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klemol [59]
3 years ago
13

As ocean waves approach shore, their velocity decreases. How does a decrease in velocity affect the frequency and wavelength of

the waves entering the shallow water?
Physics
2 answers:
Elina [12.6K]3 years ago
6 0
Because if the velocity decreases then the frequency and wavelength would be affected since frequency deals with how fast the wave is going and since shallow water is lower theres not many waves to continue the movement of the wave 
timama [110]3 years ago
3 0

Explanation:

As ocean waves approach shore, their velocity decreases. The relation between the velocity, wavelength and the frequency of a wave is given by :

v=f\times \lambda

f = frequency

\lambda = wavelength

f=\dfrac{v}{\lambda}..............(1)

So, as the velocity of the wave increases, the wavelength and frequency also increases as per equation (1). When the velocity decreases, the frequency and the wavelength also decreases.

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A block mass m (0.25 kg) is pressed against (but is not attached to) an ideal spring of force constant k (100 N/m) and negligibl
VMariaS [17]

Answer:

d. All the above choices are correct.

Explanation:

When a spring of spring constant k is compressed by distance x , the potential energy stored in it is equal to

E = 1/2 k x²

If spring constant is 2 k , potential energy stored

E = 1/2 2k x²

= k x²

which is twice the earlier potential energy.

In the first case , the energy of spring is imparted to box . The energy given to box is spent by frictional force due to which box comes to rest.

So energy of box acquired from spring = work done by frictional force.

So energy of box acquired from spring =  F X d  , F is frictional force , d is displacement .

In the second case ,

energy acquired by box becomes  two times

Work done by frictional force will also become two times to put box at rest

So displacement will be two times ( because frictional force is constant )

so option a is correct .

option b is also correct .

Because kinetic energy  of box will be twice as explained above .

So option d will be correct.

5 0
3 years ago
Which of the following describes the relationship between work and power?
Vinvika [58]

Answer: The answer is B

Explanation:

Work can be defined as the energy that is required to apply a force to an object in order to move it from one point to another. In physics, work = force x distance travelled. On the other hand, Power is the work done per time. In other words, it the rate at which work is done and is determined by using the formula, Power = Work/time. In these relationships, it can be seen that power is directly proportional to the amount of work done, hence as power increases, more work is done.

6 0
4 years ago
4
tensa zangetsu [6.8K]

Answer:

C. amount of charge on the source charge.

Explanation:

Electric field lines can be defined as a graphical representation of the vector field or electric field.

Basically, it was first introduced by Michael Faraday and it is typically a curve drawn to the tangent of a point is in the direction of the net field acting on each point.

The number, or density, of field lines on a source charge indicate the amount of charge on the source charge. Therefore, the density of field lines on a source charge is directly proportional to quantity of charge on the source.

8 0
3 years ago
The drawing shows the electrical potential as a function
Ilia_Sergeevich [38]

Answer:

a)\quad E_{ab} = 0 \quad Volts/m\\\\b)\quad E_{bc} = 10 \quad Volts/m\\\\c)\quad E_{cd} = 5 \quad Volts/m

Explanation:

As figure is not given so considering the most relevant diagram for question attached below

We can define Electric field as rate of change of electric potential with respect to space (say x-axis here). It is related in formula as

E=-\frac{\Delta V}{S}---(1)

a) For Region A to B:

As can be seen from figure, point charge moves from 0 to 0.2 along x-axis, the value of electric potential remains constant i.e 5 volts

E_{ab}=-\frac{V_{b}-V_{a}}{b-a}\\\\E_{ab}=-\frac{5-5}{0.2-0}\\\\E_{ab}=0\quad Volts/m

b) For Region B to C:

As point charge moves from B to C (0.2 to 0.4) along x-axis, the value of electric potential decreases from 5 volts to 3 volts. Electric field induced is:

E_{bc}=-\frac{V_{c}-V_{b}}{c-b}\\\\E_{bc}=-\frac{3-5}{0.4-0.2}\\\\E_{bc}=10\quad Volts/m

c) For Region C to D:

As point charge moves from C to D (0.4 to 0.8) along x-axis, the value of electric potential decreases from 3 volts to 1 volt. Electric field induced is:

E_{cd}=-\frac{V_{d}-V_{c}}{d-c}\\\\E_{cd}=-\frac{1-3}{0.8-0.4}\\\\E_{cd}=5\quad Volts/m

For Information:

Unit of Electric field is V/m or N/C

6 0
4 years ago
Disadvantages of friction
MrRa [10]
Heat, kinetic energy loss.
7 0
3 years ago
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