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Ivahew [28]
3 years ago
15

A stationary boat in the ocean is experiencing waves from a storm. The waves move at 59 km/h and have a wavelength of 145 m . Th

e 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:
krek1111 [17]3 years ago
5 0

Answer:

The time elapses until the boat is first at the trough of a wave is 4.46 seconds.

Explanation:

Speed of the wave, v = 59 km/h = 16.38 m/s

Wavelength of the wave, \lambda=145\ m

If f is the frequency of the wave. The frequency of a wave is given by :

v=f\lambda\\\\f=\dfrac{v}{\lambda}\\\\f=\dfrac{16.38\ m/s}{145\ m}\\\\f=0.112\ Hz

The time period of the wave is given by :

T=\dfrac{1}{f}\\\\T=\dfrac{1}{0.112\ Hz}\\\\T=8.92\ s

We need to find the time elapses until the boat is first at the trough of a wave. So, the time will be half of the time period of the wave.

T=\dfrac{8.92}{2}\\\\T=4.46\ s

Hence, this is the required solution.

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Using a density of air to be 1.21kg/m3, the diameter of the bottom part of the filter as 0.15m (assume circular cross-section),
salantis [7]

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     The diameter of bottom part is  d = 0.15 \ m

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let assume that the velocity is  20 m/s

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substituting values

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3 0
3 years ago
Consider a transformer. used to recharge rechargeable flashlight batteries, that has 500 turns in its primary coil, 3 turns in i
Rashid [163]

Answer:

<em>a) 0.72 V</em>

<em>b) 19.2 mA</em>

<em>c) 2.304 Watts</em>

Explanation:

A transformer is used to step-up or step-down voltage and current. It uses the principle of electromagnetic induction. When the primary coil is greater than the secondary coil, the it is a step-down transformer, and when the primary coil is less than the secondary coil, the it is a step-up transformer.

number of primary turns = N_{p} = 500 turns

input voltage = V_{p} = 120 V

number of secondary turns = N_{s} = 3 turns

output voltage = V_{s} = ?

using the equation for a transformer

\frac{V_{s} }{V_{p} }  = \frac{N_{s} }{N_{p} }

substituting values, we have

\frac{V_{s} }{120 }  = \frac{3 }{500} }

500V_{p}  = 120*3\\500V_{p} = 360

V_{p} = 360/500 =<em> 0.72 V</em>

<em></em>

b) by law of energy conservation,

I_{P}V_{p} = I_{s}V_{s}

where

I_{p} = input current = ?

I_{s} = output voltage = 3.2 A

V_{s} = output voltage = 0.72 V

V_{p} = input voltage = 120 V

substituting values, we have

120I_{p} = 3.2 x 0.72

120I_{p} = 2.304

I_{p}  = 2.304/120 = 0.0192 A

= <em>19.2 mA</em>

<em></em>

c) power input = I_{p} V_{p}

==> 0.0192 x 120 = <em>2.304 Watts</em>

7 0
3 years ago
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