A hammer of mass m falls from rest off of a roof and drops a height H onto your head.

1 answer:

6 0

For conservation of energy we have to:
mgH=mv²/2
Clearing
 v=sqrt(2gH)
Then, by definition
 F=Δp/Δt= Δ(mv)/ Δt=m Δ(v)/Δt=
 =m[sqrt(2gH)-0]/Δt= m[sqrt(2gH)]/ Δt
the answer is
F=m[sqrt(2gH)]/ Δt

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A small sphere of radius R is arranged to pulsate so that its radius varies in simple harmonic motion between a minimum of R−x a

Colt1911 [192]

Answer:

The intensity of sound wave at the surface of the sphere $I = \frac{ 2\pi^{2}R^{2} f^{2}\sqrt{\rho B}(\triangle R)^{2}}{ d^{2} }$

Explanation:

B = Bulk modulus

Intensity, $I = \frac{P_{max} ^{2} }{2\sqrt{\rho B} }$

The amplitude of oscillation of the sphere is given by:

$P_{max} = BkA\\k = \frac{2\pi }{\lambda} \\$

$A = \triangle R\\$

Substitute v and A into Pmax

$P_{max} = (2\pi f)\sqrt{\rho B} \triangle R\\ P_{max} ^{2} = 4\pi^{2} *f^{2} \rho B (\triangle R)^{2}$

$I = \frac{ 4\pi^{2} f^{2} \rho B (\triangle R)^{2}}{2\sqrt{\rho B} }$

$P_{total} = 4\pi R^{2} I$

$P_{total} =4\pi R^{2} \frac{ 2\pi^{2} f^{2} \rho B (\triangle R)^{2}}{\sqrt{\rho B} }$

The intensity of the sound wave at a distance is given by:

$I = \frac{P_{total} }{4\pi d^{2} }$

$I = 4\pi R^{2} \frac{ 2\pi^{2} f^{2} \rho B (\triangle R)^{2}}{\sqrt{\rho B} } * \frac{1}{4\pi d^{2} } \\I = \frac{ 2\pi^{2}R^{2} f^{2}\sqrt{\rho B}(\triangle R)^{2}}{ d^{2} }$