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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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In a wire, when elongation is 4 cm energy stored is E. if it is stretched by 4 cm, then what amount of elastic potential energy
myrzilka [38]
<h2>Answer:</h2>

4E

<h2>Explanation:</h2>

The elastic potential energy of an elastic material (e.g a spring, a wire), is the energy stored when the material is stretched or compressed. It is given by

U = \frac{1}{2}kx^2               --------------------(i)

Where;

U = potential energy stored

k = spring constant of the material

x = elongation (extension or compression of the material).

<em>From the first statement;</em>

<em>when elongation (x) is 4cm, energy stored (U) is E</em>

<em>Substitute these values into equation (i) as follows;</em>

E = \frac{1}{2}k(4)^2

E = 8k

<em>Make k subject of the formula</em>    

k = \frac{E}{8}   [measured in J/cm]

<em>From the second statement;</em>

<em>It is stretched by 4cm.</em>

This means that total elongation will be 4cm + 4cm = 8cm.

The potential energy stored will be found by substituting the value of x = 8cm and k = \frac{E}{8} into equation (i) as follows;

U = \frac{1}{2}\frac{E}{8} (8)^2  

U = \frac{1}{2}{8E}

U = {4E}

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an electron accelerated from rest through a voltage of 770 v enters a region of constant magnetic field. part a if the electron
finlep [7]

The magnitude of the magnetic field on which an electron accelerated from rest through a voltage of 770v enters a region of constant magnetic field is 3.744 * 10^{-4} T

From the conservation of energy, we have

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v=\sqrt{\frac{2qV}{m} }

=\sqrt{\frac{2*(1,6 * 10^{-19})(770) }{9.11 * 10^{-31} } }

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At Equilibrium, Centripetal force = magnetic force

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by putting, m=9.11* 10^-31 kg

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We get, B=3.774* 10^-4 T

Hence the magnitude of the magnetic field, B=3.774* 10^-4 T

Learn more about Magnetic Field here:

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