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3 years ago

How does running an electric current through wire cause a magnetic field?

Engineering

1 answer:

Eva8 [605]3 years ago

3 0

Answer:

When a charged particle—such as an electron, proton or ion—is in motion, magnetic lines of force rotate around the particle. Since electrical current moving through a wire consists of electrons in motion, there is a magnetic field around the wire.

Explanation:

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Waste that is generated by a business is called a _____________.

matrenka [14]

It is called a ‘Waste Stream’

This should be right! Hope this helps, have a great day!

8 0

3 years ago

A vineyard owner is planting several new rows of grapevines, and needs to know how many grapevines to plant in each row. She has

boyakko [2]

Answer:

Finally, display the number of grapevines that will fit in the row

Explanation:

A vineyard owner is planting several new rows of grapevines, and needs to know how many grapevines to plant in each row. She has determined that after measuring the length of a future row, she can use the following formula to calculate the number of vines that will fit in the row, along with the trellis end-post assemblies that will need to be constructed at each end of the row:

V=R-2E/S

The terms in the formula are:

V is the number of grapevines that will fit in the row.

R is the length of the row, in feet.

E is the amount of space, in feet, used by an end-post assembly.

S is the space between vines, in feet.

Write a program that makes the calculation for the vineyard owner. The program should ask the user to input the following：

• The length of the row, in feet

• The amount of space used by an end-post assembly, in feet

• The amount of space between the vines, in feet

Once the input data has been entered, the program should calculate and display the number of grapevines that will fit in the row.

• Declare the variables amountSpace, vinesSpace, and grapevines of double type.

• Prompt and read the length of the row, in feet, the amount of space, in feet, and the space between vines, in feet.

• Calculate the number of grapevines that will fit in the row using the below formula:

grapvines=(rowLength-2*amountSpace)/vinesSpace;

• Finally, display the number of grapevines that will fit in the row.

8 0

3 years ago

In Ohio a registered vehicle must have a valid license plate displayed on the vehicle within;

sineoko [7]

You can't legally drive it on the road without proper plates. If you will be transferring plates from a vehicle currently registered to you, you have 30 days to register the plates to the new car.

3 0

3 years ago

Read 2 more answers

Please help i will give brainilest

Eduardwww [97]

Answer:

I hot u

Explanation:

it makes things function better and more fluently than a manual version of the certain object

8 0

3 years ago

Read 2 more answers

Design a half-wave recti er which provides a peak voltage of 15 V, and anaverage voltage of 3.8 V when driven by a 120 V (rms) a

nirvana33 [79]

Answer:

You need a 120V to 24V commercial transformer (transformer 1:5), a 100 ohms resistance, a 1.5 K ohms resistance and a diode with a minimum forward current of 20 mA (could be 1N4148)

Step by step design:

Because you have a 120V AC voltage supply you need an efficient way to reduce that voltage as much as possible before passing to the rectifier, for that I recommend a standard 120V to 24V transformer. 120 Vrms = 85 V and 24 Vrms = 17V = Vin
Because 17V is not 15V you still need a voltage divider to step down that voltage, for that we use R1 = 100Ω and R2 = 1.3KΩ. You need to remember that more than 1 V is going to be in the diode, so for our calculation we need to consider it. Vf = (V*R2)/(R1+R2), V = Vin - 1 = 17-1 = 16V and Vf = 15, Choosing a fix resistance R1 = 100Ω and solving the equation we find R2 = 1.5KΩ
Finally to select the diode you need to calculate two times the maximum current and that would be the forward current (If) of your diode. Imax = Vf/R2 = 10mA and If = 2*Imax = 20mA

Our circuit meet the average voltage (Va) specification:

Va = (15)/(pi) = 4.77V considering the diode voltage or 3.77V without considering it

6 0

4 years ago