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slamgirl [31]
3 years ago
10

Consider the following hypothetical subject performing the EMG laboratory: Immediately after the subject's maximum grip strength

for the non-dominant arm is determined to be 16 kg, to what force will the subject squeeze next?
Physics
1 answer:
AnnZ [28]3 years ago
3 0

Answer:

The assessment of the muscles' health and the motor neurons <em>(the nerve cells controlling the muscles)</em> is determined by the procedure called Electromyography aka EMG, which identifyes nerves/muscles issues and also the quality of the signal transmission between them; like when the units in a contraction increase, correlates with the increase in the force of contraction. In this particular case of the EMG lab, the subject will squeeze to <em>1kg</em> of force.

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Natasha_Volkova [10]

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Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit
Ber [7]

Solution :

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$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

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$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

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Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

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Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

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Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

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3 0
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konstantin123 [22]

Answer:

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7 0
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kogti [31]

Answer:

see solution below

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Hence the voltage across the 55ohms resistor is 19.8volts

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Answer:

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