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bogdanovich [222]

3 years ago

11

What is the Coriolis Effect?

1 answer:

Crazy boy [7]3 years ago

3 0

Answer:

An effect whereby a mass moving in a rotating system experienced a force acting perpendicular to the direction of motion and to the axis of rotation.

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A _____ tide occurs when the ocean's surface reaches its maximum height in a particular place on any particular day.

Fed [463]

In one of the startling coincidences sprinkled throughout
every field of math and science, the moment of maximum
height is popularly referred to as the "high" tide.

4 0

3 years ago

Read 2 more answers

Which explanation describes the forces involved for a person to walk down the sidewalk?

Svet_ta [14]

The person walking down the sidewalk follows the newton's third law of motion.

Explanation:

A person is able to walk down the sidewalk by using the reaction forces from the ground.
In simple term, feet pushes the ground and the reaction forces makes the feet able to walk.
Another important force included in the walking mechanism is friction. With out friction one cannot walk down the sidewalks.
Hence the forces involved in the walking of a person down the sidewalk are:

Friction force
Action and reaction force between ground and person's feet.

4 0

3 years ago

Two point charges are brought closer together, increasing the force between them by a factor of 20. By what factor was their sep

olga55 [171]

The separation between them is $\frac{r}{\sqrt{20} }$

Concept :

If the force increases, distance between charges must decrease. Force is indirectly proportional to the distance squared.

Given,

Two point charges are brought closer together, increasing the force between them by a factor of 20.

Original force is

F = $\frac{kq_{1} q_{2} }{r^{2} }$ -------- ( 1 )

Here, $q_{1} , q_{2}$ are charges and r is the distance between them

New force F' = $\frac{kq_{1q_{2} } }{r'^{2} }$ ----------- (2 )

Divide ( 1 ) and ( 2 )

$\frac{F'}{F}$ = $\frac{\frac{kq_{1}q_{2} }{r'^{2} } }{\frac{kq_{1}_{2} }{r^{2} } }$

20 = $\frac{r^{2} }{r'^{2} }$

r' = $\frac{r}{\sqrt{20} }$

Given that force between them are increasing and therefore distance between them decrease by $\frac{r}{\sqrt{20} }$

Learn more about two point charges here : brainly.com/question/24206363

#SPJ4

8 0

2 years ago

Someone please answer please don’t take my points.

Anni [7]

Answer:

bro today my teacher will teach this answer only when I will be getting the answer I will copy and paste

Explanation:

wait for sometimes

5 0

3 years ago

The closest stars are 4 light years away from us. How far away must you be from a 854 kHz radio station with power 50.0 kW for t

Lubov Fominskaja [6]

Answer:

The distance from the radio station is 0.28 light years away.

Solution:

As per the question:

Distance, d = 4 ly

Frequency of the radio station, f = 854 kHz = $854\times 10^{3}\ Hz$

Power, P = 50 kW = $50\times 10^{3}\ W$

$I_{p} = 1\ photon/s/m^{2}$

Now,

From the relation:

P = nhf

where

n = no. of photons/second

h = Planck's constant

f = frequency

Now,

$n = \frac{P}{hf} = \frac{50\times 10^{3}}{6.626\times 10^{- 34}\times 854\times 10^{3}} = 8.836\times 10^{31}\ photons/s$

Area of the sphere, A = $4\pi r^{2}$

Now,

Suppose the distance from the radio station be 'r' from where the intensity of the photon is $1\ photon/s/m^{2}$

$I_{p} = \frac{n}{A} = \frac{n}{4\pi r^{2}}$

$1 = \frac{8.836\times 10^{31}}{4\pi r^{2}}$

$r = \sqrt{\frac{8.836\times 10^{31}}{4\pi}} = 2.65\times 10^{15}\ m$

Now,

We know that:

1 ly = $9.4607\times 10^{15}\ m$

Thus

$r = \frac{2.65\times 10^{15}}{9.4607\times 10^{15}} = 0.28\ ly$

5 0

3 years ago