An Olympic swimmer swims 50.0 meters in 23.1 seconds . What is his average speed

1 answer:

3 0

If you take 50 meters and divide by 23.1 seconds, you will get 2.16 meters per second.

So his average speed is 2.16 m/s.

You might be interested in

What must happen to an atom of magnesium in order to become a magnesium ion

Zielflug [23.3K]

I would assume that the magnesium would have to come into contact with an ion in order to become a magnesium ion.

6 0

3 years ago

Different between current and electrons?

BlackZzzverrR [31]

Answer:

Simply,

electrons are "PARTICLES" orbiting the atoms, where, current is the FLOW of some (free-to-move-around) electrons in a wire...

3 0

3 years ago

It is easier to see clothes with pointed needle than a blunt one

topjm [15]

Answer:

Explanation:

It is easier to see clothes with pointed needle than a blunt one because pressure exerted is more in a pointed needle as it occupies less space compared to blunt needle, A blunt has more surface area so the pressure exerted will less as compared to a pointed needle.

IF YOU FOUND MY ANSWER USEFUL THEN PLEASE MARK ME BRAINLIEST.

7 0

3 years ago

Match the indices of refraction with the corresponding effects on the waves.

Yanka [14]

Answer:

B) waves speed up

C) waves bend away from the normal

Explanation:

The index of refraction of a material is the ratio between the speed of light in a vacuum and the speed of light in that medium:

$n=\frac{c}{v}$

where

c is the speed of light in a vacuum

v is the speed of light in the medium

We can re-arrange this equation as:

$v=\frac{c}{n}$

So from this we already see that if the index of refraction is lower, the speed of light in the medium will be higher, so one correct option is

B) waves speed up

Moreover, when light enters a medium bends according to Snell's Law:

$n_1 sin \theta_1 = n_2 sin \theta_2$

where

$n_1 ,n_2$ are the index of refraction of the 1st and 2nd medium

$\theta_1,\theta_2$ are the angles made by the incident ray and refracted ray with the normal to the interface

We can rewrite the equation as

$sin \theta_2 = \frac{n_1}{n_2}sin \theta_1$

So we see that if the index of refraction of the second medium is lower ( $n_2$ ), then the ratio $\frac{n_1}{n_2}$ is larger than 1, so the angle of refraction is larger than the angle of incidence: