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2 years ago

A 65 kg woman is horizontal in a push-up position. What are the vertical forces acting on her hands?

Physics

1 answer:

Leni [432]2 years ago

8 0

Answer:

Explanation:

Sum moments about her toes, to zero.

65(9.8)[1.00] - F[1.50] = 0

F = 424.666... N

Rounding to two significant digits makes it

F = 420 N

Don't know what to tell you, None of those options seem really close and you have two options at 410.4 N so I'm guessing a typo.

If you only take a single significant digit on gravity to make it 10 m/s² you end up with answer B), but why would one assume four significant digits in an answer when only a single significant digit went into the arguments? Only thing we can say for certain is it's probably not C)

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To solve this exercise we need the concept of Kinetic Energy and its respective change: Initial and final kinetic energy.

Let's start considering that the angular velocity is given by,

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Where I is the moment of inertia previously defined.

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In the case of the final kinetic energy, we have to,

$KE_f= mgh+ \frac{1}{2} mv^2 + \frac{1}{2} I \omega^2$

$KE_f = m * 9.81 * 0.76 + \frac{1}{2} m v^2 + \frac{1}{2} (\frac{2}{5} m R^2) (\frac{v}{R})^2$

For conservation of Energy we have, that

$KE_f = KE_i$ , then (canceling the mass and the radius)

$\frac{1}{2} 3.5^2 + \frac{1}{2}(\frac{2}{5})(3.5)^2= 9.81 * 0.76 + \frac{1}{2} v^2 + \frac{1}{2} (\frac{2}{5}) (v)^2$

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3 years ago

Photoelectrons are observed when a metal is illuminated by light with a wavelength less than 383 nm. What is the metal's work fu

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Answer:

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$W=\dfrac{hc}{\lambda}$

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Since, $1\ eV=1.6\times 10^{-19}\ J$

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3 years ago

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Consider a series LRC-circuit in which C-120.0 uF. When driven at a frequency w = 200.0 rad s-1 the com ples impedance is given

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Answer:

(a). (i). The reactants are $X_{L} =31.66\ \Omega$ .

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Explanation:

Given that,

Capacitor = 120.0 μC

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$X_{C}=41.66\ \Omega$

(II). We know that,

Formula of impedance is

$Z=\sqrt{R^2+X_{L}^2+X_{C}^2}$ ...(I)

Given equation of impedance is

$Z=(100-10j)$ ...(II)

On Comparing of equation (I) and (II)

$R = 100$

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Using formula of the resonant angular frequency

$angular\ frequency =\dfrac{1}{\sqrt{L\times C}}$

$angular\ frequency =\dfrac{1}{\sqrt{0.1583\times120\times10^{-6}}}$

$angular\ frequency =229.4\ rad/s$

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(b). The resonant angular frequency is 229.4 rad/s.

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3 years ago

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Answer:

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Explanation:

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3 years ago

A 65 kg woman is horizontal in a push-up position. What are the vertical forces acting on her hands?​

A 65 kg woman is horizontal in a push-up position. What are the vertical forces acting on her hands?