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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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What process is a method of heat transfer but does NOT contribute significantly to heating the surface or atmosphere of the eart
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conduction

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A cubic box has a 0.6 cubic decimetre side. Find the mass of air in the box in gram. (1 Liter of air has a mass of 1.3 g).
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Answer:

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Explanation:

A litre equals a cubic decimetre. By definition of density (\rho), in grams per litre, we calculate the mass of air inside the cube (m), in grams:

m = \rho \cdot V (1)

Where V is the volume occupied by air within the cube, in litres.

If we know that V = 0.6\,L and \rho = 1.3\,\frac{g}{L}, then the mass of air is:

m = \left(1.3\,\frac{g}{L} \right)\cdot (0.6\,L)

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4 0
3 years ago
Niobium metal becomes a superconductor when cooled below 9 K. Its superconductivity is destroyed when the surface magnetic field
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To develop the problem it is necessary to apply the concepts related to Magnetic Field.

The magnetic field is defined as

B = \frac{\mu_0 I}{2\pi r}

Where,

\mu_0 = Permeability constant in free space

r = Radius

I = Current

Our values are given as,

B = 0.1T

d = 4.5mm

r = 2.25mm

If the maximum current that the wire can carry is I, then

B = \frac{\mu_0 2I}{4\pi r}

I = \frac{Br}{2\frac{\mu_0}{4\pi}}

I = \frac{(0.1T)(2.25*10^{-3}m)}{2(1*10^{-7}N/A^2)}}

I = 1125A

Therefore the maximum current is 1125A

4 0
4 years ago
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