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4 years ago

8

ASAP! What examples of resources can you give that can be classified as both material resources and energy resources? if u answe

r i'll give u a thanks

1 answer:

natulia [17]4 years ago

4 0

Wood because it is natural and we use it for furniture but we also burn it for energy.

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You are preparing a performance review and have the following measurement at hand: pv = 300; ac = 200; and ev = 250. what is cpi

Vilka [71]

CPI of the project is 1.25 so the correct answer is B

EV / AC is the formula used to calculate CPI. Here, 250 divided by 200 equals 1.25.The worth of the work that has been finished thus far in comparison to the budget is referred to as earned value.A. Schedule performance index equals (EV / PV) and schedule variance equals (EV - PV).The CPI is regarded as the most important EVM metric. It gauges the project work's cost effectiveness as of the measurement date.

To learn more about CPI:

brainly.com/question/26682248

#SPJ4

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2 years ago

If a leaf falls from a tree, has work been done on the leaf? Explain.

mihalych1998 [28]

Answer:

Hope this helps!

5 0

3 years ago

A hollow cylinder with an inner radius of 5 mm and an outer radius of 26 mm conducts a 4-A current flowing parallel to the axis

bearhunter [10]

Answer:

B = 38.2μT

Explanation:

By the Ampere's law you have that the magnetic field generated by a current, in a wire, is given by:

$B=\frac{\mu_o I_r}{2\pi r}$ (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

r: distance from the center of the cylinder, in which B is calculated

Ir: current for the distance r

In this case, you first calculate the current Ir, by using the following relation:

$I_r=JA_r$

J: current density

Ar: cross sectional area for r in the hollow cylinder

Ar is given by $A_r=\pi(r^2-R_1^2)$

The current density is given by the total area and the total current:

$J=\frac{I_T}{A_T}=\frac{I_T}{\pi(R_2^2-R_1^2)}$

R2: outer radius = 26mm = 26*10^-3 m

R1: inner radius = 5 mm = 5*10^-3 m

IT: total current = 4 A

Then, the current in the wire for a distance r is:

$I_r=JA_r=\frac{I_T}{\pi(R_2^2-R_1^2)}\pi(r^2-R_1^2)\\\\I_r=I_T\frac{r^2-R_1^2}{R_2^2-R_1^2}$ (2)

You replace the last result of equation (2) into the equation (1):

$B=\frac{\mu_oI_T}{2\pi r}(\frac{r^2-R_1^2}{R_2^2-R_1^2})$

Finally. you replace the values of all parameters:

$B=\frac{(4\pi*10^{-7}T/A)(4A)}{2\PI (12*10^{-3}m)}(\frac{(12*10^{-3})^2-(5*10^{-3}m)^2}{(26*10^{-3}m)^2-(5*10^{-3}m)^2})\\\\B=3.82*10^{-5}T=38.2\mu T$

hence, the magnitude of the magnetic field at a point 12 mm from the center of the hollow cylinder, is 38.2μT

8 0

3 years ago

What is shown in the Diagram?

Alexandra [31]

Trust me, i'm a k12 student and its motor

4 0

3 years ago

Read 2 more answers

Which of the following measure is more accurate? 500.00kg, 0.0005kg,6.00kg

gavmur [86]

Answer: 500.00kg

Explanation:

3 0

3 years ago