Answer:
L' = 1.231L
Explanation:
The transmission coefficient, in a tunneling process in which an electron is involved, can be approximated to the following expression:
L: width of the barrier
C: constant that includes particle energy and barrier height
You have that the transmission coefficient for a specific value of L is T = 0.050. Furthermore, you have that for a new value of the width of the barrier, let's say, L', the value of the transmission coefficient is T'=0.025.
To find the new value of the L' you can write down both situation for T and T', as in the following:
Next, by properties of logarithms, you can apply Ln to both equations (1) and (2):
Next, you divide the equation (3) into (4), and finally, you solve for L':
hence, when the trnasmission coeeficient has changes to a values of 0.025, the new width of the barrier L' is 1.231 L
Answer:
Part a)
Induced EMF when length vector is along Z direction is 0.72 V
Part b)
Induced EMF when length vector is along Y direction is ZERO
Explanation:
As we know that the motional EMF induced in the wire is given as
1)
As we know that
now we have
so we have
2)
If the length vector is along Y direction then we have
so again we have
so we have
EMF = 0
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Answer:
v = 38.73 m/s
Explanation:
Given
Extension of the bow, x = 50 cm = 0.5 m
Force of the arrow, F = 150 N
Mass of the arrow, m = 50 g = 0.05 kg
speed of arrow, v = ? m/s
We start by finding the spring constant
Remember, F = kx, so
k = F/x
k = 150 / 0.5
k = 300 N/m
the potential energy if the bow when pulled back is
E = 1/2kx²
E = 1/2 * 300 * 0.5²
E = 0.5 * 300 * 0.25
E = 37.5 J
The speed of the arrow will now be found by using the law of conservation of energy
1/2kx² = 1/2mv²
kx² = mv²
v² = kx²/m, on substituting, we have
v² = (300 * 0.5²) / 0.05
v² = 75 / 0.05
v² = 1500
v = √1500
v = 38.73 m/s
Answer:
ULTIMATE CORRECT ANSWER
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Explanation:
