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yaroslaw [1]
2 years ago
14

Examine the circuit. Pretend you are an electron flowing through this circuit and you are with a group of other electrons. Sudde

nly you come to the ‘fork in the road’ and you and some of the other electrons are temporarily separated. A short while later you are back together with the other electrons. This type of circuit is called a(n) ____________ circuit.
PLEASE HELP!!!
Physics
2 answers:
melamori03 [73]2 years ago
7 0

You knew that this question is ridiculously easy.  So, just to
make it harder, you decided not to let us see the picture, so
that we could not "examine the circuit".

The description is talking about a parallel circuit.  The other
kind is a series circuit, and that one has no forks in the road.
valentinak56 [21]2 years ago
5 0

Answer:

parallel

Explanation:

In an electric circuit a ‘fork in the road’ appears in a parallel configuration. This kind of circuit is characterized by the different currents each 'road' take after the bifurcation, depending on the different resistance each one has. The voltage remains the same for every 'road'.

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Tex, an 85.0 kilogram rodeo bull rider is thrown from the bull after a short ride. The 520. kilogram bull chases after Tex at 13
Julli [10]
The question above can be answered through using the concept of Conservation of Momentum which may be expressed as,
                 m1v1 + m2v2 = mTvT
where m1 and v1 are mass and initial velocity of Tex, 2s are that of the bull, and the Ts are the total. Then substituting,
                    (85 kg)(3 m/s) + (520 kg)(13 m/s) = (520 + 85)(vT)
The value of vT obtained from above equation is 11.6 m/s
3 0
2 years ago
A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±qseparated by a distance s. T
atroni [7]

Answer:

The magnitude of the force on the dipole due to the charge Q = \rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi }\dfrac{2qQs}{r^3}.

The magnitude of the torque on the dipole = \rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi}\dfrac{2qQs^2}{r^3}.

Explanation:

Given that a point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q, separated by a distance s and the charge Q is located in the plane that bisects the dipole.

The magnitude of the electric field that the dipole exerts at the position where the charge Q is held is given by

\rm E = \dfrac{k2qs}{(r^2+s^2)^{3/2}}.

<em>where</em>,

k is the Coulomb's constant, having value = \dfrac{1}{4\pi \epsilon_o}

\epsilon_o is the electrical permittivity of free space.

Also, r>>s, therefore, \rm r^2+s^2\approx r^2.

\rm E = \dfrac{k2qs}{(r^2)^{3/2}}=\dfrac{k2qs}{r^3}.

The magnitude of the electric force F on a charge q placed at a point and the magnitude of the electric field E at that point are related as

\rm F=qE

Therefore, the electric force on the charge Q due to the dipole is given by

\rm F=Q\dfrac{k2qs}{r^3}=\dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs}{r^3}.

According to Newton's third law of motion, the magnitude of the force exerted by the dipole on the charge Q is same as the magnitude of the force exerted by the charge on the dipole.

Thus, the magnitude of the force on the dipole due to the charge Q = \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi }\dfrac{2qQs}{r^3}.

The magnitude of the torque on the dipole is given by

\rm \tau = Fs\ \sin\theta

Since the charge Q is placed in the plane that bisects the dipole, therefore, \theta = 90^\circ.

\rm \tau = \dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs}{r^3}\cdot s\cdot 1=\dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs^2}{r^3}.

4 0
2 years ago
7) A heavy dart and a light dart are launched horizontally on a frictionless table by identical ideal springs. Both springs were
Elenna [48]

Answer:

A, B, and E

Explanation:

The springs are identical, and are compressed the same amount, so they have the same initial elastic potential energy. (E is true)

Energy is conserved, so the darts have the same amount of kinetic energy. (A is true, C is false)

The lighter dart has the same energy as the heavier dart.  Since it has less mass, it must have a greater speed. (B is true, D is false)

6 0
2 years ago
A 1.0 104 kg spacecraft is traveling through space with a speed of 1200 m/s relative to Earth. A thruster fires for 2.0 min, exe
aniked [119]
We are given information:
m=1.0* 10^{4} kg \\ v=1200m/s \\ t=2min=120s \\ F = 25kN = 25000N

If we apply Newton's second law we can calculate acceleration:
F = m * a
a = F / m
a = 25000 / 10000
a = 2.5 m/s^2

Now we can use this information to calculate change of speed.
a = v / t
v = a * t
v = 2.5 * 120
v = 300 m/s

Force is being applied in direction that is opposite to a direction in which space craft is moving. This means that final speed will be reduced.
v = 1200 - 300
v = 900 m/s

Formula for momentum is:
p = m * v
Initial momentum:
p = 10000 * 1200
p = 12 000 000
p = 12 *10^6 kg*m/s
Final momentum:
p = 10000 * 900
p = 9 000 000
p = 9 *10^6 kg*m/s

7 0
2 years ago
Jane puts some water into an electric kettle and then she connects it to the power source. She observes that after some time the
Setler [38]
C) electrical energy is transformed into heat energy
4 0
2 years ago
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