Engineering High School

Everyday materials that do a better job of filtering microscopic particles
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Answers

Answer 1

The materials such as high-quality furnace filters, tightly woven cotton fabric, electrostatic cloth, and activated carbon are everyday options that provide effective filtration of microscopic particles.

There are several everyday materials that can effectively filter microscopic particles.

One commonly used material is high-quality furnace filters, which are designed to capture small particles and allergens. These filters consist of dense fibers that create a barrier, preventing microscopic particles from passing through.

Another effective material is tightly woven cotton fabric, such as quilting cotton or flannel. The tight weave of these fabrics can trap small particles, making them suitable for homemade face masks or air filtration systems.

Electrostatic cloth, like microfiber towels, also has excellent filtering capabilities. The fibers in these cloths carry an electric charge that attracts and captures particles, including microscopic ones.

Activated carbon is another material used for filtering microscopic particles. It works by adsorbing particles onto its surface through a process called adsorption. Activated carbon is commonly found in air purifiers, water filters, and respirators.

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Related Questions

Explain the problem with the initial step in the implementation plan in the following scenario.

Situation: Frank, the project lead, was assigned two programmers and a writer for the project team. He decides that is not enough, and he hires two additional programmers and a second writer.

Answers

Answer:

The problem with the initial implementation plan is that Frank, the project lead, made changes to the team composition without proper justification or evaluation.

Explanation:

Question 5 (5 points)
A Vex motor has a has a voltage of 9 volts running through it, and a
current of 15 mA. What is the power?
(Precision of 0.000)

Answers

The power of the Vex motor is 0.135 watts.

How to solve

The power of the Vex motor can be calculated using the formula:

Power (P) = Voltage (V) * Current (I)

Substituting the given values:

P = 9 volts * 15 mA

To simplify the calculation, we convert milliamperes (mA) to amperes (A):

P = 9 volts * 0.015 amperes

P = 0.135 watts

Therefore, the power of the Vex motor is 0.135 watts.


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Which of the following statements is not true about proper tire care?

Answers

Answer:

The answer is 'Tire tread depth can only be checked by a properly trained tire expert.'

Explanation:

Yooooooooooooooooooooooo

Answers

1. Dependability

2. Commission

3. PRAM

Energy problem formulas
Potential Energy = mgh
v = velocity or speed
Kinetic energy = mv²
9 = 9.8 m/s²
m = mass in kg
(Precision of 1)
h = height in meters
You serve a volleyball with a mass of 3.1 kg. The ball leaves your hand
with a speed of 20 m/s.
a) Calculate Kinetic Energy
(Precision of 1)
b) How much work is being done to the system to create this kinetic
energy?


I need this asap

Answers

a. The kinetic energy is 620 J

b. The amount of work done is equal to the kinetic energy. In this case, the work done is 620 J.

How to calculate the value

a. The formula for kinetic energy is:

KE = 1/2mv²

where:

KE is the kinetic energy in joules (J)

m is the mass in kilograms (kg)

v is the velocity in meters per second (m/s)

In this case, we have:

m = 3.1 kg

v = 20 m/s

So, the kinetic energy is:

KE = 1/2(3.1 kg)(20 m/s)²

= 620 J

b) How much work is being done to the system to create this kinetic energy?

Work is done to the system to create kinetic energy. The amount of work done is equal to the kinetic energy. In this case, the work done is 620 J.

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Amy uses 30N of force to push a lawn mower 5 meters. How much
work does she do?
(Precision of 1)
Really need this

Answers

Amy does 150 Joules (J) of work using 30N of force to push a lawn mower 5 meters.

Here, we use Joules (J) to represent the amount of energy spent when a force of one newton is applied over a distance of one meter.

How to the work done by Amy?

To calculate the work done by Amy, we will multiply the force she applies (30 N) by the distance she moves the lawn mower (5 m).

Work = Force × Distance

Given:

Force = 30 N

Distance = 5 m

Plugging in the values:

Work = 30 N × 5 m

Work = 150 N·m

Therefore, Amy does 150 Joules (J) of work.

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Explain why the following scenario fails to meet the definition of an implementation plan.

Situation: Ian, the project lead for flight simulator software, has identified his project team, including three programmers, two junior programmers, and a writer. He has requested a budget of $250,000 for six months and is preparing to begin coding.

Answers

Explanation:

based on the information provided, it is difficult to determine why this scenario would fail to meet the definition of an implementation plan. However, it is possible that additional information is needed to fully understand the situation and evaluate whether or not it meets the criteria for an implementation plan. An implementation plan typically includes specific steps and timelines for completing a project, as well as a budget and resources needed to carry out those steps. It may also include contingency plans and risk management strategies. Without more detailed information about Ian's plan and how he intends to execute it, it is unclear if this scenario meets the criteria for an implementation plan.

Question 6 (5 points)
A Tesla car weighs 250,000 Newtons Travels a distance of 64 M in 2
seconds (115 km / hour). The Voltage of the Battery of the car is 375
Volts. The Current is 32 kA.
What is the % Efficiency of the power

Answers

The efficiency of the power is 70%. This means that 70% of the energy supplied by the battery is used to move the car, and the remaining 30% is lost as heat and other forms of energy loss.

How to calculate the value

The efficiency of the power is calculated as follows:

Efficiency = Work output / Work input

The force is the weight of the car, which is 250,000 N, and the distance is 64 m. So the work output is:

Work output = 250,000 N * 64 m

= 16,000,000 J

The voltage is 375 V, the current is 32 kA, and the time is 2 seconds. So the work input is:

Work input = 375 V * 32 kA * 2 s = 23,040,000 J

The efficiency is then:

Efficiency = 16,000,000 J / 23,040,000 J = 70%

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