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jonny [76]
3 years ago
8

HELP ASAP PLEASE - Which best describes how the arrangement of the sun, moon, and the earth affect the range of the tides during

a spring tide?
A. Because the earth, moon, and sun are all out of alignment, the pull of the moon and sun act in the opposite direction causing the highest high tides and the lowest low tides.


B. Because the earth, moon, and sun are in alignment, the pull of the moon and sun act in the same direction, causing the highest high tides and the lowest low tides.


C. Because the earth, moon, and sun are all out of alignment, the pull of the moon and sun act in the opposite direction, causing the lowest high tides and the highest low tides.


D. Because the earth, moon, and sun are all in alignment, the pull of the moon and sun act in the same direction causing the lowest high tides and the highest low tides.
Physics
2 answers:
koban [17]3 years ago
3 0
B. When the sun and the moon are aligned, their combined gravitational force attracts the water, causing super high tides on one side of the planet, and super low tides on the opposite side.
Simora [160]3 years ago
3 0

Answer: B. Because the earth, moon, and sun are in alignment, the pull of the moon and sun act in the same direction, causing the highest high tides and the lowest low tides.

Explanation:

The rise in level of waves due to gravitational pull of the sun and the moon are tides. Spring tides occur during new moon or full moon when the Sun, the Moon and The Earth all align in one straight line. The strong gravitational pull causes the highest high tides and the lowest low tides. Thus, the correct option is B.

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Answer:

537 N

Explanation:

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F = GMm / r²

where G is the universal gravitational constant

M is the mass of the planet

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and r is the distance between the object and the center of the planet

On Earth, you weigh 716 N, so:

716 N = GMm / r²

On planet X:

F = G (3M) m / (2r)²

F = 3/4 GMm / r²

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Answer:

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Explanation:

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A circuit consists of a series combination of 6.50 −kΩ and 4.50 −kΩ resistors connected across a 50.0-V battery having negligibl
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Answer:

Part A: 16.1 V

Part B: 20.5 V

Part C: 21.5%

Explanation:

The voltmeter is in parallel with the 4.5-kΩ resistor and the combination is in series with the 6.5-kΩ resistor. The equivalent resistance of the parallel combination is given as

\dfrac{1}{R_E}=\dfrac{1}{4.50}+\dfrac{1}{10.0}

R_E=\dfrac{4.50\times10.0}{4.50+10.0} = 3.10

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The voltmeter reading is the potential difference across the parallel combination. This is found by using the voltage-divider rule.

V_1 = \dfrac{3.10}{3.10+6.50}\times50.0 = \dfrac{3.10}{9.60}\times50.0 = 16.1 \text{ V}

Part B

Without the voltmeter, the potential difference across the 4.5-kΩ resistor is found using the same rule as above:

V_2 = \dfrac{4.50}{4.50+6.50}\times50.0 = \dfrac{4.50}{11.0}\times50.0 = 20.5 \text{ V}

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The error in % is given by

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