rate air section A and the mass flow rated section B are equivalent, according to the continuity equation. Mass flow rate in section B is therefore 0.02, or five kilograms per second.

To know more about continuity equation, click on the link below:

brainly.com/question/19566865

#SPJ4

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A 110 V power line is protected by a 15 A fuse. What is the maximum number of 400 W lamps that can be simultaneously operated in
Nitella [24]

Answer:

Total number of lamps will be 4            

Explanation:

We have given power of the lamp W = 400 watt

Potential difference across the lamp V=110 volt

We know that power is equal to P=VI

So 400=110\times I

I=3.636A

Total current is given 15 A

As it is given that lamps are connected in parallel so total current is the sum of current through each lamp

So number of lamp will be n=\frac{15}{3.636}=4.125

As the lamp can not be in negative

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How does light from a headlamp use a lens and a mirror to produce a narrow beam?
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2 years ago
Read 2 more answers
A box of mass 60 kg is at rest on a horizontal floor that has a static coefficient of friction of 0.6 and a kinetic coefficient
gavmur [86]

Answer:

a) The minimum force required to start moving the box is 352.86 N

b) i) The friction force for the box in motion is 147.025 N

ii) The acceleration of box is 4.21625 m/s²

Explanation:

The parameters of the box at rest and the floor are;

The mass of the box = 60 kg

The static coefficient friction of the floor = 0.6

The kinetic coefficient friction of the floor = 0.25

Frictional force = Normal force × Friction coefficient

For an horizontal floor and the box laying on the floor, we have;

The normal force = The weight of the box = Mass of the box × Acceleration due to gravity, g

The acceleration due to gravity, g = 9.81 m/s²

The weight of the box  = 60 × 9.81 = 588.6 N

a) The static coefficient gives the frictional force observed by the box and which must be surpassed to bring about motion

Therefore;

The minimum force required to start moving the box = The static frictional force = Weight of the box × The static coefficient of friction

The minimum force required to start moving the box = 588.1 × 0.6 = 352.86 N

The minimum force required to start moving the box = 352.86 N

b) i) When an horizontal force of 400 N is applied, the applied force is larger than the static friction force, and the box will be in motion with the kinetic coefficient of friction being the source of friction

The friction force for the box in motion = 588.1 × 0.25 = 147.025 N

ii) The force, F with the box is in motion, is given as follows;

F = Mass of box × Acceleration of box, a = Applied force - Kinematic friction force

F = 60 × a = 400 - 147.025 = 252.975 N

60 × a = 252.975 N

a = 252.975 N/(60 kg) = 4.21625 m/s²

Acceleration of box, a = 4.21625 m/s².

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The correct answer is C.)

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