Elaborate on the reason why heating water up will increase the rate of dissolving for a solute.
2 answers:
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Answer:
aₓ = 0 , ay = -6.8125 m / s²
Explanation:
This is an exercise that we can solve with kinematics equations.
Initially the rabbit moves on the x axis with a speed of 1.10 m / s and after seeing the predator acceleration on the y axis, therefore its speed on the x axis remains constant.
x axis
vₓ = v₀ₓ = 1.10 m / s
aₓ = 0
y axis
initially it has no speed, so v₀_y = 0 and when I see the predator it accelerates, until it reaches the speed of 10.6 m / s in a time of t = 1.60 s. let's calculate the acceleration
= v_{oy} -ay t
ay = (v_{oy} -v_{y}) / t
ay = (0 -10.9) / 1.6
ay = -6.8125 m / s²
the sign indicates that the acceleration goes in the negative direction of the y axis
Answer:
the branch currents are as follows:
top left: I2 = 0.625 A
middle left: I1 = 2.500 A
bottom left: I1-I2 = 1.875 A
top center: I2+I3 = 2.500 A
bottom center: I2+I3-I1 = 0 A
right: I3 = 1.875 A
Explanation:
You can write the KVL equations:
Top left loop:
I2(4) +(I2 +I3)(2) +I1(1) = 10
Bottom left loop:
(I1-I2)(4) +(I1-I2-I3)(2) +I1(1) = 10
Right loop:
(I2+I3)(2) +(I2+I3-I1)(2) = 5
In matrix form, the equations are ...
These equations have the solution ...
This means the branch currents are as follows:
top left: I2 = 0.625 A
middle left: I1 = 2.500 A
bottom left: I1-I2 = 1.875 A
top center: I2+I3 = 2.500 A
bottom center: I2+I3-I1 = 0 A
right: I3 = 1.875 A
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This can be worked almost in your head by using the superposition theorem. When the 5V source is shorted, the 10V source is supplying (I1) to a circuit that is the 4 Ω and 2 Ω resistors in parallel with their counterparts, and that 2+1 Ω combination in series with 1 Ω for a total of a 4Ω load on the 10 V source. That is, I1 due to the 10V source is 2.5 A, and it is nominally split in half through the upper and lower branches of the circuit. There is no current flowing through the (shorted) 5 V source branch.
When the 10V source is shorted, the 5V source is supplying a 4 +4 Ω branch in parallel with a 2 +2 Ω branch, a total load of 8/3 Ω. This makes the current from that source (I3) be 5/(8/3) = 15/8 = 1.875 A. There is zero current from this source through the 1 Ω resistor.
Nominally, the current from the 5V source splits 2/3 through the 2 Ω branch and 1/3 through the 4 Ω branch.
Using superposition, I2 = I1/2 -I3/3 = (2.5 A/2) -(1/3)(15/8 A) = 0.625 A. This is the same answer as above, without any matrix math.
(I1, I2, I3) = (2.5 A, 0.625 A, 1.875 A)
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It helps to be familiar with the formulas for resistors in series and parallel.