1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
kolezko [41]
3 years ago
7

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?
A
frequency increases and wavelength decreases
B. frequency decreases and wavelength increases
C. frequency stays the same and wavelength increases
D.
frequency stays the same and wavelength decreases
Physics
2 answers:
miskamm [114]3 years ago
5 0

As ocean waves approach shore, their velocity decreases, the frequency stays the same and the wavelength decreases.

Explanation:

The velocity, wavelength, and frequency are related by the following mathematical equation:

 Velocity = wavelength × frequency

v=\lambda \times f  

When the waves reach the shore with reduced velocity, the frequency of the waves can be kept constant, only by reducing the wavelength. A direct relationship is established between velocity and wavelength.  

Therefore, reduction in the velocity leads to a proportional reduction in the wavelength of the wave, keeping the frequency the same.  

Thus, option (D) is the correct answer.

Leona [35]3 years ago
4 0

Answer:

D .Frequency stays the same and wavelength decreases .

Explanation:

Due to friction between shore and waves ,  the velocity of waves decreases while reaching the shores . As v = fλ, either frequency or wavelength should decrease. As frequency is the property of the source , it will stay constant and wavelength only decreases .

You might be interested in
Consider a concave spherical mirr or that has focal length f = +19.5 cm.
lidiya [134]

The distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.

<h3>What is concave mirror?</h3>

A concave mirror has a reflective surface that is curved inward and away from the light source.

Concave mirrors reflect light inward to one focal point and it usually form real and virtual images.

<h3>Object distance of the concave mirror</h3>

Apply mirrors formula as shown below;

1/f = 1/v + 1/u

where;

  • f is the focal length of the mirror
  • v is the object distance
  • u is the image distance

when image height = object height, magnification = 1

u/v = 1

v = u

Substitute the given parameters and solve for the distance of the object from the mirror's vertex

1/f = 1/v + 1/v

1/f = 2/v

v = 2f

v = 2(19.5 cm)

v = 39 cm

Thus, the distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.

Learn more about concave mirror here: brainly.com/question/27841226

#SPJ1

7 0
2 years ago
Question 4
Minchanka [31]

Answer:

C

Explanation:

3 0
3 years ago
A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is
olga nikolaevna [1]

Explanation:

It is given that,

Spring constant of the spring, k = 15 N/m

Amplitude of the oscillation, A = 7.5 cm = 0.075 m

Number of oscillations, N = 31

Time, t = 15 s

(a) Let m is the mass of the ball. The frequency of oscillation of the spring is given by :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

Total number of oscillation per unit time is called frequency of oscillation. Here, f=\dfrac{31}{15}=2.06\ Hz

m=\dfrac{k}{4\pi^2f^2}

m=\dfrac{15}{4\pi^2\times 2.06^2}

m = 0.0895 kg

or

m = 89 g

(b) The maximum speed of the ball that is given by :

v_{max}=A\times \omega

v_{max}=A\times 2\pi f

v_{max}=0.075\times 2\pi \times 2.06

v_{max}=0.970\ m/s

v_{max}=97\ cm/s

Hence, this is the required solution.

5 0
3 years ago
The plum pudding model of the atom states that
seropon [69]

Answer:

According to this model, the atom is a sphere of positive charge, and negatively charged electrons are embedded in it to balance the total positive charge.

Explanation:

Hope this helps you

8 0
3 years ago
Read 2 more answers
A force of 40N is applied to a 28 g mass, what is the acceleration? (round to the hundredths place)
Leno4ka [110]

Answer:

1428.6m/s²

Explanation:

Given parameters:

Force applied on the body  = 40N

Mass of the body  = 28g

                    1000g  = 1kg

                      28g will therefore be 0.028kg

Unknown:

Acceleration  = ?

Solution:

To solve this problem, we use the expression derived from Newton's second law of motion.

         Force  = mass x acceleration

Insert the parameters and solve;

            40  = 0.028 x acceleration

           Acceleration  = \frac{40}{0.028}   = 1428.6m/s²

7 0
3 years ago
Other questions:
  • Friction removes energy from objects in motion. Which statement best describes how this works?
    13·1 answer
  • Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates
    14·1 answer
  • List five situations in which temporary wiring would be used
    12·1 answer
  • Which experiment showed that the atom has a dense positive center?
    9·1 answer
  • Think about your displacement at three different times throughout your day and compare it with e distance you traveled. (6 point
    15·1 answer
  • The unit of work is called derived unit.Why​
    12·1 answer
  • 8. Contrast. The energy that can be released during a nuclear fission reaction with the energy that can be released during a nuc
    14·1 answer
  • A superhero flies 285 m from the top of a tall building at an angle of 35◦ below the horizontal.
    5·1 answer
  • A rifle is aimed horizontally at a target 50.0 m away. The bullet hits the target 2.90 cm below the aim point. . . Whats the bul
    14·1 answer
  • A) What magnitude point charge creates a 12596.37 N/C electric<br> held at a distance of 0.593 m?
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!