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avanturin [10]
3 years ago
5

Need help with physics.

Physics
1 answer:
GenaCL600 [577]3 years ago
8 0

The centripetal force : F = 293.3125 N

<h3> Further explanation </h3>

Given

mass = 65 kg

v = 9.5 m/s

r = 20 m

Required

the centripetal force

Solution

Centripetal force is a force acting on objects that move in a circle in the direction toward the center of the circle  

\large {\boxed {\bold {F = \frac {mv ^ 2} {R}}}

F = centripetal force, N  

m = mass, Kg  

v = linear velocity, m / s  

r = radius, m  

Input the value :

F = 65 x 9.5² / 20

F = 293.3125 N

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What are tides? the regular daily rises and falls in sea level caused by the gravitational attraction of the Moon and Sun on Ear
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6 0
3 years ago
A 1.80-m -long uniform bar that weighs 531 N is suspended in a horizontal position by two vertical wires that are attached to th
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Answer:

(a) 498.4 Hz

(b) 442 Hz

Solution:

As per the question:

Length of the wire, L = 1.80 m

Weight of the bar, W = 531 N

The position of the copper wire from the left to the right hand end, x = 0.40 m

Length of each wire, l = 0.600 m

Radius of the circular cross-section, R = 0.250 mm = .250\times 10^{- 3}\ m

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Applying the equilibrium condition at the left end for torque:

T_{Al}.0 + T_{C}(L - x) = W\frac{L}{2}

T_{C}(1.80 - 0.40) = 531\times \frac{1.80}{2}

T_{C} = 341.357\ Nm

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W = T_{Al} + T_{C}

531 = T_{Al} + 341.357

T_{Al} = 189.643\ Nm

Now,

The fundamental frequency is given by:

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

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(a) For the fundamental frequency of Aluminium:

f = \frac{1}{2L}\sqrt{\frac{T_{Al}}{\mu}}

f = \frac{1}{2L}\sqrt{\frac{T_{Al}}{\pi R^{2}\rho_{Al}}}

where

\rho_{l} = 2.70\times 10^{3}\ kg/m^{3}

f = \frac{1}{2\times 0.600}\sqrt{\frac{189.643}{\pi 0.250\times 10^{- 3}^{2}\times 2.70\times 10^{3}}} = 498.4\ Hz

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f = \frac{1}{2L}\sqrt{\frac{T_{C}}{\mu}}

f = \frac{1}{2L}\sqrt{\frac{T_{C}}{\pi R^{2}\rho_{C}}}

where

\rho_{C} = 8.90\times 10^{3}\ kg/m^{3}

f = \frac{1}{2\times 0.600}\sqrt{\frac{341.357}{\pi 0.250\times 10^{- 3}^{2}\times 2.70\times 10^{3}}} = 442\ Hz

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