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never [62]
1 year ago
14

Why is there a delay for the tides to react to the Moon's and Sun's gravitational effect?

Chemistry
1 answer:
Sloan [31]1 year ago
7 0

There is no delay because the effects of gravity are constantly acting on bodies of water. Hence, option D is correct.

<h3>What is the gravitational effect?</h3>

Gravitation or just gravity is the force of attraction between any two bodies.

The Moon exerts over twice the gravitational pull of the Sun on Earth's tides because of its proximity to Earth. -The lunar tidal bulges are about twice the size of the solar tidal bulges. -The Sun's contribution to the tides is less than that of the Moon.

When the sun and moon are perfectly unaligned (they form a 90-degree angle relative to the earth), there are still tides because the moon's gravitational gradient is stronger than the sun's. The sun's gravitational gradient never completely cancels out the moon's.

Hence, option D is correct.

Learn more about the gravitational effect here:

brainly.com/question/14433868

#SPJ1

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Matt's cube, after 5 trial, had an average destiny of 7.40. g/cm what is it made of ?​
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Explanation:

5 trial, had an average destiny of 7.40. g/cm

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3 years ago
A container has a mixture of NO2 gas and N2O4 gas in equilibrium. The chemical reaction between the two gases is described by th
kondaur [170]

Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄

Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.

For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:

Kp = \frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }

where:

P(N₂O₄) and P(NO₂) are the partial pressure of each gas.

Calculating constant:

Kp = \frac{38.8}{61.2^{2} }

Kp = 0.0104

After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.

P(N₂O₄) + P(NO₂) = 200

P(N₂O₄) = 200 - P(NO₂)

Kp = \frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }

0.0104 = \frac{200 - P(NO_{2})  }{[P(NO_{2} )]^{2}}

0.0104[P(NO_{2} )]^{2} + P(NO_{2} ) - 200 = 0

Resolving the second degree equation:

P(NO_{2} ) = \frac{-1+\sqrt{9.32} }{0.0208}

P(NO_{2} ) = 98.7

Find partial pressure of N₂O₄:

P(N₂O₄) = 200 - P(NO₂)

P(N₂O₄) = 200 - 98.7

P(N₂O₄) = 101.3

The partial pressures are P(NO_{2} ) = 98.7 MPa and P(N₂O₄) = 101.3 MPa

3 0
3 years ago
A student designed an experiment to test the effect on the magnitude of the magnetic field that is generated around the wire loo
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Answer: C ) the student’s dependent variable is the magnitude of the magnetic field that is generated

E ) the student’s independent variable is the amount of current that is being passed through the wire

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On combustion of 0.2 gm of organic compound gave 0.147g of CO2, 0.12g of H2O and 74.6 ml of nitrogen gas at STP. Find emperical
zhuklara [117]

Answer:

0.2 is conpound Co2 STP.

Explanation:

on combustion of 0.2 gm of organic compound gave 0.147g of CO2, 0.12g of H2O and 74.6 ml of nitrogen gas at STP. Find emperical formula of compound.?

​

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Which of the following is the quantum number set for an electron in the 3rd energy level, dumb-bell shaped orbital, on the z-axi
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Answer:

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