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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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<h2>Answer:</h2>

Phytochemicals are compounds that are produced by plants ("phyto" means "plant"). They are found in fruits, vegetables, grains, beans, and other plants. Some of these phytochemicals are believed to protect cells from damage that could lead to cancer.

8 0
3 years ago
A tank contains gas at 13.0°C pressurized to 10.0 atm. The temperature of the gas is increased to 95.0°C, and half the gas is re
fomenos

Answer:

The pressure of the remaining gas in the tank is 6.4 atm.

Explanation:

Given that,

Temperature T = 13+273=286 K

Pressure = 10.0 atm

We need to calculate the pressure of the remaining gas

Using equation of ideal gas

PV=nRT

For a gas

P_{1}V_{1}=nRT_{1}

Where, P = pressure

V = volume

T = temperature

Put the value in the equation

10\times V=nR\times286....(I)

When the temperature of the gas is increased

Then,

P_{2}V_{2}=\dfrac{n}{2}RT_{2}....(II)

Divided equation (I) by equation (II)

\dfrac{P_{1}V}{P_{2}V}=\dfrac{nRT_{1}}{\dfrac{n}{2}RT_{2}}

\dfrac{10\times V}{P_{2}V}=\dfrac{nR\times286}{\dfrac{n}{2}R368}

P_{2}=\dfrac{10\times368}{2\times286}

P_{2}= 6.433\ atm

P_{2}=6.4\ atm

Hence, The pressure of the remaining gas in the tank is 6.4 atm.

4 0
3 years ago
PLEASE HELP ASAP...TIMED TEST. PLEASE ANSWER WITH AT LEAST A PARAGRAPH!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
charle [14.2K]

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The best I could do.

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snow_tiger [21]

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7 0
3 years ago
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