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Paladinen [302]
3 years ago
6

) A friend of yours complains that he often has lower back pain. One day while he picks up a package, you notice he bends at his

hips rather than keeping his back straight and bending his knees. Assume he pivots at the hips so that his spine is horizontal. His back muscles then exert a force at 48 cm away from the hips at 13.5 degrees above the horizontal (directed slightly up and to the left in the picture shown). The weight of his upper body is 560 N, and his center of mass is 38 cm from the hips. Draw a force diagram of the spine. What total force is exerted by the back muscle when he picks up a 10 kg package at 82 cm from the hips? Extra credit: Find the force exerted by the hips on the spine/pelvis
Physics
1 answer:
max2010maxim [7]3 years ago
3 0

Answer:

Explanation:

Please see the figure attached for free body diagram .

Taking torque about the hip joint ,

560 x 38 + 100 x 82 = F sin13.5 x 48

21280 + 8200 = 11.2 F

F = 2632.14 N .

Force exerted by the hip on the spine = 2632.14 N .

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Consider the power dissipated in a series R–L–C circuit with R=280Ω, L=100mH, C=0.800μF, V=50V, and ω=10500rad/s. The current an
ki77a [65]

Answer:

0.28802

2.57162 W

14.28 W

53.55 W

6.07142 W

Explanation:

R = 280Ω

L = 100 mH

C = 0.800 μF

V = 50 V

ω = 10500rad/s

For RLC circuit impedance is given by

Z=\sqrt{R^2+(X_L-X_C)^2}\\\Rightarrow Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}\\\Rightarrow Z=\sqrt{280^2+(10500\times 100\times 10^{-3}-\dfrac{1}{10500\times 0.8\times 10^{-6}})^2}\\\Rightarrow Z=972.1483\ \Omega

Power factor is given by

F=\dfrac{R}{Z}\\\Rightarrow F=\dfrac{280}{972.1483}\\\Rightarrow F=0.28802

The power factor is 0.28802

The average power to the circuit is given by

P=\dfrac{V^2}{Z}\\\Rightarrow P=\dfrac{50^2}{972.1483}\\\Rightarrow P=2.57162\ W

The average power to the circuit is 2.57162 W

Power to resistor

P_R=IR\\\Rightarrow P_R=5.1\times 10^{-2}\times 280\\\Rightarrow P_R=14.28\ W

Power to resistor is 14.28 W

Power to inductor

P_L=IX_L\\\Rightarrow P_L=5.1\times 10^{-2}\times 10500\times 100\times 10^{-3}\\\Rightarrow P_L=53.55\ W

Power to the inductor is 53.55 W

Power to the capacitor

P_C=IX_C\\\Rightarrow P_C=5.1\times 10^{-2}\times \dfrac{1}{10500\times 0.8\times 10^{-6}}\\\Rightarrow P_C=6.07142\ W

The power to the capacitor is 6.07142 W

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3 years ago
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The normal force which the path exerts on a particle is always perpendicular to the _________________ tangent to the path. trans
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A pail in a water well is hoisted by means of a frictionless winch, which consists of a spool and a hand crank. When Jill turns
Katena32 [7]

Answer:

166 W

Explanation:

Power is the rate at which work is done.

\text{Power} = \dfrac{\text{Work done}}{\text{time}}

The work done by Jill is the product of the weight of the pail and the height it moves.

The weight is the product of the mass and acceleration of gravity, <em>g</em>. Taking <em>g</em> as 9.81 m/s², the weight is

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Work done = (67.689 N)(27.0 m) = 1827.603 J

Power = (1827.603 J) ÷ (11.0 s) = 166 W

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4 years ago
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