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Sveta_85 [38]
4 years ago
14

11.76 cm² into two significant figures and include the appropriate units

Physics
1 answer:
PilotLPTM [1.2K]4 years ago
8 0
You will need to turn this into scientific notation, so:

1.176 × 10^1 cm^2

To two significant figures means to two digits so since 7 is greater than 5, you will need to round up.

Final answer is:

1.2×10^1 cm^2
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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

Now,

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

The weight of the wire balances the tension in both the wires collectively:

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}}

where

\mu = A\rho = \pi R^{2}\rho

(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

(b)  For the fundamental frequency of Copper:

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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