Let R₁ and R₂ be the two Resistance
R₁ - Resistance of 1st resistor
R₂ - Resistance of 2nd resistor
V = Voltage = 12V
I = O.31A ( in series)
I = 1.6 A ( in parallel)
when two resistance connected in series
Rs = R₁ + R₂
V = IRs = 12 /0.31 V
R₁+R₂ = 38.700Ω (equation1.)
When two resistance connected in parallel
V = I Rp
V= I (R₂ R₁ /R₁ +R₂)
R₁ R₂ = 290.25 Ω
As we know
(R₁ - R₂ ) ² = (R₁ + R₂ ) ² - 4R1R2
Using above values we get,
(R₁ - R₂ ) ² = (38.70) ² - 4x 290.25
(R₁ - R₂ ) ² = 336.69 Ω
R₁ - R₂ = 18.34 (equation 2 )
Using equation 1 and 2 we get
R₁+ R₂ = 38.70 52
R₁ - R₂ = 18.34
R₁ = 28.535 Ω
Using the value of R₁ in equation 1 we get
R₂ = 38·700 - 28.535
RR₂ = 10.125 Ω
R₁ = 28. 535 v and R₂ 10.125 Ω
To know more about resistance in series and parallel:
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The magnitude of force acting on the body is 36N
I'll bite:
-- Since the sled's mass is 'm', its weight is 'mg'.
-- Since the coefficient of kinetic friction is μk, the force acting opposite to the direction it's sliding is (μk) times (mg) .
-- If the pulling force is constant 'F', then the horizontal forces on the sled
are 'F' forward and (μk · mg) backwards.
-- The net force on the sled is (F - μk·mg).
(I regret the visual appearance that's beginning to emerge,
but let's forge onward.)
-- The sled's horizontal acceleration is (net force) / (mass) = (F - μk·mg) / m.
This could be simplified, but let's not just yet.
-- Starting from rest, the sled moves a distance 's' during time 't'.
We know that s = 1/2 a t² , and we know what 'a' is. So we can write
s = (1/2 t²) (F - μk·mg) / m .
Now we have the distance, and the constant force.
The total work is (Force x distance), and the power is (Work / time).
Let's put it together and see how ugly it becomes. Maybe THEN
it can be simplified.
Work = (Force x distance) = F x (1/2 t²) (F - μk·mg) / m
Power = (Work / time) = <em>F (t/2) (F - μk·mg) / m </em>
Unless I can come up with something a lot simpler, that's the answer.
To simplify and beautify, make the partial fractions out of the
2nd parentheses:
<em> F (t/2) (F/m - μk·m)</em>
I think that's about as far as you can go. I tried some other presentations,
and didn't find anything that's much simpler.
Five points,ehhh ?
Answer:A piece of driftwood moves up and down as water waves pass beneath it. However, it does not move toward the shore with the waves. What does this demonstrate about the propagation of waves through a medium?
A) Waves transmit energy but not matter as they progress through a medium.
B) Waves transmit matter but not energy as they progress through a medium.
C) Waves do not transmit matter or energy as they progress through a medium.
D) Waves transmit energy as well as matter as they progress through a medium.
Explanation:
A piece of driftwood moves up and down as water waves pass beneath it. However, it does not move toward the shore with the waves. What does this demonstrate about the propagation of waves through a medium?
A) Waves transmit energy but not matter as they progress through a medium.
B) Waves transmit matter but not energy as they progress through a medium.
C) Waves do not transmit matter or energy as they progress through a medium.
D) Waves transmit energy as well as matter as they progress through a medium.
You are correct because nothing is being done to the cake
