1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
WARRIOR [948]
3 years ago
9

Please help I am give brainiliest

Engineering
1 answer:
Tanzania [10]3 years ago
5 0

Answer:

A. Original energy source is converted to AC power, then transported through conductors to your house.

Explanation:

The correct option is - A. Original energy source is converted to AC power, then transported through conductors to your house.

You might be interested in
Which equation can be used to find x, the length of the hypotenuse of the right triangle? A triangle has side lengths 63, 16, x.
Kryger [21]

Answer: 16 squared + 63 squared = x squared

Explanation:

Hi, since we have a right triangle we have to apply the Pythagorean Theorem:  

c^2 = a^2 + b^2  

Where c is the hypotenuse of the triangle (the longest side) and a and b are the other sides.  

Replacing with the values given:  

x^2 = 63^2 + 16^2  

So, the correct option is  

16 squared + 63 squared = x squared

Feel free to ask for more if needed or if you did not understand something.  

3 0
4 years ago
Read 2 more answers
I need help!!!!!!!!!
katovenus [111]

Answer:

buy a new one

Explanation:

go on newegg and you will most likely find one there

8 0
3 years ago
Suppose Q1 (0.0, - 3.0 M, 0.0) = 4.0 nC, Q2 (0.0, 3.0 m, 0.0) = 4.0 nC, and Q3 (4.0 m, 0.0, 0.0) = 1.0 nC.
frozen [14]

Answer:

Explanation:

Force on Q₃ due to charge Q₁

= 9 x 10⁹x 4 x 10⁻⁹ x1 x 10⁻⁹ /  5²

= 1.44 x 10⁻⁹ N

Force due to Q₂ will also be 1.44 x 10⁻⁹ N

component of these forces along x axis

-= 2 x 1.44 x 10⁻⁹ cosθ

= 2.88 x  10⁻⁹  x 4/5

= 2.30x10⁻⁹  N along x axis.

The y-component will calcel out.

b ) In this case , Q₁ will repel and Q₂ will attract.

In this case Q₁ will repel and Q₂ will attract. Component along y - axis will be same as earlier one or 2.30x10⁻⁹  N . Component along x axis will cancel out.

c ) Electric field in case 1 and case 2 will be

= 2.30x10⁻⁹  / 1 x 10⁻⁹

= 2.3 N / C , because field is force per unit charge. The sane field will be in case 2 .

5 0
4 years ago
Find three examples of good websites and three examples of bad websites. List them below, and in a
Fudgin [204]

Answer: its c

Explanation:

4 0
3 years ago
What parts are needed to complete a loop for an LED?
nika2105 [10]

Answer:

it depends on what LED's your using... if you have one long strip of them you cant make a loop unless you want it to be bent

Explanation:

3 0
3 years ago
Read 2 more answers
Other questions:
  • Madison and Oxford are the same distance from the equator and they are both near the ocean. Use the information on the map to an
    11·1 answer
  • Kkhghghglgghklghghghlk
    6·2 answers
  • Calculate the thermal efficiency (ηth) for the actual cycle using pump efficiency (ηpump) = 0.85. You’ll need to find the high-p
    13·2 answers
  • A socket can be driven using any of the following except for (A) a socket ratchet
    14·1 answer
  • What additional information would make the following problem statement stronger? Select all that apply.
    8·1 answer
  • When using a Hammer I should always keep an eye on may fingers so as not to hit them.?
    14·2 answers
  • a circular pile, 19 m long is driven into a homogeneous sand layer. The piles width is 0.5 m. The standard penetration resistanc
    6·1 answer
  • Conclusion. What process is responsible for the bubbling action of the organism? What is your evidence?
    13·1 answer
  • Tech A says that wiring diagrams are essentially a map of all of the electrical components and their connections. Tech B says th
    7·1 answer
  • It is given that 50 kg/sec of air at 288.2k is iesntropically compressed from 1 to 12 atm. Assuming a calorically perfect gas, d
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!