Consider the products you use and the activities you perform on a daily basis. Describe three examples that use both SI units and customary units for measurement.
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Three activities that I can do on a daily basis that involve both metric units (SI units) and customary units are: measuring the length of a door with a tape measure, which includes both SI units and customary units (like feet, inches, and centimeters); baking a cake that calls for one teaspoon (customary unit) of baking soda, which can also be converted to four grams (SI unit); and weighing myself on a weighing scale, which can be measured in pound and kilogram (metric unit).
Answer: Three examples of activities that I can perform on a daily basis that involves both metric units (SI units) and customary units include: measuring the length of a door using a tape measure, which includes both SI units and customary units (like feet, inches, and centimeters); baking a cake that requires one teaspoon (customary unit) of baking soda, which could also be converted into four grams (SI unit); weighing myself on a weighing scale, which can be measured by pounds (customary unit) or kilograms (metric unit).
Explanation:I big brain :) (Not Really I Just Wanted To Help) I hope this helped! ;)
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At its current short-run level of production, a firm's average variable costs equal $20 per unit, and its average fixed costs equal $30 per unit. Its total costs at this production level equal $2,500. A. What is the firm's current output level? B. What are its total variable costs at this output level? C. What are its total fixed costs?
An aquifer has three different formations. Formation A has a thickness of 8.0 m and hydraulic conductivity of 25.0 m/d. Formation B has a thickness of 2.0 m and a conductivity of 142 m/d. Formation C has a thickness of 34 m and a conductivity of 40 m/d. Assume that each formation is isotropic and homogeneous. Compute both the overall horizontal and vertical conductivities.
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An inflatable structure has the shape of a half-circular cylinder with hemispherical ends. The structure has a radius of 40 ft when inflated to a pressure of 0.60 psi. A longitudinal seam runs the entire length of the structure. The seam fails in tension when the load is 600 pounds per inch of seam. What is the factor of safety with respect to longitudinal seam failure?
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Answer: c) 450 kPa
Explanation:
Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.
(At constant temperature and number of moles)
where,
= initial pressure of gas = 150 kPa
= final pressure of gas = ?
= initial volume of gas = v L
= final volume of gas =
Therefore, the new pressure of the gas will be 450 kPa.
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Answer:
I want to believe the program is to be written in java and i hope your question is complete. The code is in the explanation section below
Explanation:
import java.util.Date;
public interface Downloadable {
//abstract methods
public String getUrl();
public Date getLastDownloadDate();
}
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Answer:
The pressure rise in the tire when the air temperature in the tire rises to 50°C is 337.43 KPa.
The amount of air that must be bled off to restore pressure 0.007 Kg
Explanation:
Knowing
T1 = 25°C = 298 K
T2 = 50°C = 323 K
volume of the tire = V = 0.025
P = 210 kPa (gage)
Pabs = 210 + 101 = 311 KPa
Before the temperature rise
P1 V1 = m1 R1 T1
m1 =
After the temperature rise
P2 =
after bleeding the pressure and the volume returns to its first value
P1 = P2 and V1 = V2
then
m2 =
m2 =
mbleed = m1 - m2 --> mbleed = 0.91 - 0.84 = 0.007 Kg
P2 = 337.43 KPa
mbleed = 0.007 Kg
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Answer:
The air heats up when being compressed and transefers heat to the barrel.
Explanation:
When a gas is compressed it raises in temperature. Assuming that the compression happens fast and is done before a significant amount of heat can be transferred to the barrel, we could say it is an adiabatic compression. This isn't exactly true, it is an approximation.
In an adiabatic transformation:
For air k = 1.4
SO
SInce it is compressing, the fraction P1/P0 will always be greater than one, and raised to a positive fraction it will always yield a number greater than one, so the final temperature will be greater than the initial temperature.
After it was compressed the hot air will exchange heat with the barrel heating it up.
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Answer:
il(t) = e^(-100t)
Explanation:
The current from the source when the switch is closed is the current through an equivalent load of 15 + 50║50 = 15+25 = 40 ohms. That is, it is 80/40 = 2 amperes. That current is split evenly between the two parallel 50-ohm resistors, so the initial inductor current is 2/2 = 1 ampere.
The time constant is L/R = 0.20/20 = 0.01 seconds. Then the decaying current is described by ...
il(t) = e^(-t/.01)
il(t) = e^(-100t) . . . amperes
Determine the temperature at the center plane of the brass plate after 3 minutes of cooling.
Answers
Answer:
809.98°C
Explanation:
STEP ONE: The first step to take in order to solve this particular Question or problem is to find or determine the Biot value.
Biot value = (heat transfer coefficient × length) ÷ thermal conductivity.
Biot value = (220 × 0.1)÷ 110 = 0.2.
Biot value = 0.2.
STEP TWO: Determine the Fourier number. Since the Biot value is greater than 0.1. Tis can be done by making use of the formula below;
Fourier number = thermal diffusivity × time ÷ (length)^2.
Fourier number = (3 × 60 × 33.9 × 10^-6)/( 0.1)^2 = 0.6102.
STEP THREE: This is the last step for the question, here we will be calculating the temperature of the center plane of the brass plate after 3 minutes.
Thus, the temperature of the center plane of the brass plane after 3 minutes = (1.00705) (0.89199) (900- 15) + 15.
= > the temperature of the center plane of the brass plane after 3 minutes = 809.98°C.