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yaroslaw [1]
3 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]3 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]3 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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If the current through a 20-Ω resistor is 8.0 A , how much energy is dissipated by the resistor in 1.0 h ? Express your answer w
marshall27 [118]

Answer:

P(3600)=593.247W

Explanation:

First, let's find the voltage through the resistor using ohm's law:

V=IR=20*8=160V

AC power as function of time can be calculated as:

P(t)= V*I*cos(\phi)-V*I*cos(2 \omega t-\phi)  (1)

Where:

\phi=Phase\hspace{3}angle\\\omega= Angular\hspace{3}frequency

Because of the problem doesn't give us additional information, let's assume:

\phi=0\\\omega=2 \pi f=2*\pi *(60)=120\pi

Evaluating the equation (1) in t=3600 (Because 1h equal to 3600s):

P(3600)=160*8*cos(0)-160*8*cos(2*120\pi*3600-0)\\P(3600)=1280-1280*cos(2714336.053)\\P(3600)=1280-1280*0.5365255751\\P(3600)=1280-686.7527361=593.2472639\approx=593.247W

5 0
3 years ago
A 0.540-kg bucket rests on a scale. Into this bucket you pour sand at the constant rate of 56.0 g/s. If the sand lands in the bu
romanna [79]

Answer:

a) 12.8212 N

b) 12.642 N

Explanation:

Mass of bucket = m = 0.54 kg

Rate of filling with sand  = 56.0 g/ sec = 0.056 kg/s

Speed of sand = 3.2 m/s

g= 9.8 m/sec2

<u>Condition (a);</u>

Mass of sand = Ms = 0.75 kg

So total mass becomes = bucket mass + sand mass = 0.54 +0.75=1.29 kg

== > total weight = 1.29 × 9.8 = 12.642 N

Now impact of sand = rate of filling × velocity = 0.056 × 3.2 =  0.1792 kg. m /sec2=0.1792 N

Scale reading is sum of impact of sand and weight force ;

i-e

scale reading = 12.642 N+0.1792 N = 12.8212 N

<u>Codition (b);</u>

bucket mass + sand mass = 0.54 +0.75=1.29 kg

==>weight = mg = 1.29 × 9.8 = 12.642 N (readily calculated above as well)

6 0
3 years ago
Chinook salmon are able to move upstream faster by jumping out of the water periodically; this behavior is called porpoising. Su
deff fn [24]

Answer:he formula for average speed is (total distance/total time)

the y-component does not matter in this problem. so do 6.26(cos45)=4.43m/s to find the x-component velocity which is constant throughout the duration of the flight. the total distance is 2L because he travels distance L twice.

the total time is ((time in water)+(time out of water)) since you dont have time you must eliminate it. to do this you need (distance)/(time)=velocity

solve for time and you get T=D/V

time in water is L/3.52 and time out of water is L/4.43

add them together and you get (4.43L+3.52L)/(15.59) = 7.95L/15.59

that value is your total time

divide you total distance (2L) by total time (7.95L/15.59) and the Ls cancel out and you get

(31.18)/(7.95) = 3.92 m/s = Average Speed

Explanation:

7 0
3 years ago
Heat energy is:
Archy [21]
I claim that it's d moving electrons.
6 0
3 years ago
A playground merry-go-round has a radius of 4.6 m and a moment of inertia of 200 kg-m2 and turns with negligible friction about
tankabanditka [31]

Answer:

8050 J

Explanation:

Given:

r = 4.6 m

I = 200 kg m²

F = 26.0 N

t = 15.0 s

First, find the angular acceleration.

∑τ = Iα

Fr = Iα

α = Fr / I

α = (26.0 N) (4.6 m) / (200 kg m²)

α = 0.598 rad/s²

Now you can find the final angular velocity, then use that to find the rotational energy:

ω = αt

ω = (0.598 rad/s²) (15.0 s)

ω = 8.97 rad/s

W = ½ I ω²

W = ½ (200 kg m²) (8.97 rad/s)²

W = 8050 J

Or you can find the angular displacement and find the work done that way:

θ = θ₀ + ω₀ t + ½ αt²

θ = ½ (0.598 rad/s²) (15.0 s)²

θ = 67.3 rad

W = τθ

W = Frθ

W = (26.0 N) (4.6 m) (67.3 rad)

W = 8050 J

6 0
3 years ago
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