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MatroZZZ [7]
3 years ago
13

A stationary boat in the ocean is experiencing waves from a storm. the waves move at 56 km/h and have a wavelength of 160 m, bot

h typical values. the boat is at the crest of a wave. how much time elapses until the boat is first at the trough of a wave?
Physics
1 answer:
Lera25 [3.4K]3 years ago
5 0
The wavelength \lambda of the wave is 160 m, and this is the distance between two consecutive crests. The boat is located at a crest of the wave, this means that the first trough is located 80 meters from the boat (because the distance between a crest and a trough is half the wavelength).

The speed of the wave is 
v=56 km/h = 15.6 m/s
so the time the boat takes to reach the first trough is
t= \frac{S}{v} = \frac{80 m}{15.6 m/s}=5.1 s
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g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart
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The speed of q₂ is 4\sqrt{10}\ m/s

Explanation:

Given that,

Distance = 0.4 m apart

Suppose, A small metal sphere, carrying a net charge q₁ = −2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = −8μC and mass 1.50g, is projected toward q₁. When the two spheres are 0.800m apart, q₂ is moving toward q₁ with speed 20m/s.

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Using conservation of energy

E_{i}=E_{f}

\dfrac{1}{2}mv_{i}^2+\dfrac{kq_{1}q_{2}}{r_{i}}=\dfrac{kq_{1}q_{2}}{r_{f}}+\dfrac{1}{2}mv_{f}^2

\dfrac{1}{2}m(v_{i}^2-v_{f}^2)=kq_{1}q_{2}(\dfrac{1}{r_{f}}-\dfrac{1}{r_{i}})

Put the value into the formula

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0.00075(400-v_{f}^2)=0.18&#10;

400-v_{f}^2=\dfrac{0.18}{0.00075}

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v_{f}^2=160

v_{f}=4\sqrt{10}\ m/s

Hence, The speed of q₂ is 4\sqrt{10}\ m/s

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