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4 years ago

What happens to the density of a given substance if you increase the amount of the substance that you have?

2 answers:

7 0

Have you ever looked up the density of a substance ? You ought to try it. Go ahead. Pick a substance, then go online or open up an actual book and find its density. You will never see any particular volume mentioned along with the density . . . because it doesn't matter. The whole idea of density is that it describes the substance, no matter how much or how little you have of it. The density of a tiny drop of water under a microscope is the same as the density of a supertanker-ful of water.

puteri [66]4 years ago

5 0

The density stays the same because it's all one substance

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A parallel-plate capacitor is constructed of two square plates, size L×L, separated by distance d. The plates are given charge ±

gladu [14]

Answer:

E(final)/E(initial)=2

Explanation:

Applying the law of gauss to two parallel plates with charge density equal σ:

$E=\sigma/\epsilon_{o}=Q/(L^{2}*\epsilon_{o})\\$

So, if the charge is doubled the Electric field is doubled too

E(final)/E(initial)=2

3 0

3 years ago

A mercury thermometer reads 10oC when dipped into melting ice and 90oC

Crank

Answer:

Thermometer will read 26 degrees Celsius.

Please vote for Brainliest and I hope this helps!

3 0

2 years ago

What is the electric potential v due to the nucleus of hydrogen at a distance of 5.292×10−11 m ?

viktelen [127]

Answer: 27.21 V

Explanation:

The electric potential $V_{E}$ due to a point charge is expressed as:

$V_{E}=k\frac{q}{r}$

Where:

$k=9(10)^{9}\frac{Nm^{2}}{C^{2}}$ is the electric constant

$q=1.6(10)^{-19}C$ is the electric charge of the hydrogen nucleus, which is positive

$r=5.292(10)^{-11}m$ is the distance

Rewritting the equation with the known values:

$V_{E}=9(10)^{9}\frac{Nm^{2}}{C^{2}}\frac{1.6(10)^{-19}C}{5.292(10)^{-11}m}$

Finally:

$V_{E}=27.21 V$

5 0

3 years ago

If you know this please help

Ainat [17]

Answer:

Explanation:

It's b I sis thebksqjwnsx0qkqnsnd991isnd

3 0

3 years ago

Read 2 more answers

Se lanza verticalmente una esfera con una rapidez de 30m/se. Determinar la rapidez de la esfera a una altura de 40m (g=10m/s2)

sergeinik [125]

$v^2-{v_0}^2=2a(x-x_0)$

dónde $v$ es la velocidad de la esfera, $v_0$ es suya velocidad inicial, $a=-g$ la aceleración debida a la gravedad, $x$ la posición, y $x_0$ la posición inicial. Tomamos $x_0=0\,\mathrm m$ a referirse a la posición de la esfera en el momento que la esfera fue lanzada.

Entonces

$v^2-\left(30\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-10\,\dfrac{\mathrm m}{\mathrm s^2}\right)(40\,\mathrm m)$

$\implies v^2=100\,\dfrac{\mathrm m^2}{\mathrm s^2}\implies v=\pm10\,\dfrac{\mathrm m}{\mathrm s}$

Esto nos dice que la esfera alcanza una altura de 40 m en dos momentos - una vez hacia arriba y una vez hacia abajo. Sin embargo, independientemente del signo de la velocidad, sabemos que suya magnitud es 10 m/s, y así tenemos una rapidez de 10 m/s también en ambos momentos.

4 0

3 years ago