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2 years ago

Valleys have dense population but plateaus don't have dense population.why?

Physics

1 answer:

Juliette [100K]2 years ago

6 0

Answer:

Another very important reason why people prefer to live in valleys instead of plateaus is because of available sources of water. This conducive nature of valleys in turn accommodate factors like economic growth,social and cultural, which in turn attract more people, hence making these areas densely populated.

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Four football players are running down the field at the same speed. Player 1 weighs 180 lbs and is running toward the south goal

Serggg [28]

Player 2 because moment is mass times acceleration and since they are all going the same speed. Speed doesn't matter so the only thing that is left is mass/ weight and he has the most

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3 years ago

Apply Pascal's law to a ketchup packet. Explain what will happen if you tear a small

vivado [14]

Answer:

its will explode because it's like a soda with baking soda added together and you shake it the small hole can make a huge explosion

6 0

2 years ago

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The central star of a planetary nebula emits ultraviolet light with wavelength 104nm. This light passes through a diffraction gr

Gala2k [10]

Answer: 31.33 degrees

Explanation:

The diffraction angles $\theta_{n}$ when we have a slit divided into $n$ parts are obtained by the following equation:

$dsin\theta_{n}=n\lambda$ (1)

Where:

$d$ is the width of the slit

$\lambda$ is the wavelength of the light

$n$ is an integer different from zero.

Now, the first-order diffraction angle is given when $n=1$ , hence equation (1) becomes:

$dsin\theta_{1}=\lambda$ (2)

Now we have to find the value of $\theta_{1}$ :

$sin\theta_{1}=\frac{\lambda}{d}$

$\theta_{1}=arcsin(\frac{\lambda}{d})$ (3)

We know:

$\lambda=104nm=104(10)^{-9}m$

In addition we are told the diffraction grating has 5000 slits per mm, this means:

$d=\frac{1mm}{5000}=\frac{1(10)^{-3}m}{5000}$

Substituting the known values in (3):

$\theta_{1}=arcsin(\frac{104(10)^{-9}m}{\frac{1(10)^{-3}m}{5000}})$

$\theta_{1}=arcsin(0.52)$

<u>Finally:</u>

$\theta_{1}=31.33\º$ >>>This is the first-order diffraction angle

4 0

2 years ago

Two point charges of equal magnitude q are held a distance d apart. Consider only points on the line passing through both charge

NemiM [27]

let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.

the mathematical formula for potential is $V=\frac{1}{4\pi\epsilon} \frac{q}{d}$

for positive charges the potential is positive and is negative for negative charges.

the formula for electric field is given as- $E=\frac{1}{4\pi\epsilon} \frac{q}{r^2}$

for positive charges,the line filed is away from it and for negative charges the filed is towards it.

we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.

here the net electric field due to the dipole can not be zero between the two charges,but we can find the points situated on the axial line but outside of charges where the electric field is zero.

now let the two charges of same nature.let these are positively charged.

here we can not find a point between two charges and on the line joining two charges where the potential is zero.

but at the mid point of the line joining two charges the filed is zero.

5 0

3 years ago

An object is moving with uniform speed in a circle of radius r. Calculate the distance and displacement

dimaraw [331]

Answer and Explanation:

distance will be 2×3.14 (pie)×r

displacement will be 2r (diameter)

the motion is uniform circular motion as the object is moving in a circular path with uniform motion

8 0

2 years ago

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