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Mashcka [7]
3 years ago
13

In the context of the loop and junction rules for electrical circuits, a junction is: Group of answer choices where a wire is be

nt. where three or more wires are joined. where only two wires are joined. where a wire is connected to a resistor. where a wire is connected to a battery.
Physics
2 answers:
vitfil [10]3 years ago
6 0

Answer:

In the context of the loop and junction rules for electrical circuits, a junction is where three or more wires are joined.

Explanation:

A point where at least three circuit paths meet i.e wires, is referred to as a junction.

Kirchhoff’s circuit laws are two(2) equations first published by Gustav Kirchhoff in 1845. Fundamentally, they address conservation of energy and charge in the context of electrical circuits. One of the laws known as Kirchoff's Current Law deals with the principle of application of conserved energy in electrical circuits. Kirchoff's Current Law states that the sum of all currents entering a junction must equal the sum of all currents leaving the junction.

This basically means, the algebraic sum of currents in a network of conductors(wires) meeting at a point is equal to zero

Solnce55 [7]3 years ago
4 0

A junction is a point where at least three circuit paths meet.

A branch is a path connecting two junctions .

Kirchhoff's circuit laws

This law states that, for any node (junction) in an electrical circuit ,

the sum of currents flowing into that node is equal to the sum of

currents flowing out of that node.

Kirchhoff's circuit laws are two equalities that deal with the current

and potential difference in the lumped

element model of electrical circuits .

Some terms associated with Kirchhoff's circuit laws

resistor : An electric component that transmits

current in direct proportion to the voltage

across it.

electromotive force: (EMF)—The voltage

generated by a battery or by the magnetic force

according to Faraday’s Law. It is measured in

units of volts (not newtons, N; EMF is not a

force).

capacitor : An electronic component consisting

of two conductor plates separated by empty

space (sometimes a dielectric material is

instead sandwiched between the plates), and

capable of storing a certain amount of charge.

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Force of 10N down,10N to the right,and 5N to the left are acting on a ball .it acceleration horizontally to the right.what other
denpristay [2]

Answer:

A 10 N force pointing up

Explanation:

If the net acceleration of the object is horizontal pointing to the right, that means that all vertical forces must have canceled out, and the only ones "unbalanced" are the horizontal ones (10 N to the right minus 5 N to the left giving a net force of 5 N to the right).

Since they mentioned only one vertical force pointing down (10 N), there must be another one of same magnitude but pointing in opposite direction (up).

Then there must also be a 10 N force pointing up acting on the object.

3 0
3 years ago
A baseball is thrown horizontally at a rate of 40m/s toward a home plate 18.4 m away. How far below the launch height is the bal
sammy [17]

Answer:

<h3>1.03684m</h3>

Explanation:

Using the formula for calculating range expressed as;

R = U√2H/g where

R is the distance moves in horizontal direction = 18.4m

H is the height

U is the velocity of the baseball = 40m/s

g is the acceleration due to gravity = 9.8m/s²

Substitute the given parameters into the formula and calculate H as shown;

18.4 = 40√2H/9.8

18.4/40 = √2H/9.8

0.46 = √2H/9.8

square both sides;

(0.46)² = (√2H/9.8)²

0.2116 = 2H/9.8

2H = 9.8*0.2116

2H = 2.07368

H = 2.07368/2

H = 1.03684m

Hence the ball is 1.03684m below the launch height when it reached home plate.

8 0
3 years ago
Help me quick!!! please!!
Rudik [331]

Answer:

39.240 W

Explanation:

Let's start by calculating the work done by the engine. We can assume that it is the same work done by the weight of the object to bring it from 40m to the surface: as much energy it takes to bring it up, the same ammount it takes to bring it down. Said work is w= \vec F\cdot \vec{h} = mg h = 1000 \times 9.81\times 40 = 392.400 J

At this point we can simply apply the definition of power, that is P = \frac wt, to get the power of the engine is 39.240 W

4 0
3 years ago
Read 2 more answers
If you wanted to detect x rays coming from the sun, where would you place the detector? Why?
Damm [24]
You would have to place your sensor above earth's atmosphere because it blocks out nearly all x-rays. this is why we have the Chandra observatory

hope this helps
4 0
3 years ago
Read 2 more answers
A snowball with a mass of 85 g hits the top hat of a 1.5 m tall snowman and sticks to it. the hat and the snowball, with a combi
LenaWriter [7]
Part (a): Velocity of the snowball
By conservation of momentu;
m1v1 + m2v2 = m3v3,

Where, m1 = mass of snowball, v1, velocity of snowball, m2 = mass of the hat, v2 = velocity of the hat, m3 = mass of snowball and the hat, v3 = velocity of snowball and the hut.

v2 = 0, and therefore,
85*v1 + 0 = 220*8 => v1 = 220*8/85 = 20.71 m/s

Part (b): Horizontal range
x = v3*t
But,
y = vy -1/2gt^2, but y = -1.5 m (moving down), vy =0 (no vertical velocity), g = 9.81 m/s^2

Substituting;
-1.5 = 0 - 1/2*9.81*t^2
1.5 = 4.905*t^2
t = Sqrt (1.5/4.905) = 0.553 seconds

Then,
x = 8*0.553 = 4.424 m
7 0
3 years ago
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