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aleksandr82 [10.1K]
2 years ago
5

Applying Kirchhoff’s Junction Rule, what happens to the power source and current when the parallel circuit has two branches, eac

h with a resistor, R? In one to two sentences, answer the question and explain your reasoning.
Physics
1 answer:
slamgirl [31]2 years ago
7 0

The power source voltage remains the same in a  parallel circuit,

And we'll have equal current in both lines

<h3>Kirchhoff's junction rule</h3>

Generally, Kirchhoff's junction rule states that when there is current flow at any junction of a circuit, the total sum of this current rushing into the junction amount to the same amount of current out of the Node.

Therefore, when the parallel circuit has two branches

i=i1+12

Since we have an equal resistor therefore we'll have equal current in both lines i.e i1=i2

And Voltage remains the same in a  parallel circuit

More on Voltage

brainly.com/question/14883923

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Kyle is flying a helicopter at 125 m/s on a heading of 325 o . If a wind is blowing at 25 m/s toward a direction of 240.0 o , wh
frosja888 [35]

Answer:

The resultant velocity of the helicopter is \vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right).

Explanation:

Physically speaking, the resulting velocity of the helicopter (\vec v_{H}), measured in meters per second, is equal to the absolute velocity of the wind (\vec v_{W}), measured in meters per second, plus the velocity of the helicopter relative to wind (\vec v_{H/W}), also call velocity at still air, measured in meters per second. That is:

\vec v_{H} = \vec v_{W}+\vec v_{H/W} (1)

In addition, vectors in rectangular form are defined by the following expression:

\vec v = \|\vec v\| \cdot (\cos \alpha, \sin \alpha) (2)

Where:

\|\vec v\| - Magnitude, measured in meters per second.

\alpha - Direction angle, measured in sexagesimal degrees.

Then, (1) is expanded by applying (2):

\vec v_{H} = \|\vec v_{W}\| \cdot (\cos \alpha_{W},\sin \alpha_{W}) +\|\vec v_{H/W}\| \cdot (\cos \alpha_{H/W},\sin \alpha_{H/W}) (3)

\vec v_{H} = \left(\|\vec v_{W}\|\cdot \cos \alpha_{W}+\|\vec v_{H/W}\|\cdot \cos \alpha_{H/W}, \|\vec v_{W}\|\cdot \sin \alpha_{W}+\|\vec v_{H/W}\|\cdot \sin \alpha_{H/W} \right)

If we know that \|\vec v_{W}\| = 25\,\frac{m}{s}, \|\vec v_{H/W}\| = 125\,\frac{m}{s}, \alpha_{W} = 240^{\circ} and \alpha_{H/W} = 325^{\circ}, then the resulting velocity of the helicopter is:

\vec v_{H} = \left(\left(25\,\frac{m}{s} \right)\cdot \cos 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \cos 325^{\circ}, \left(25\,\frac{m}{s} \right)\cdot \sin 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \sin 325^{\circ}\right)\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)

The resultant velocity of the helicopter is \vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right).

8 0
3 years ago
A cyclical heat engine, operating between temperatures of 450º C and 150º C produces 4.00 MJ of work on a heat transfer of 5.00
gogolik [260]

Answer:

(a) Heat transfer to the environment is: 1 MJ and (b) The efficiency of the engine is: 41.5%

Explanation:

Using the formula that relate heat and work from the thermodynamic theory as:W=Q=Q_{in}-Q_{out} solving to Q_out we get:Q_{out}=Q_{in}-W=5(MJ)-4(MJ)=1(MJ) this is the heat out of the cycle or engine, so it will be heat transfer to the environment. The thermal efficiency of a Carnot cycle gives us: n=1-\frac{T_{Low} }{T_{High}} where T_Low is the lowest cycle temperature and T_High the highest, we need to remember that a Carnot cycle depends only on the absolute temperatures, if you remember the convertion of K=°C+273.15 so T_Low=150+273.15=423.15 K and T_High=450+273.15=723.15K and replacing the values in the equation we get:n=1-\frac{423.15}{723.15} =0.415=41.5%

5 0
3 years ago
Why is acceleration of an object moving at a constant velocity always zero?
k0ka [10]

Answer:

If an object is moving with a constant velocity, then by definition it has zero acceleration. So there is no net force acting on the object. The total work done on the object is thus 0 (that's not to say that there isn't work done by individual forces on the object, but the sum is 0 ).

Explanation:

In the middle, when the object was changing position at a constant velocity, the acceleration was 0. This is because the object is no longer changing its velocity and is moving at a constant rate.

8 0
3 years ago
What is the oldest animal on earth?
Lilit [14]

Answer:

e3f3ewfeewfewgwgewggegegeggegeggege

Explanation:

6 0
3 years ago
Read 2 more answers
Two balls undergo a perfectly elastic head-on collision, with one ball initially at rest. if the incoming ball has a speed of 20
melamori03 [73]
what is the final speed of the incoming ball if it is much more massive than the stationary ball? express your answer using two significant figures. v1 = 200 m / s submitprevious answers correct
 Perfectly elastic collisions means that both mechanical energy and
momentum are conserved.
 Therefore, for this case, we have the equation to find the final velocity of the incoming ball is given by
 v1f = ((m1-m2) / (m1 + m2)) v1i
 where,
 v1i: initial speed of ball 1.
 v1f: final speed of ball 1.
 m1: mass of the ball 1
 m2: mass of the ball 2
 Since the mass of the ball 1 is much larger than the mass of the ball 2 m1 >> m2, then rewriting the equation:
 v1f = ((m1) / (m1) v1i
 v1f = v1i
 v1f = 200 m / s
 answer
 200 m / s
 part b part complete what is the final direction of the incoming ball with respect to the initial direction if it is much more massive than the stationary ball? forward submitprevious answers correct

 Using the equation of part a, we can include in it the directions:
 v1fx = ((m1-m2) / (m1 + m2)) v1ix
 v1i: initial velocity of ball 1 in the direction of the x-axis
 v1f: final speed of ball 1 in the direction of the x-axis
 like m1 >> m2 then
 v1fx = v1ix
 v1fx = 200 m / s (positive x direction)
 So it is concluded that the ball 1 continues forward.
 answer:
 forward


 part c part complete what is the final speed of the stationary ball if the incoming ball is much more massive than the stationary ball ?.
 The shock is perfectly elastic. For this case, we have that the equation to find the final velocity of the stationary ball is given by
 v2f = ((2m1) / (m1 + m2)) v1i
 where,
 v1i: initial speed of ball 1.
 v2f: final speed of ball 2.
 m1: mass of the ball 1
 m2: mass of the ball 2
 Then, as we know that m1 >> m2 then
 v2f = ((2m1) / (m1) v1i
 v2f = 2 * v1i
 v2f = 2 * (200 m / s)
 v2f = 400 m / s
 answer
 400m / s
7 0
3 years ago
Read 2 more answers
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