Two metallic terminals protrude from a device. The terminal on the left is the positive reference for a voltage called vx (the o
ther terminal is the negative reference). The terminal on the right is the positive reference for a voltage called vy (the other terminal being the negative reference). If it takes 1 mJ of energy to push a single electron into the left terminal, determine the voltages vx and vy.
2 answers:
Answer:
vx=6.242PV
vy=-6.242PV
Explanation:
Two metallic terminals protrude from a device. The terminal on the left is the positive reference for a voltage called vx (the other terminal is the negative reference). The terminal on the right is the positive reference for a voltage called vy (the other terminal being the negative reference). If it takes 1 mJ of energy to push a single electron into the left terminal, determine the voltages vx and vy.
Here 1mj of energy is needed to an electron into the other terminal
voltage of an electron
Recall that E=IVt(product of power and time
but Q=quantity of charge is it (measured in coulombs)
E/Q=V
Ve=
0.6242*10^16
6.242*10^15
6.242PV(P=Peta=10^15)
voltage ve=6.242PV
Ve is used as Vy
The voltage vf is
Vf=
Vf=-6.242*10^15
vf=-6.242PV
therefore, vf is designated as vy
the voltages are of the same magnitude but in different direction
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Answer:
1.176
Explanation:
When the bullets impact the mass they become embedded on it, it is a plastic collision, therefore momentum is conserved.
v2 * (M + mb) = v1 * mb
Where
v1: muzzle velocity of the bullet
M: mass of the bob
mb: mass of the bullet
v2: mass of the bob with the bullet after being hit
v2 = v1 * mb / (M + mb)
Upon being impacted the bob will acquire speed v2, this implies a kinetic energy. The bob will then move and raise a height h. Upon acheiving the maximum height it will have a speed of zero. At that point all kinetic energy will be converted into potential energy.
Ek = 1/2 (M + mb) * v2^2
Ep = (M + mb) * g * h
Ek = Ep
1/2 (M + mb) * v2^2 = (M + mb) * g * h
1/2 * (v1 * mb / (M + mb))^2 = g * h
1/2 * v1^2 * mb^2 / (M + mb)^2 = g * h
v1^2 = g *h * (M+ mb)^2 / (1/2 * mb^2)
The height h that it reaches is related to the length L of the pendulum arm and the angle it forms with the vertical.
h = L * (1 - cos(a))
For the 9 mm:
For the 0.44 caliber:
The ratio is 460 / 391 = 1.176