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White raven [17]
3 years ago
6

Consider the accuracy of recording muscle electrical activity using electrodes placed on the skin surface, compared to directly

in the muscle. What are the differences between surface and intramuscular EMG recordings
Physics
1 answer:
OverLord2011 [107]3 years ago
8 0

Answer: The common difference between surface EMG and intramuscular EMG is that that former is non-invasive while the later is an invasive method

Explanation:

Electromyography (EMG) is used clinically for the examination of muscle excitations (muscle electrical activity) in both normal or abnormal conditions. There are two forms of EMG includes:

--> Surface EMT and

--> Intramuscular EMT

Surface EMT is a non invasive method of examination of muscle excitations for superficial and easily accessible muscles.

Intramuscular EMT is the invasive method of examination of muscle excitations usually for deep muscles.

The difference between the two forms of EMT includes:

- surface EMT is non- invasive while intramuscular EMT is invasive

- surface EMT is used to access superficial muscle while intramuscular EMT is used to access deep muscles.

- surface EMT requires less skill and time to carry out while intramuscular EMT requires special skills and takes more time while carrying out the procedure.

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Push a notebook or some other object so that it slides across a table or desktop. What force causes the notebook to start moving
Tju [1.3M]

Answer:

Kinetic force

Explanation:

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3 years ago
Ignoring reflection at the air-water boundary, if the amplitude of a 10 GHz incident wave in air is 20 V/m at the water surface,
dimaraw [331]

Answer:

0.80267 m

Explanation:

E(z) = Electric field = 1 µV/m

E_0 = 20 V/m

z = Depth

\sigma = Conductivity = 0.1 S/m

\epsilon_r = 81

\mu = Impedance of free space = 120\pi\ \Omega

Frequency is given by

E(z)=E_0e^{-\alpha z}

Parameter is given by

\alpha=\dfrac{\sigma}{2}\sqrt{\dfrac{\mu}{\epsilon_r}}\\\Rightarrow \alpha=\dfrac{1}{2}\sqrt{\dfrac{(120\pi)^2}{81}}\\\Rightarrow \alpha=20.94395\ N_p/m

From the first equation

1\times 10^{-6}=20e^{-20.94395z}\\\Rightarrow ln\dfrac{1\times 10^{-6}}{20}=-20.94395z\\\Rightarrow z=\dfrac{ln\dfrac{1\times 10^{-6}}{20}}{-20.94395}\\\Rightarrow z=0.80267\ m

The depth is 0.80267 m

7 0
3 years ago
. How far apart must two point charges of 75.0 nC (typical of static electricity) be to have a force of 1.00 N between them?
o-na [289]

Answer:

0.00712 m

Explanation:

Given:

Charge on first particle (q₁) = 75 nC = 75\times 10^{-9}\ C

Charge on second particle (q₂) = 75 nC = 75\times 10^{-9}\ C

Force (F) = 1.00 N

Separation (d) = ?

The magnitude of force is given by Coulomb's law which states that, the magnitude of force acting between two charged particles separated by a distance is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.

Therefore, the magnitude of force is given as:

F=\dfrac{k|q_1||q_2|}{d^2}

Where, k=9\times 10^9\ N\cdot m^2/ C^2 is the coulomb's constant.

Plug in the given values and solve for 'd'. This gives,

1.00=\frac{9\times 10^9\times 75.0\times 10^{-9}\times 75.0\times 10^{-9} }{d^2}\\\\d^2=\frac{9\times 10^9\times 75.0\times 10^{-9}\times 75.0\times 10^{-9} }{1.00}\\\\d=\sqrt{\frac{9\times 10^9\times 75.0\times 10^{-9}\times 75.0\times 10^{-9} }{1.00}}\\\\d=0.00712\ m

Therefore, the distance between the charges is 0.00712 m.

6 0
3 years ago
Sketch a position-time graph for a bear starting
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Explanation:

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