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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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Two identical charges, 2 m apart, exert forces of magnitude 4 N on each other. The value of each charge is: 1. 9 × 105 C 2. 4.2
lesya692 [45]

Answer:

The value of each charge is 4.22 x 10⁻⁵ C

Explanation:

Given;

distance between the two identical charges, d = 2 m

the force of repulsion between these two charges, F = 4N

Apply Coulomb's law;

F = \frac{kq_1q_2}{r^2} \\\\but \ q_1 =q_2,then \ let \ q_1 =q_2 = q\\\\F = \frac{kq^2}{r^2}\\\\q^2 = \frac{Fr^2}{k}\\\\q^2 = \frac{4*2^2}{9*10^9} \\\\q ^2 = 1.7778*10^{-9}\\\\q = \sqrt{1.7778*10^{-9}}\\\\q =4.22 *10^{-5} C\\\\q= q_1=q_2= 4.22 *10^{-5} C

Therefore, the value of each charge is 4.22 x 10⁻⁵ C

7 0
3 years ago
A tourist drops (from rest) a ping pong ball from the top of the tower, which has a height of 324 meters. Assuming no air resist
Elanso [62]

Answer:

8.13secs

Explanation:

From the question weal are given

Height H =324m

Required

time it takes to drop t

Using the equation of motion

H = ut + 1/2gt²

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324 = 0(t)+1/2(9.8)t²

324 = 1/2(9.8)t²

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t² =324/4.9

t² = 66.12

t = √66.12

t = 8.13secs

Hence the time taken to drop is 8.13secs

4 0
2 years ago
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lana66690 [7]

hi <3

i believe i explained this answer properly in my last answer but it would be 4kg and 2400m as these are the SI units for these values.

hope this helps :)

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6 0
3 years ago
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