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3 years ago

During spring tide, the sun and moon are aligned. Which statement below best describes the tidal range during a spring tide?

Physics

2 answers:

Dima020 [189]3 years ago

6 0

Answer:

Explanation:

e-lub [12.9K]3 years ago

4 0

Answer:

Explanation:

A ) The smallest tidal ranges are less than 1 m (3 feet). The highest tides, called spring tides, are formed when the earth, sun and moon are lined up in a row. This happens every two weeks during a new moon or full moon. Smaller tides, called neap tides, are formed when the earth, sun and moon form a right angle.

C ) The most extreme tidal range occurs during spring tides, when the gravitational forces of both the Moon and Sun are aligned (syzygy), reinforcing each other in the same direction (new moon) or in opposite directions (full moon).

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3 years ago

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3 years ago

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The loop is in a magnetic field 0.20 T whose direction is perpendicular to the plane of the loop. At t = 0, the loop has area A

love history [14]

Answer:

Part a)

$EMF = 14 \times 10^{-3} V$

Part b)

$EMF = 15.67 \times 10^{-3} V$

Explanation:

As we know that magnetic flux through the loop is given as

$\phi = B.A$

now we have

$\phi = B\pi r^2$

now rate of change in flux is given as

$\frac{d\phi}{dt} = B(2\pi r)\frac{dr}{dt}$

now we know that

$A = \pi r^2$

$0.285 = \pi r^2$

$r = 0.30 m$

Now plug in all data

$EMF = (0.20)\times 2\pi\times (0.30) \times (0.037)$

$EMF = 14 \times 10^{-3} V$

Part b)

Now the radius of the loop after t = 1 s

$r_1 = r_0 + \frac{dr}{dt}$

$r_1 = 0.30 + 0.037$

$r_1 = 0.337 m$

Now plug in data in above equation

$EMF = (0.20)\times 2\pi\times (0.337) \times (0.037)$

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3 years ago

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3 years ago

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A bulldozer does 4,500 J of work to push a mound of soil to the top of a ramp that is 15 m high. The ramp is at an

Andreas93 [3]

Answer:

170 N

Explanation:

Given in the question that, work a bulldozer can do = 4500 J

We will use trigonometry identity to find the distance bulldozer will travel up the hill

sin(35) = opp/hypo

sin(35) = 15/hypo

hypo = 15/sin(35)

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Formula to use

work done = force × distance

Plug values in the above formula

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3 years ago

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