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grin007 [14]
2 years ago
14

A coastline on Earth experiences tides according to the graph below.

Physics
2 answers:
zloy xaker [14]2 years ago
6 0

Answer:

Its C,  Mixed semidiurnal

Explanation:

Salsk061 [2.6K]2 years ago
5 0

Answer:

Most coastal areas, with some exceptions, experience two high tides and two low tides every lunar day (Ross, D.A., 1995). Almost everyone is familiar with the concept of a 24-hour solar day, which is the time that it takes for a specific site on the Earth to rotate from an exact point under the sun to the same point under the sun. Similarly, a lunar day is the time it takes for a specific site on the Earth to rotate from an exact point under the moon to the same point under the moon. Unlike a solar day, however, a lunar day is 24 hours and 50 minutes. The lunar day is 50 minutes longer than a solar day because the moon revolves around the Earth in the same direction that the Earth rotates around its axis. So, it takes the Earth an extra 50 minutes to “catch up” to the moon (Sumich, J.L., 1996; Thurman, H.V., 1994).

the answer is C

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_____ can alter wind flow and create a rain shadow effect. Beaches Lakes Mountains Oceans
Ket [755]
The correct answer is Mountains. Mountains can alter wind flow and create a rain shadow effect. A mountain is a large landform and is high surrounding a limited area. It is generally steeper and higher than a hill.
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2 years ago
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By what factor will the Electrostatic Force between two charged objects change when the amount of charge on both objects doubles
Mademuasel [1]

Answer:

F' = (4/9)F

Explanation:

The electrostatic force between two charged objects is given by Coulomb's Law:

F = kq₁q₂/r²   -------------------- equation (1)

where,

F = Electrostatic Force

k = Coulomb's Constant

q₁ = magnitude of first charge

q₂ = magnitude of second charge

r = distance between charges

Now, when the charges and distance altered as follows:

q₁' = 2q₁

q₂' = 2q₂

r' = 3r

Then,

F' = kq₁'q₂'/r'²

F' = k(2q₁)(2q₂)/(3r)²

F' = (4/9)kq₁q₂/r²

using equation (1):

<u>F' = (4/9)F</u>

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2 years ago
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Nataly_w [17]

(6) Wagon B is at rest so it has no momentum at the start. If <em>v</em> is the velocity of the wagons locked together, then

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(7) False. If you double the time it takes to perform the same amount of work, then you <u>halve</u> the power output:

<em>E</em> <em>/</em> (2<em>t </em>) = 1/2 × <em>E/t</em> = 1/2 <em>P</em>

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3 0
3 years ago
horizontal clothesline is tied between 2 poles, 14 meters apart. When a mass of 3 kilograms is tied to the middle of the clothes
lbvjy [14]

Answer:

The tension is  T =  103.96N

Explanation:

The free body diagram of the question is shown on the first uploaded image From the question we are told that

           The distance between the two poles is D =14 m

          The mass tied between the two cloth line is  m = 3Kg

         The distance it sags is d_s = 1m

The objective of this solution is to obtain the magnitude of the tension on the ends of the  clothesline

Now the sum of the forces on the y-axis is zero assuming  that the whole system is at equilibrium

       And this can be mathematically represented as

                             \sum F_y = 0

 To obtain \theta we apply SOHCAHTOH Rule

 So    Tan \theta = \frac{opp}{adj}

          \theta = tan^{-1} [\frac{opp}{adj} ]

            = tan^{-1} [\frac{1}{7}]

          =8.130^o

=>  \  \ \ \ \ \ \ \ 2T sin\theta -mg =0

=>  \  \ \ \ \ \ \ \ T =\frac{mg}{2 sin\theta}

=>  \  \ \ \ \ \ \ \ T = \frac{3 * 9.8 }{2 sin \theta }

=>  \  \ \ \ \ \ \ \  T =\frac{29.4}{2sin(8.130)}

=>  \  \ \ \ \ \ \ \  T = 103.96N

             

                 

5 0
2 years ago
Read 2 more answers
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