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3 years ago

12

Introduces una masa de 80g en una probeta con 60cm cúbicos de agua y el nivel sube hasta 75cm cúbicos. qué densidad tiene la mas

a?

1 answer:

ExtremeBDS [4]3 years ago

4 0

Answer:

$d=5.33\ gr/cm^3$

Explanation:

<u>Densidad</u>

La densidad de una masa m que ocupa un volumen v se define como:

$\displaystyle d=\frac{m}{V}$

Tenemos que una masa de 80 gr hace que un volumen de agua se desplace desde 60 cm3 hasta 75 cm3, es decir, el volumen propio de la masa es m = 75 - 60 = 15 cm3.

Con esta información se calcula la densidad:

$\displaystyle d=\frac{80\ gr}{15\ cm^3}$

Operando:

$\boxed{d=5.33\ gr/cm^3}$

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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:

$\theta_2>\theta_1$

This means that the wave will bend away from the normal. So the other correct option is

C) waves bend away from the normal

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3 years ago

A star is located at a distance of about 100 million light years from Earth. An astronomer plans to measure the distance of the

nlexa [21]

<span>d. The parallaxes beyond a few thousand light years are
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I'm not sure that parallax can even be used out to a few
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The NEAREST star to Earth has the BIGGEST parallax.
The star is Alpha Centauri. It's only 4 light years away
from us, and its parallax is 0.000206 of a degree !
I have no idea how astronomers can measure angles
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2 years ago

When the ambulance passes the person (switching from moving towards him to moving away from him), the perceived frequency of the

dlinn [17]

To develop this problem we will apply the considerations made through the concept of Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. At first the source is moving towards the observer. Than the perceived frequency at first

$F_1 = F \frac{{343}}{(343-V)}$

Where F is the actual frequency and v is the velocity of the ambulance

Now the source is moving away from the observer.

$F_2 = F\frac{343}{(343+V)}$

We are also so told the perceived frequency decreases by 11.9%

$F_2 = F_1 - 9.27\% \text{ of } F_1$

$F_2 = F_1-0.0927F_1$

$F_2 = 0.9073F_1$

Equating,

$F\frac{343}{(343+V)}= 0.9073(F\frac{343}{(343-V)})$

$\frac{1}{(343+V)}= 0.9073\frac{1}{(343-V)}$

$0.9073(343+V) = 343-V$

$(0.9073)(343)+(0.9073)V = 343-V$

$V+0.9073V = 343-(0.9073)(343)$

Solving for V,

$V = 16.67 m/s$

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2 years ago

Name two elements that have similar properties. How can you tell using the periodic table?

tangare [24]

Helium and nitrogen because they are both from the noble gas group.

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3 years ago

A 2.0 kg otter starts from rest at the top of a muddy incline 85 cm long and slides down to the bottom in 0.50 s. What net force

Julli [10]

Answer:

The net force acting on the otter along the incline is 13.96 N.

Explanation:

It is given that,

Mass of the otter, m = 2 kg

Distance covered by otter, d = 85 cm = 0.85 m

It takes 0.5 seconds.

We need to find the net force acts on the otter along the incline. If a is the acceleration of the otter. It can be calculated using second equation of motion as :

$d=ut+\dfrac{1}{2}at^2$

Here, u = 0 (at rest)

$d=\dfrac{1}{2}at^2$

$a=\dfrac{2d}{t^2}$

$a=\dfrac{2\times 0.85}{0.5^2}$

$a=6.8\ m/s^2$

The net force acting on the otter along the incline is given by :

F = ma

$F=2\ kg\times 6.8\ m/s^2$

F = 13.6 N

So, the net force acting on the otter along the incline is 13.96 N. Hence, this is the required solution.

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3 years ago