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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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What happens to the atomic number of an atom when the number of neutrons in the nucleus of that atom increases? a It decreases b
kupik [55]

Answer:

It remains the same

Explanation:

It remains the same. This is because the number of protons doesn't change and the number of protons determines the atomic number.

8 0
2 years ago
Relay of thermostate is not working why
Colt1911 [192]

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7 0
3 years ago
A mass oscillates up and down on a vertical spring with an amplitude of 3 cm and a period of 2 s. What total distance does the m
harina [27]

Explanation:

It is given that,

A mass oscillates up and down on a vertical spring with an amplitude of 3 cm and a period of 2 s. It is a case of simple harmonic motion. If the amplitude of a wave is T seconds, then the distance cover by that object is 4 times the amplitude.

In 2 seconds, distance covered by the mass is 12 cm.

In 1 seconds, distance covered by the mass is 6 cm

So, in 16 seconds, distance covered by the mass is 96 cm

So, the distance covered by the mass in 16 seconds is 96 cm. Hence, this is the required solution.

6 0
3 years ago
A 4.00-m-long, 470 kg steel beam extends horizontally from the point where it has been bolted to the framework of a new building
adell [148]

Answer:

12164.4 Nm

Explanation:

CHECK THE ATTACHMENT

Given values are;

m1= 470 kg

x= 4m

m2= 75kg

Cm = center of mass

g= acceleration due to gravity= 9.82 m/s^2

The distance of centre of mass is x/2

Center of mass(1) = x/2

But x= 4 m

Then substitute, we have,

Center of mass(1) = 4/2 = 2m

We can find the total torque, through the summation of moments that comes from both the man and the beam.

τ = τ(1) + τ(2)

But

τ(1)= ( Center of m1 × m1 × g)= (2× 470× 9.81)

= 9221.4Nm

τ(2)= X * m2 * g = ( 4× 75 × 9.81)= 2943Nm

τ = τ(1) + τ(2)

= 9221.4Nm + 2943Nm

= 12164.4 Nm

Hence, the magnitude of the torque about the point where the beam is bolted into place is 12164.4 Nm

6 0
3 years ago
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mylen [45]

Answer:

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8 0
3 years ago
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