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melamori03 [73]
3 years ago
14

A student is considering doing a complete repeated measures design experiment involving motor skills. The student's advisor has

told him that people show a large initial improvement on the task followed by slow steady improvement after this initial change. The student must choose a technique for balancing practice effects. Which technique should the student NOT use?
a. block randomization
b. latin square
c. ABBA counterbalancing
d. all possible orders
Physics
1 answer:
Lelu [443]3 years ago
3 0

Answer:

<em>c. ABBA counterbalancing </em>

Explanation:

The student should not use the method because it is a progressive error management technique for each subject by introducing all <em>treatment circumstances twice, first in one sequence, then in the other (AB, BA) by subject counterbalancing.</em>

If participants experience conditions more than once, they experience the conditions first in one order, then the opposite order.

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2 years ago
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Problem #3: 3.6. A diode has wdo = 0.4 µm and φj = 0.85 V. (a) What reverse bias is required to triple the depletion-layer width
Alchen [17]

Answer:

a) 6.8 Volt

b) 1.21Цm

Explanation:

We are given from the question that

   The zero -bias depletion layer width(W_{do}) is 0.4Цm

  The built in voltage φj  is 0.85V

Now to calculate the reverse voltage( V_{R}) that would be required  to triple the depletion - layer width.

  The depletion - layer width (W_{d}) of the diode has the formula

                          W_{d} = W_{do} \sqrt{1 + \frac{V_{R} }{Qj} }

    For three times of   W_{d} we have

        3W_{d} = W_{do} \sqrt{1 +\frac{V_{R} }{Qj} }

         =>      \frac{V_{R} }{Qj} = 3^{2} -1

        => V_{R} = 8Qj

        Substituting value of φj

       We have

                         V_{R} = 8(0.85V)

                               =   6.8 V

The required bias voltage  V_{R} is  6.8 V

   The solution for the b part of the question is uploaded on first image

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D)LT^-1 speed=distance(L)/time(T)——>L/T
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Explanation:

Given:

m = 1.673 × 10^-27 kg

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r = 0.75 nm

= 0.75 × 10^-9 m

A.

Energy, U = (kQq)/r

Ut = 1/2 mv^2 + 1/2 mv^2

1.673 × 10^-27 × v^2 = (8.99 × 10^9 × (1.602 × 10^-19)^2)/0.75 × 10^-9

v = 1.356 × 10^4 m/s

B.

F = (kQq)/r^2

F = m × a

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4 0
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So,       V = \frac{10^{-3}}{19.3 \times 10^{3} kg/m^{3}}

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