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d1i1m1o1n [39]
3 years ago
13

A fisherman fishing from a pier observes that the float on his line bobs up and down, taking 2.4 s to move from its highest poin

t to its lowest point. He also estimates that the distance between adjacent wave crests is 48 m. What is the speed of the waves going past the pier?Immersive Reader
Physics
1 answer:
goblinko [34]3 years ago
4 0

Answer:

  v = 10 m / s

Explanation:

For this exercise we will use the relationship between the speed of a wave and its frequency and wavelength

         v = λ f

the wavelength is the distance at which the wave repeats, in this case the distance between the two ridges λ = 48 m.

the frequency is the number of oscillations per unit of time, it is also the inverse of the period which is the time in a complete oscillation, in this case they give us the time of half a period, ½ T = 2.4 s

   T = 4.8 s

the frequency is

        f = 1 / T

        f = 1 / 4.8

        f = 0.2083 Hz

let's calculate

        v = 0.2083 48

        v = 10 m / s

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Answer:

Elements in the same period have the same number of electron shells; moving across a period (so progressing from group to group), elements gain electrons and protons and become less metallic. This arrangement reflects the periodic recurrence of similar properties as the atomic number increases.

Explanation:

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3 years ago
A carmaker has designed a car that can reach a maximum acceleration of 12 meters/second2. The car’s mass is 1,515 kilograms. Ass
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2 years ago
A 15x10^-6c charge is placed at the origin and a 9x10^-6C charge is placed on the x-axis at x=1.00m. where, on the x-axis is the
Harlamova29_29 [7]

The Electric field is zero at a distance 2.492 cm from the origin.

Let A be point where the charge 15\times10^-6 C is placed which is the origin.

Let B be the point where the charge 9\times 10^-6 C is placed. Given that B is at a distance 1 cm from the origin.

Both the charges are positive. But charge at origin is greater than that of B. So we can conclude that the point on the x-axis where the electric field = 0 is after B on x - axis.

i.e., at distance 'x' from B.

Using Coulomb's law, \frac{kQ_A^2}{d_A^2} = \frac{kQ_B^2}{d_B^2},

Q_A = 15\times 10^-6 C

Q_B=9\times10^-6C

d_A = 1+x cm

d_B=x cm

k is the Coulomb's law constant.

On substituting the values into the above equation, we get,

\frac{(15\times10^-6)^2}{(1+x)^2} =\frac{(9\times10^-6)^2}{x^2}

Taking square roots on both sides and simplifying and solving for x, we get,

1.67x = 1+x

Therefore, x = 1.492 cm

Hence the electric field is zero at a distance 1+1.492 = 2.492 cm from the origin.

Learn more about Electric fields and Coulomb's Law at brainly.com/question/506926

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1 year ago
Light travels at the same speed at all times. True False
topjm [15]

Answer:

true hope this help :}

Explanation:

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