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valentina_108 [34]
2 years ago
9

The average depth of water in a place near the coast of the Pacific ocean is 5.0 The largest tidal range in that place is 2.0m C

alculate the maximum depth of water at that place
Physics
1 answer:
GREYUIT [131]2 years ago
7 0

A sinusoidal function is a function based on the sine function.

The maximum depth of the water at the place, is <u>6.0 meters</u>.

Reason:

The known parameters are;

The average depth of water at a place, d_{ave} = 5.0

The largest tidal range in the place, R = 2.0 m.

Required;

To calculate the maximum depth of water at the given place.

Solution:

The tidal range is the consecutive difference between highs and lows of

the tides, which is a sinusoidal motion.

Therefore, the tidal range is twice the amplitude of the tide level.

The amplitude of the tide, <em>a</em> = \dfrac{2.0}{2} m = 1 m

The maximum depth of water at the place, d_{max} = a + d_{ave}

∴ d_{max} = 1.0 + 5.0 = 6.0

The maximum depth, d_{max} = <u>6.0 meters</u>

Learn more about sinusoidal functions here:

brainly.com/question/16300816

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The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in
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Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

T_1+V_1=T_2+V_2

0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2}  \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}

Also;

\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}

Thus;

k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}

where;

\delta = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}

k = 104.82\ \  N/m

Thus; the spring constant = 104.82 N/m

b

The angular velocity can be calculated by also using the conservation of energy;

T_1+V_1 = T_3 +V_3  \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3}  \omega_3^2 +  \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0

\frac{m(a+b)^2}{3} \omega_3^2  + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0

\frac{1.53(0.6+0.6)^2}{3} \omega_3^2  + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0

0.7344 \omega_3^2 = 2.128

\omega _3 = \sqrt{\frac{2.128}{0.7344} }

\omega _3 =1.70 \ rad/s

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s

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