Answer:
what do the choices look like
Explanation:
Answer:
C. The motor tasks performed in the experiment were too simple.
On this case is the best option since the student wants to explain the effect of motor imagery and action observation together into the excitability. And maybe is too simple, since we need to cover other possibilities in order to analyze the excitability.
Explanation:
A. The procedure used did not include MEP recordings prior to each task.
Not true, is not a requisite record MEP prior to the task to evaluate the variable of interest on this case.
B. MEP amplitudes in an individual are typically highly consistent.
The Motor evoked potentials (MEP) "are electrical signals recorded from neural tissue or muscle after activation of central motor pathways". But on this case that's a technical aspect related to the topic and this not would be the reason why we need to withhold the presentation
C. The motor tasks performed in the experiment were too simple.
On this case is the best option since the student wants to explain the effect of motor imagery and action observation together into the excitability. And maybe is too simple, since we need to cover other possibilities in order to analyze the excitability.
D. The six different conditions were run in random order.
That's not true the student are not analyzing 6 different conditions, just 2.
The correct option is out of the screen.
As the motion of positive charge is the direction of current in the wire. From the right-hand curl rule, the magnetic field direction will be outside the paper or the screen. As the <span>wire runs left to right and carries a current in the direction from left to right, the magnetic field lines will be outside the screen.</span>
Answer:
2.4 mm
Explanation:
Given that:
Initial Original length of the wire L = 3 mm
The stretch of the first wire ΔL= 1. 2 mm
The length of the second wire L'' = 6 mm
The stretch of the second wire ΔL'' = ???
Considering the Tension of the system; the Young modulus and the cross sectional remains constant ; as such:
Thus, the same material under the same tension stretches 2.4 mm
Answer:
frequency is 195.467 Hz
Explanation:
given data
length L = 4.36 m
mass m = 222 g = 0.222 kg
tension T = 60 N
amplitude A = 6.43 mm = 6.43 × 
m
power P = 54 W
to find out
frequency f
solution
first we find here density of string that is
density ( μ )= m/L ................1
μ = 0.222 / 4.36
density μ is 0.050 kg/m
and speed of travelling wave
speed v = √(T/μ) ...............2
speed v = √(60/0.050)
speed v = 34.64 m/s
and we find wavelength by power that is
power = μ×A²×ω²×v / 2 ....................3
here ω is wavelength put value
54 = ( 0.050 ×(6.43 × )²×ω²× 34.64 ) / 2
0.050 ×(6.43 × )²×ω²× 34.64 = 108
ω² = 108 / 7.160 × 
ω = 1228.16 rad/s
so frequency will be
frequency = ω / 2π
frequency = 1228.16 / 2π
frequency is 195.467 Hz
