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Elena L [17]
2 years ago
5

What is a electric curcit

Physics
2 answers:
Anastaziya [24]2 years ago
8 0

\large \gray { \boxed{{ \colorbox{g}{ \tt{ answer \:}}}}}

<h2><u>ELECTRIC</u><u> </u><u>CIRCUIT</u></h2>

  • <u>electric circuit, path for transmitting electric current. An electric circuit includes a device that gives energy to the charged particles constituting the current, such as a battery or a generator; devices that use current, such as lamps, electric motors, or computers; and the connecting wires or transmission lines.</u>

<u>\tt \color {green}{hope \:  \: it \:  \: helps}</u>

loris [4]2 years ago
4 0

Hey there! I'll try to provide you with my best answer.

Answer: Electric circuit is a path for transmitting electric current. An electric circuit includes a device that gives energy to the charged particles constituting the current, such as a battery or a generator; devices that use current, such as lamps, electric motors, or computers; and the connecting wires or transmission lines.

Hope it helps!

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What is the amount of heat required to raise the temperature of 200.0 g of aluminum by 10 c?
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Q = 175.8J = 1.8• 10 2 J
5 0
3 years ago
A car moving has a speed of 35 m/s. What acceleration would it have if it took 5.0 s to come to a complete stop?
Pepsi [2]

Answer:

7

Explanation:

7 0
3 years ago
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A car of mass 1,000 kilograms is moving initially at the speed of 22 meters/second. When the brakes are applied, it takes the ca
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A ship leaves a harbor and sails at 23.5 ∘ to the north of due west. After traveling 575 km, how far west is the ship from the h
solniwko [45]

Answer:

D = 527.31 Km

Explanation:

given,

angle of ship, θ = 23.5° N of W

distance travel in the direction = 575 Km

Distance of ship in west from harbor = ?

now,

Distance of the ship in the west direction

D = d cos θ

d = 575 Km

θ = 23.5°

inserting all the values

D = 575 x cos 23.5°

D = 575 x 0.91706

D = 527.31 Km

Hence, the distance travel by the ship in west from harbor is equal to D = 527.31 Km  

6 0
3 years ago
This is a velocity versus time graph of a car starting from rest. If the area under the line is 10 meters, what is the correspon
adelina 88 [10]

So the area under a velocity time graph is distance or displacement, if you have done calculus yet you will understand that if you take the integral of a velocity function then you end up with displacement. Thats for later understanding however.

So this appears to be a right triangle so we can find the area of a triangle as:

0.5bh = A

Since our area is 10 meters lets alter our formula a bit to fit the situation:

Our base here is time and our height is velocity so:

0.5tv = Δx

So we can read off the graph that our velocity at the end, or our final velocity appears to be near 2.0 m/s

So we have v, and Δx so lets isolate for time by dividing by v and 0.5

t = Δx / 0.5v

Now lets plug all that in:

t = 10 / 0.5(2)

t = 10 seconds

Hope this helped!

8 0
3 years ago
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