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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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nataly862011 [7]

Answer:

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b) v_c=0.0566\ m.s^{-1}

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Explanation:

Given:

mass of the book, m_b=1.35\ kg

combined mass of the student and the skateboard, m_c=97\ kg

initial velocity of the book, v_b=4.61\ m.s^{-1}

angle of projection of the book from the horizontal, \theta=28^{\circ}

a)

velocity of the student before throwing the book:

Since the student is initially at rest and no net force acts on the student so it remains in rest according to the Newton's first law of motion.

u_c=0\ m.s^{-1}

where:

u_c= initial velocity of the student

velocity of the student after throwing the book:

Since the student applies a force on the book while throwing it and the student standing on the skate will an elastic collision like situation on throwing the book.

m_c.v_c=m_b.v_b\times \cos\theta

where:

v_c= final velcotiy of the student after throwing the book

b)

m_c.v_c=m_b.v_b\times \cos\theta

97\times v_c=1.35\times 4.61\cos28

v_c=0.0566\ m.s^{-1}

c)

Since there is no movement of the student in the vertical direction, so the total momentum transfer to the earth will be equal to the momentum of the book in vertical direction.

p_e=m_b.v_b\sin\theta

p_e=1.35\times 4.61\times \sin28^{\circ}

p_e=2.9218\ kg.m.s^{-1}

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Two identical small charged spheres hang in equilibrium with equal masses (0.02kg). The length of the strings is equal (0.18m) a
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Answer:

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Explanation:

From the question we are told that

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=>     q = tan \theta * \frac{m * g * r^2 }{k}

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