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gizmo_the_mogwai [7]
3 years ago
14

What forces control the strength of the tides

Physics
2 answers:
Alona [7]3 years ago
8 0

Answer: gravity is one major force that creates tides

Explanation:

in 1687, sir Isaac newton explained that ocean tides result from the gravitational attraction of the sun and the moon on the oceans of the earth

Colt1911 [192]3 years ago
7 0

Answer:

Gravity is one major force that creates tides. In 1687, Sir Isaac Newton explained that ocean tides result from the gravitational attraction of the sun and moon on the oceans of the earth (Sumich, J.L., 1996).

Explanation:

I hope this helps.

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HELP OR I WILL DIEJSNCSDKKS.
iogann1982 [59]

Answer:

Increase

Explanation:

because it makes it makes sense lol

6 0
3 years ago
Read 2 more answers
A certain comet has a parallax 1/40 as large as the moon. How far away is it (compared to the Moons distance)? Show Work
Ad libitum [116K]
The distance and parallax are inversely related. We can find the distance using the following equation:

d= \frac{1}{p}

where d is distance and p is parallax.

We are given the parallax of the comet relative to the moon, and we are looking for the distance to the comet relative to the moon's distance, so wee can plug in the following value:

d= \frac{1}{ \frac{1}{40} }

The distance is 40 times as far away as the moon.
3 0
4 years ago
If velocity is positive, which woul
shusha [124]

Answer:

C. An inital volocity that is faster than the final volocity

Explanation:

.

5 0
3 years ago
A record of travel along a straight path is as follows: 1. Start from rest with constant acceleration of 2.45 m/s^2 for 20.0 s.
Anton [14]

Answer:

a) The total displacement of the trip was 5.32 × 10³ m

b) The average speeds were:

leg 1: 24.5 m/s

leg 2: 49 m/s

leg 3: 23.9 m/s

Complete trip: 43.8 m/s

Explanation:

The position and velocity equations for an object moving along a straight line are as follows:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

If the velocity is constant, then a = 0 and x = x0 + v · t where "v" is the velocity.

a) To calculate the total displacement of the trip, let´s calculate the distance traveled in each phase.

Phase 1:

x = x0 + v0 · t + 1/2 · a · t²

x = 0 m + 0 m/s · t + 1/2 · 2.45 m/s² · (20.0 s)²

x = 490 m

The velocity reached in that phase is:

v = v0 + a · t

v = 0 m/s +  2.45 m/s² · 20.0 s

v = 49.0 m/s

Phase 2:

x = x0 + v · t

x = 490 m + 49.0 m/s · 96.0 s

x = 5.19 × 10³ m

Phase 3:

x = x0 + v0 · t + 1/2 · a · t²

x =  5.19 × 10³ m + 49 m/s · 5.44 s - 1/2 · 9.00 m/s² · (5.44 s)²

x = 5.32 × 10³ m

The total displacement of the trip was 5.32 × 10³ m

b) The average speed is calculated as the traveled distance divided by the elapsed time:

average speed v = final position - initial position / (final time- initial time)

Phase 1:

v = 490 m - 0 m / 20.0 s = 24.5 m/s

Phase 2:

v = 5.19 × 10³ m - 490 / 96.0 s

v = 48.9 m/s   (without rounding the final position the result is 49.0 m/s)

Phase 3:

v =  5.32 × 10³ m -  5.19 × 10³ m / 5.44 s = 23.9 m/s

For the complete trip:

v =  5.32 × 10³ m  - 0 m / (20.0 s + 96.0 s + 5.44 s)

v = 43.8 m/s

7 0
4 years ago
Calculate the hang time of a person who moves 3 m horizontally during a 1.25-m high jump. What is the hang time when the person
jekas [21]

The duration of time for which an object stays in air is called the hang time.

For an athlete who moves 3m horizontally during a 1.25m high jump, the hang time will be the sum of the time taken by the athlete to reach the maximum height and the time taken for the athlete to reach the ground from maximum height.

Calculate the time taken t_1 by the athlete to reach the maximum height

t_1 = \sqrt{\frac{2h}{g}}

t_1 = \sqrt{\frac{2(1.25)}{9.8}}

t_1 = 0.5s

The athlete takes same time to reach the ground from the maximum height, so t_2 = 0.5s

Calculate the hang time will be

t =t_1+t_2

t = 0.5+0.5

t = 1s

Therefore the hang time of the athlete when he moves a horizontal distance of 3m is 1s.

Similarly, when the athlete runs 6m horizontally, then also there will not be a change in the hang time of the athlete as the hang time is independent of the horizontal distance covered.

4 0
3 years ago
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