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Andru [333]
3 years ago
6

A sailor strikes the side of his ship just below the surface of the sea. he hears the echo of the wave reflected from the ocean

floor directly below 5.0 s later. how deep is the ocean at this point
Physics
1 answer:
Luden [163]3 years ago
3 0

Answer: 3750 m


Explanation:


1) You can find in the literature or internet that the speed of sound in the ocean is about 1500 m/s.


2) Then, you can find the distance travel by the sound wave in 5.0s with the equation distance = speed × time


⇒ distance = 1500m/s × 5.0 s = 7500 m


3) That sound went down until the ocean floor below and cameback all the way until the sailor heard it.


Then, the deep of the ocean fllor is half that distance:


deep = 7500m / 2 = 3750m

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The force of repulsion that two like charges exert on each other 5N. what will be if the distance between the charge is decrease
GalinKa [24]

The new force becomes One Ninth (1/9) of the original force.

The force between two point charges (let's say \mathrm{Q} 1$ and $\mathrm{Q} 2$ ) is given by the following formula:

Force $=k \times Q 1 \times Q 2$divided by ( $r$ squared)

Here r is the distance.

If we multiply r by three then after squaring it will become $9 \times r$ squared.

Let's rewrite the formula and call it new Force:

New Force =\mathrm{K} \times \mathrm{Q} 1 \times \mathrm{Q} 2 divided by $(9 \times \mathrm{r}$ squared )

Now just separate the 9 :

New Force $=1 / 9(\mathrm{~K} \times \mathrm{Q} 1 \times \mathrm{Q} 2$divided by $(\mathrm{r}$ Squared ))

New Force $=1 / 9$ (Force)

So turns out that the new force becomes One Ninth (1/9) of the original force.

A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to define force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

Learn more about force brainly.com/question/13191643

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6 0
1 year ago
An airplane flies horizontally with a constant speed of 155.0 m/s at an unknown altitude. A package is released out of the airpl
vladimir1956 [14]

Answer:

 y₀ = 1020.3 m

Explanation:

This is a projectile launching exercise, in this case as the package is released its initial vertical velocity is zero.

            y = y₀ + v_{oy} t - ½ g t²

when it reaches the ground its height is zero

           0 = y₀ + 0 - ½ g t²

           y₀ = ½ g t²

           

let's calculate

         y₀ = ½ 9.8 14.43²

         y₀ = 1020.3 m

8 0
3 years ago
A photoelectric effect experiment finds a stopping potential of 1.93 V when light of wavelength 200 nm is used to illuminate the
GenaCL600 [577]

a) Zinc (work function: 4.3 eV)

The equation for the photoelectric effect is:

E=\phi + K (1)

where

E=\frac{hc}{\lambda} is the energy of the incident photon, with

h = Planck constant

c = speed of light

\lambda = wavelength

\phi = work function of the metal

K = maximum kinetic energy of the photoelectrons emitted

The stopping potential (V) is the potential needed to stop the photoelectrons with maximum kinetic energy: so, the corresponding electric potential energy must be equal to the maximum kinetic energy,

eV=K

So we can rewrite (1) as

E=\phi + eV

where we have:

\lambda=200 nm = 2\cdot 10^{-7} m

V = 1.93 V

e is the electron charge

First of all, let's find the energy of the incident photon:

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{2\cdot 10^{-7}m}=9.95\cdot 10^{-19} J

Converting into electronvolts,

E=\frac{9.95\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=6.22 eV

And now we can solve eq.(1) to find the work function of the metal:

\phi = E-eV=6.22 eV-1.93 eV=4.29 eV

so, the metal is most likely zinc, which has a work function of 4.3 eV.

b) The stopping potential is still 1.93 V

Explanation:

The intensity of the incident light is proportional to the number of photons hitting the surface of the metal. However, the energy of the photons depends only on their frequency, so it does not depend on the intensity of the light. This means that the term E in eq.(1) does not change.

Moreover, the work function of the metal is also constant, since it depends only on the properties of the material: so \phi is also constant in the equation. As a result, the term (eV) must also be constant, and therefore V, the stopping potential, is constant as well.

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3 years ago
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bearhunter [10]

Answer:

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Explanation:

5 0
3 years ago
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Answer:

s= 64m I'm not 100% sure

Explanation: like 97.9%

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