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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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What does Newton's first law of motion state? (3 points) a Every action has an equal and opposite reaction. b A body at rest wil
kirza4 [7]

Answer:

B) A body at rest will stay at rest

Explanation:

A body in motion will stay in motion unless acted on by another force.

3 0
2 years ago
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If block D weighs 300 lb and block B weighs 275 lb determine the required weight of block C and the angle theta for equilibrium?
adoni [48]
Set up a free body diagram. 

<span>and by reason, Tcd = Tbd </span>

<span>Tbd y = 275 - 300*sinθ </span>
<span>Tcd y = Tc - 300*sin30 </span>

<span>Tbd x = 300*cosθ </span>
<span>Tcdx = 300 * cos30 </span>

<span>Tbd^2 = (275 - 300*sinθ)^2 + (300*cosθ)^2 </span>
<span>Tcd^2 = (300*sin30)^2 + (300 * cos30)^2 </span>

<span>(275 - 300*sinθ)^2 + (300*cosθ)^2 = (300*sin30)^2 + (300 * cos30)^2 </span>
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5 0
3 years ago
PLEASEEEE HEEEEELP!!!!!
8090 [49]

Answer:

Before:

p_{truck}=16400\ kg.m/s

p_{car}=10000\ kg.m/s

After:

p_{truck}=8000\ kg.m/s

p_{car}=8400\ kg.m/s

v_{fcar}=8.4\ m/s

F=9333.33 \ Nw

Explanation:

<u>Conservation of Momentum</u>

Two objects of masses m1 and m2 moving at speeds v1o and v2o respectively have a total momentum of

p_1=m_1v_{1o}+m_2v_{2o}

After the collision, they have speeds of v1f and v2f and the total momentum is

p_2=m_1v_{1f}+m_2v_{2f}

Impulse J is defined as

J=F.t

Where F is the average impact force and t is the time it lasted

Also, the impulse is equal to the change of momentum

J=\Delta p

As the total momentum is conserved:

p_1=p_2

m_1v_{1o}+m_2v_{2o}=m_1v_{1f}+m_2v_{2f}

We can compute the speed of the second object by solving the above equation for v2f

\displaystyle v_{2f}=\frac{ m_1v_{1o}+m_2v_{2o} -m_1v_{1f} }{  m_2 }

The given data is

m_1=2000\ kg

m_2=1000\ kg\\v_{1o}=8.2\ m/s\\v_{2o}=0\ m/s\\v_{1f}=4\ m/s

a) The impulse will be computed at the very end of the answer

b) Before the collision

p_{truck}=2000\cdot 8.2=16400\ kg.m/s

p_{car}=1000\cdot 0=0\ kg.m/s

c) After collision

p_{truck}=2000\cdot 4=8000\ kg.m/s

Compute the car's speed:

\displaystyle v_{2f}=\frac{ 16400+0 -8000 }{ 1000 }

v_{2f}=8.4\ m/s

And the car's momentum is

p_{car}=1000\cdot 8.4=8400\ kg.m/s

The Impulse J of the system is zero because the total momentum is conserved, i.e. \Delta p=0.

We can compute the impulse for each object

J_1=\Delta p_1=2000(4-8.2)=-8400 \N.s

The force can be computed as

\displaystyle F=\frac{J}{t}=-\frac{8400}{0.9}=-9333.33 \ Nw

The force on the car has the same magnitude and opposite sign

7 0
3 years ago
Please help solve this inelastic collision question please use GUESS to solve it if you can’t it’s fine
uranmaximum [27]

Answer:

srfscfsd<vfrz<gfgrfdfsf<fesrf<sdfsre<f

Explanation:

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