Option D: 2 m/s is the average velocity of the water flowing through a pipe with a cross-sectional area of 0.002 m2 at a mass flow rate of 4 kg/s.
According to the question:
cross-sectional area of the pipe = 0.002m²
Mass flowrate = 4 kg/s
Density of water = 1000 kg/m³
We are asked to find, average velocity =?
Average velocity is the net or total displacement covered by a body in a given time. The mass flow rate divided by the pipe's cross-sectional area and density ratio is the formula for calculating a fluid's average velocity.
As a result, the water's average flow rate through the pipe is provided by:
v = m / (ρ × A)
where, v is the average velocity, m is the mass flow rate, ρ is the density of water, and A is the cross-sectional area of the pipe. Substituting the values in the above equation we get:
v = 4 / (1000 × 0.002)
v = 2m/s
Therefore, the average velocity of water flowing through a pipe of cross-sectional area of 0.002m² is 2m/s.
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Correct question is:
Water flows through a pipe with a cross-sectional area of 0.002 m2 at a mass flow rate of 4 kg/s. The density of water is 1 000 kg/m3. Determine its average velocity. Multiple choice question.
20 m/s
200 m/s
0.02 m/s
2 m/s
0.2 m/s
a tired worker pushes a heavy (100-kg) crate that is resting on a thick pile carpet. the coefficients of static and kinetic friction are 0.6 and 0.4, respectively. the worker pushes with a force of 600 n. the frictional force exerted by the surface is
When a tired worker pushes a heavy (100-kg) crate that is resting on a thick pile carpet, the frictional force exerted by the surface on the crate is 588 N.
When a tired worker pushes a heavy (100-kg) crate that is resting on a thick pile carpet, the frictional force exerted by the surface can be calculated as follows:
The weight of the crate = m × g = 100 kg × 9.8 m/s² = 980 N
Force applied by the worker = F = 600 N
The force of friction acting on the crate is given by the following formula:
Ff = μF
Where, μ is the coefficient of friction, F is the normal force acting on the crate.
Notes: The normal force is equal and opposite to the weight of the crate. i.e., N = 980 N1. The frictional force exerted by the surface on the crate is the static frictional force initially. Hence, we use the coefficient of static friction for our calculation.
2. If the force applied by the worker is not enough to overcome the static frictional force, then the crate will not move and the frictional force will remain static friction.
3. Once the crate starts moving, the static friction will convert to kinetic friction. Hence, we will use the coefficient of kinetic friction if the force applied by the worker is greater than the force of static friction. Initially, the force applied by the worker is less than the force of static friction, hence the frictional force exerted on the crate will be the static frictional force.
Frictional force = Ff = μN
The normal force acting on the crate = Weight of the crate = 980 N
Frictional force =
Ff = μN
= 0.6 × 980 N
= 588 N
Therefore, the frictional force exerted by the surface on the crate is 588 N.
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given two identical iron bars, one of which is a permanent magnet and the other unmagnetized, how could you tell which is which by using only the two bars?
There are two identical iron bars, one of which is a permanent magnet and the other unmagnetized. We can identify that: when the magnetized bar is brought near the other bar, it will stick to it, indicating that it is magnetized. The bar that does not stick is unmagnetized.
Iron bars are used to make permanent magnets by a process called magnetization. Permanent magnets are composed of atoms and aligned electrons that have magnetic properties. The other bar that is not magnetized does not have aligned electrons, so it will not attract other magnets as a magnetized bar would.
The direction of a magnetic field will change when a magnet is brought near it. The North Pole will attract the South Pole, and they will come together. The North Pole will repel the North Pole, and the South Pole will repel the South Pole. The magnetized bar will be attracted to the unmagnetized bar, and the unmagnetized bar will not be attracted to the magnetized bar.
As a result, when the magnetized bar is brought near the other bar, it will stick to it, indicating that it is magnetized. The bar that does not stick is unmagnetized. Thus, with the aid of two bars, one magnetized and the other unmagnetized, we can determine which is which.
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pry on the power steering reservoir to adjust the tension of the power steering belt. true or false?
The statement "pry on the power steering reservoir to adjust the tension of the power steering belt" is: false.
The tension of the power steering belt is adjusted by adjusting the position of the power steering pump. There is a tension adjustment bolt on the power steering pump that is used to adjust the tension of the power steering belt. The adjustment bolt should be turned clockwise or counterclockwise to adjust the tension of the belt.
A belt tension gauge may be used to ensure that the belt is properly tensioned. A pry bar should not be used on the power steering reservoir to adjust the tension of the power steering belt. This could cause damage to the reservoir or other components of the power steering system. The reservoir should be inspected for damage or leaks, but it should not be used to adjust the tension of the belt.
In summary, the tension of the power steering belt should be adjusted by adjusting the position of the power steering pump, not by prying on the power steering reservoir.
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the maximum horizontal distance from the center of the robot base to the end of its end effector is known as .
The maximum horizontal distance from the center of the robot base to the end of its end effector is known as reach.
The maximum horizontal distance from the center of the robot base to the end of its end effector is known as reach.
A robot is a machine that is programmable to execute tasks autonomously or semi-autonomously. Robots are usually electro-mechanical systems that are driven by a computer program or an electronic controller. They are frequently used in factories and manufacturing to automate production and perform tasks that are too dangerous, time-consuming, or repetitive for humans to perform.
Robotics is a branch of technology that deals with the design, construction, operation, and application of robots. In robotics, reach is a term used to describe the distance between the robot's base and the farthest point on its end effector that it can physically reach. It is usually given in three dimensions:
horizontal reach, vertical reach, and depth reach. In robotics, reach is critical because it determines the size of the work envelope (the region that the robot can reach).The maximum horizontal distance from the center of the robot base to the end of its end effector is known as reach.
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calculate the force required to stop a car of mass 1400 kg in 2 seconds if it is moving with a velocity of 10 m/s.
The force required to stop a car of mass 1400 kg in 2 seconds if it is moving with a velocity of 10 m/s is 7000 N in the opposite direction to the car's motion.
Calculate the force required to stop a car of mass 1400 kg in 2 seconds if it is moving with a velocity of 10 m/s.
To solve the given problem, we can use the equation:
F = (m * Δv) / Δt
where F = force
required to stop the carm = mass of the car Δv = change in velocity = final velocity - initial velocityΔt = time taken to stop the car.
Given, mass of the car, m = 1400 kg Initial velocity, u = 10 m/s Final velocity, v = 0 m/s Time taken to stop, t = 2 seconds Therefore, Δv = v - u = 0 - 10 = -10 m/s
Substituting the given values in the above equation, we get:
F = (m * Δv) / Δt = (1400 kg * (-10 m/s)) / (2 s) = -7000 N
Here, the negative sign indicates that the force required to stop the car is acting in the opposite direction to the car's motion.
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the intensity of the sound of a television commercial is 10 times greater than the intensity of the television program it follows. by how many decibels does the loudness increase?
The television commercial loudness increases by 10 decibels.
Increase in the Intensity of soundThe decibel (dB) scale is a logarithmic measure of sound intensity. The intensity of a sound is measured in watts per square meter and the decibel scale is a way to express the relative loudness of a sound, compared to a reference level.
A 10 dB increase in intensity is a 10-fold increase in sound power. This means that a sound with an intensity of 10 watts per square meter is 10 times louder than a sound with an intensity of 1 watt per square meter.
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an object falls freely from rest on a planet where the acceleration due to gravity is 20 m/s2. after 5 seconds, the object will have a speed of
Answer : If an object falls freely from rest on a planet where the acceleration due to gravity is 20 m/s2 then after 5 seconds, the object will have a speed of 100 m/s
This can be calculated using the equation v = a*t, where v is the velocity, a is the acceleration due to gravity, and t is the time elapsed. Therefore, in this case, v = 20 m/s2 * 5 s = 100 m/s. These values are given in question, so we just have to put them in equation.
Since the object is falling freely, its acceleration remains constant and it follows a uniform acceleration motion. Therefore, the velocity of the object will increase linearly with time. After 10 seconds, the velocity will double to 200 m/s, and so on.
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measurements show a certain star has a very high luminosity (100,000 x the sun's) while its temperature is quite cool (3500 k). how can this be?
The star might be quite large in size, with a much larger surface area than the sun. This would increase its luminosity despite its cooler temperature.
The star has a high luminosity (100,000 x the sun's) and a cool temperature (3500 K) because of its size.
A star's luminosity is proportional to its size, so if a star is very large, it can have a high luminosity even if it is relatively cool.
Another possibility is that the star is in a phase of its life cycle where it has expanded and cooled, such as a red giant or supergiant, but still retains a high luminosity due to its large size.
These stars have relatively low surface temperatures, but their large sizes give them very high luminosities.
Therefore, this star is likely very large and thus has a very high luminosity despite its low temperature.
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