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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?

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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3 years ago

A skier is skiing down a slope on the mountain.

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The mechanical energy isn't conserved. Some energy is lost to friction.

Option A.

<h3><u>Explanation:</u></h3>

The mechanical energy is defined as the energy of a body which it achieves by virtue of its position and velocity. The mechanical energy are of two types - potential energy and kinetic energy. The potential energy is the energy of the body which it achieves by means of its relative position and is directly proportional to the height of the body from its relative plane. Whereas the kinetic energy of the body is achieved by virtue of its velocity and is directly proportional to the square of velocity of the body.

As the mountaineer is skiing down the slope of a mountain, the potential energy of the person is gradually changing into his kinetic energy. Had it been in an ideal situation, the potential energy lost would have been just equal to the kinetic energy gained by the person. But there's friction which opposes the speed of the body and reduces the velocity. Thus the kinetic energy will be lost to some extent and the energy won't be conserved.

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3 years ago

What change in entropy occurs when a 0.15 kg ice cube at -18 °C is transformed into steam at 120 °c 4.

Studentka2010 [4]

<u>Answer:</u> The change in entropy of the given process is 1324.8 J/K

<u>Explanation:</u>

The processes involved in the given problem are:

$1.)H_2O(s)(-18^oC,255K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\3.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(100^oC,373K)\\4.)H_2O(l)(100^oC,373K)\rightarrow H_2O(g)(100^oC,373K)\\5.)H_2O(g)(100^oC,373K)\rightarrow H_2O(g)(120^oC,393K)$

Pressure is taken as constant.

To calculate the entropy change for same phase at different temperature, we use the equation:

$\Delta S=m\times C_{p,m}\times \ln (\frac{T_2}{T_1})$ .......(1)

where,

$\Delta S$ = Entropy change

$C_{p,m}$ = specific heat capacity of medium

m = mass of ice = 0.15 kg = 150 g (Conversion factor: 1 kg = 1000 g)

$T_2$ = final temperature

$T_1$ = initial temperature

To calculate the entropy change for different phase at same temperature, we use the equation:

$\Delta S=m\times \frac{\Delta H_{f,v}}{T}$ .......(2)

where,

$\Delta S$ = Entropy change

m = mass of ice

$\Delta H_{f,v}$ = enthalpy of fusion of vaporization

T = temperature of the system

Calculating the entropy change for each process:

<u>For process 1:</u>

We are given:

$m=150g\\C_{p,s}=2.06J/gK\\T_1=255K\\T_2=273K$

Putting values in equation 1, we get:

$\Delta S_1=150g\times 2.06J/g.K\times \ln(\frac{273K}{255K})\\\\\Delta S_1=21.1J/K$

<u>For process 2:</u>

We are given:

$m=150g\\\Delta H_{fusion}=334.16J/g\\T=273K$

Putting values in equation 2, we get:

$\Delta S_2=\frac{150g\times 334.16J/g}{273K}\\\\\Delta S_2=183.6J/K$

<u>For process 3:</u>

We are given:

$m=150g\\C_{p,l}=4.184J/gK\\T_1=273K\\T_2=373K$

Putting values in equation 1, we get:

$\Delta S_3=150g\times 4.184J/g.K\times \ln(\frac{373K}{273K})\\\\\Delta S_3=195.9J/K$

<u>For process 4:</u>

We are given:

$m=150g\\\Delta H_{vaporization}=2259J/g\\T=373K$

Putting values in equation 2, we get:

$\Delta S_2=\frac{150g\times 2259J/g}{373K}\\\\\Delta S_2=908.4J/K$

<u>For process 5:</u>

We are given:

$m=150g\\C_{p,g}=2.02J/gK\\T_1=373K\\T_2=393K$

Putting values in equation 1, we get:

$\Delta S_5=150g\times 2.02J/g.K\times \ln(\frac{393K}{373K})\\\\\Delta S_5=15.8J/K$

Total entropy change for the process = $\Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4+\Delta S_5$

Total entropy change for the process = $[21.1+183.6+195.9+908.4+15.8]J/K=1324.8J/K$

Hence, the change in entropy of the given process is 1324.8 J/K

4 0

3 years ago

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0

3 years ago

Compare and contrast electric potential energy and electric potential difference? Explain.

forsale [732]

Answer:

<u><em>Electric Potential Energy:</em></u>

The energy that is needed to move a charge against an electric firld is called Electric Potential Energy

<u><em>Electric Potential Difference:</em></u>

The amount of work done in carrying a unit charge from one point to an other in an electric field is called Electric Potential Difference.

<u><em>Relation:</em></u>

Relation between Electric potential and electrical potential energy is given by

$\delta V=\frac{PE}{q}$

Here PE represents Electric potential energy

and $\delta V$ is Electric potential difference

it means electric potential difference is the difference in electric potential energy divided by the charge.

6 0

3 years ago