Two electrons, each with mass mmm and charge qqq, are released from positions very far from each other. With respect to a certai
n reference frame, electron A has initial nonzero speed vvv toward electron B in the positive x direction, and electron B has initial speed 3v3v toward electron A in the negative x direction. The electrons move directly toward each other along the x axis (very hard to do with real electrons). As the electrons approach each other, they slow due to their electric repulsion. This repulsion eventually pushes them away from each other.
What is the minimum separation r_min that the electrons reach?
What is the minimum separation r_min that the electrons reach?
1 answer:
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Answer:
35mA
Explanation:
Hello!
To solve this problem we must use the following steps
1. Find the electrical resistance of the metal rod using the following equation
WHERE
α=
metal rod resistivity=2x10^-4 Ωm
l=leght=2m
A= Cross-sectional area
solving
2. Now we model the system as a circuit with parallel resistors, where we will call 1 the metal rod and 2 the man(see attached image)
3.we know that the sum of the currents in 1 and 2 must be equal to 5A, by the law of conservation of energy
I1+I2=5
4.as the voltage on both nodes is the same we can use ohm's law in resitance 1 and 2 (V=IR)
V1=V2
(0.14I1)=2000(i2)
solving for i1
I1=14285.7i2
5.Now we use the equation found in step 3
14285.7i2+i2=5
Answer:
Volume of gasoline that expands and spills out is 1.33 ltr
Explanation:
As we know that when temperature of the liquid is increased then its volume will expand and it is given as
here we know that
volume expansion coefficient of the gasoline is given as
change in temperature is given as
Now we have
Answer:
Plot ln K vs 1/T
(a) -0.5004; (b) 0.002 539 K⁻¹; (c) -197.1 K⁻¹; (d) 1.64 kJ/mol
Explanation:
This is an example of the Arrhenius equation:
Thus, if we plot ln k vs 1/T, we should get a straight line with slope = -Eₐ/R and a y-intercept = lnA
Data:
Calculations:
(a) Rise
Δy = y₂ - y₁ = -0.9545 - (-0.4541) = -0.9545 + 0.4541 = -0.5004
(b) Run
Δx = x₂ - x₁ = 0.004 444 - 0.001 905 = 0.002 539 K⁻¹
(c) Slope
Δy/Δx = -0.5004/0.002 539 K⁻¹ = -197.1 K⁻¹
(d) Activation energy
Slope = -Eₐ/R
Eₐ = -R × slope = -8.314 J·K⁻¹mol⁻¹ × (-197.1 K⁻¹) = 1638 J/mol = 1.64 kJ/mol
