# What is the speed of an ocean wave if it’s wavelength is 5.0 m and it’s frequency is 3/s?

15 m/s

Explanation:

We know that where f = frequency & d = wavelength .

So here.

Wavelength = 5 m

Frequency = 3 s⁻¹

Hence Speed = 5 * 3 = 15 m/s

## Related Questions

In a two-slit experiment, the slit separation is 3.00 × 10-5 m. The interference pattern is recorded on a flat screen-like detector that is 2.00 m away from the slits. If the seventh bright fringe on the detector is 10.0 cm away from the central fringe, what is the wavelength of the light passing through the slits? (The central bright fringe is zeroth one).

The wavelength of the light passing through the slit is 214 nm.

### What is the wavelength?

The wavelength is the distance between identical points in the adjacent cycles of a waveform.

Given that the separation between two slits d is 3.00 × 10^-5 m and the distance from the slit to screen r is 2 m. The distance from the central spot to fringe s is 10.0 m and the bright bands of the spectrum m are 7 for the seventh bright fringe.

The wavelength of the light passing through the slit is calculated as given below.

Hence we can conclude that the wavelength of the light passing through the slit is 214 nm.

To solve this problem, the concepts related to destructive and constructive Interference of light spot and dark spot are necessary.

By definition in the principle of superposition, light interference is defined as

Where,

d = Separation of the two slits

R = Distance from slit to screen

m= Any integer, which represents the repetition of the spectrum. The order of m equal to 1,2,3,4,5 represent bright bands and the order of m equal to 1.5,2.5,3.5 represent the dark bands.

Y = Distance from central spot to fringe.

Re-arrange the equation to find \lambda we have that

Our values are gives according the problem as,

m = 7 (The seventh bright fringe)

R = 2m

Therefore the wavelength of the light passing through the slits is 214nm

5. (Serway 9th ed., 7-3) In 1990, Walter Arfeuille of Belgium lifted a 281.5-kg object through a distance of 17.1 cm using only his teeth. (a) How much work was done on the object by Arfeuille in this lift, assuming the object was lifted at constant speed? (b) What total force was exerted on Arfeuille’s teeth during the lift? (Ans. (a) 472 J; (b) 2.76 kN)

Para resolver este problema es necesario aplicar los conceptos de Fuerza, dados en la segunda Ley de Newton y el concepto de Trabajo, como expresión de la fuerza necesaria para realizar una actividad en una distancia determinada.

El trabajo se define como

W = F*d

Where,

F = Force

d = Distance

At the same time we have that the Force by second's Newton law is equal to

F = mg

Where,

m = mass

g = Gravitational acceleration

PART A) Using our values and replacing we have that

PART B) Using Newton's Second law we have that,

What happens at night- describing air circulation

The environment is warmed by the light throughout the day, such that the temperature increases. The weather is decreasing and the temperature decreases in the night as the sun falls. There was a misunderstanding. Thanks to the density, the atmosphere becomes densest on the earth. The air becomes colder and colder when you move up.

Explanation:

Hope this helps.

A person is pushing a lawnmower of mass m D 38 kg and with h D 0:75 m, d D 0:25 m, A D 0:28 m, and B D 0:36 m. Assuming that the force exerted on the lawnmower by the person is completely horizontal and that the mass center of the lawnmower is at G, and neglecting the rotational inertia of the wheels, determine the minimum value of this force that causes the rear wheels (labeled A) to lift off the ground. In addition, determine the corresponding acceleration of the mower.

The acceleration of the mower will be "4.7 m/s²".

Explanation:

Balance of vertical force will be:

⇒

For wheel to take off at A,

⇒

Hence,

Balancing moments about G will be:

⇒

As we know,

Force, F =

On putting the values, we get

⇒           =

⇒           =

Now,

Acceleration, a =

⇒                       =

⇒                       =

A 50 kg woman and an 80 kg man stand 12.0 m apart on frictionless ice.(a) How far from the woman is their CM?
m

(b) If each holds one end of a rope, and the man pulls on the rope so that he moves 1.3 m, how far from the woman will he be now?
m

(c) How far will the man have moved when he collides with the woman?
m

Given that

m₁ = 50 kg

m₂=80 kg

d= 12 m

a)

We know that center of mass given as

X = (x₁m₁+x₂m₂)/(m₁+m₂)

Lets take distance of CM from woman is X

So now by putting the value

X = (0 x 50+12 x 80)/(50+80)

x=7.38 m

b)

There is no any external force so the CM  will not move.

So we can say that

x₁m₁+x₂m₂ = 0

50(x) - 80(1.3)=0

x=2.08

So the distance move by woman d=12-2.08-1.3=8.62 m

d=8.62 m

c) lets take distance move by man is x

50 (x) - 80 (12-x) =0

x=7.38

So the distance move by woman d=12-7.38

d=4.62 m

What best describes a societal law