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AveGali [126]
3 years ago
15

Approximately where is it currently high tide on Earth? Group of answer choices wherever it is currently noon anywhere that ocea

n water laps upon the shore on the portion of Earth facing directly toward the Moon and on the portion of Earth facing directly away from the Moon only on the portion of the Earth facing directly toward the Moon
Physics
1 answer:
Wewaii [24]3 years ago
7 0

Answer:

Option D, only on the portion of the Earth facing directly toward the Moon

Explanation:

Tides are caused by the gravitational pull of moon. The part of earth that faces the moon experiences the highest gravitational force and hence the high tides will occur in this regions only. The regions that do not faces the moon experiences low tides. It is the gravity of moon that attracts the ocean water towards itself.

Hence, Option D is correct

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A place kicker must kick a football from a point 36.0 m (about 40 yards) from the goal. half the crowd hopes the ball will clear
trapecia [35]
<span>The ball clears by 11.79 meters Let's first determine the horizontal and vertical velocities of the ball. h = cos(50.0)*23.4 m/s = 0.642788 * 23.4 m/s = 15.04 m/s v = sin(50.0)*23.4 m/s = 0.766044 * 23.4 m/s = 17.93 m/s Now determine how many seconds it will take for the ball to get to the goal. t = 36.0 m / 15.04 m/s = 2.394 s The height the ball will be at time T is h = vT - 1/2 A T^2 where h = height of ball v = initial vertical velocity T = time A = acceleration due to gravity So plugging into the formula the known values h = vT - 1/2 A T^2 h = 17.93 m/s * 2.394 s - 1/2 9.8 m/s^2 (2.394 s)^2 h = 42.92 m - 4.9 m/s^2 * 5.731 s^2 h = 42.92 m - 28.0819 m h = 14.84 m Since 14.84 m is well above the crossbar's height of 3.05 m, the ball clears. It clears by 14.84 - 3.05 = 11.79 m</span>
4 0
4 years ago
Suppose you have a pendulum clock which keeps correct time on Earth(acceleration due to gravity = 1.6 m/s2). For ever hour inter
kaheart [24]

The moon clock is A) (9.8/1.6)h compared to 1 hour on Earth

Explanation:

The period of a simple pendulum is given by the equation

T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration of gravity

In this problem, we want to compare the period of the pendulum on Earth with its period on the Moon. The period of the pendulum on Earth is

T_e=2\pi \sqrt{\frac{L}{g_e}}

where

g_e = 9.8 m/s^2 is the acceleration of gravity on Earth

The period of the pendulum on the Moon is

T_m=2\pi \sqrt{\frac{L}{g_m}}

where

g_m = 1.6 m/s^2 is the acceleration of gravity on the Moon

Calculating the ratio of the period on the Moon to the period on the Earth, we find

\frac{T_m}{T_e}=\frac{g_e}{g_m}=\frac{9.8}{1.6}

Therefore, for every hour interval on Earth, the Moon clock will display a time of

A) (9.8/1.6)h

#LearnwithBrainly

6 0
3 years ago
What is free fall? Explain it in brief.​
xxTIMURxx [149]
 · free fall is any motion of a body where gravity is the only forceacting upon it. In the context of general relativity, where gravitation is reduced to a space-time curvature, a body in free fall has no force acting on it.

3 0
3 years ago
An airplane went from 120 m/s to 180 m/s in 4.0 seconds. What was its acceleration?
scoundrel [369]
<span>15 m/s^2 The first thing to calculate is the difference between the final and initial velocities. So 180 m/s - 120 m/s = 60 m/s So the plane changed velocity by a total of 60 m/s. Now divide that change in velocity by the amount of time taken to cause that change in velocity, giving 60 m/s / 4.0 s = 15.0 m/s^2 Since you only have 2 significaant figures, round the result to 2 significant figures giving 15 m/s^2</span>
3 0
3 years ago
True or false the refraction of a wave is how many wavelengths pass a fixed point each second
Ksenya-84 [330]
False. That description fits the wave's 'frequency'. 
It has nothing to do with refraction.
8 0
3 years ago
Read 2 more answers
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