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3 years ago

5

- Una sustancia posee un punto de fusión de -110ºC y un punto de ebullición de 6ºC. Explica en qué estado físico se encontrarán

10 g de esta sustancia en un recipiente que está a :
a) -80ºC

b) 0 ºC

c) 14 ºC

1 answer:

Burka [1]3 years ago

3 0

Answer:

B

Explaination:

...

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The graph shows the number of beans eaten by a

il63 [147K]

Answer:

You didn't show a graph

Explanation:

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4 years ago

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How would the magnetic field lines appear for a bar magnet cut at the midpoint, with the two pieces placed end to end with a spa

Zarrin [17]

Answer:

cutting the magnet in two parts each part has a North and South pole,

Explanation:

In magnetism the magnetic mono-poles are not found, this means that we do not have magnetic charges alone, therefore when cutting the magnet in two parts each part has a North and South pole, the magnetic lines go from the North pole to the South pole, see attached.

The density of the lines is approximately the intensity of the magnetic field.

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3 years ago

The properties of element 17 are most similar to which of the following elements?

DedPeter [7]

Halogens are highly reactive nonmetal elements in group 17 of the periodic table. ... Halogens are among the most reactive of all elements. They have seven valence electrons, so they are very “eager” to gain one electron to have a full outer energy level.

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3 years ago

A horizontal 745 N merry-go-round of radius

Arturiano [62]

Answer:

The kinetic energy of the merry-goround after 3.62 s is 544J

Explanation:

Given :

Weight w = 745 N

Radius r = 1.45 m

Force = 56.3 N

To Find:

The kinetic energy of the merry-go round after 3.62 = ?

Solution:

Step 1: Finding the Mass of merry-go-round

$m = \frac{ weight}{g}$

$m = \frac{745}{9.81 }$

m = 76.02 kg

Step 2: Finding the Moment of Inertia of solid cylinder

Moment of Inertia of solid cylinder I = $0.5 \times m \times r^2$

Substituting the values

Moment of Inertia of solid cylinder I

=> $0.5 \times 76.02 \times (1.45)^2$

=> $0.5 \times 76.02\times 2.1025$

=> $79.91 kg.m^2$

Step 3: Finding the Torque applied T

Torque applied T = $F \times r$

Substituting the values

T = $56.3 \times 1.45$

T = 81.635 N.m

Step 4: Finding the Angular acceleration

Angular acceleration , $\alpha = \frac{Torque}{Inertia}$

Substituting the values,

$\alpha = \frac{81.635}{79.91}$

$\alpha = 1.021 rad/s^2$

Step 4: Finding the Final angular velocity

Final angular velocity , $\omega = \alpha \times t$

Substituting the values,

$\omega = 1.021 \times 3.62$

$\omega = 3.69 rad/s$

Now KE (100% rotational) after 3.62s is:

KE = $0.5 \times I \times \omega^2$

KE = $0.5 \times 79.91 \times 3.69^2$

KE = 544J

6 0

3 years ago

If a 2 x 10^-4C test charge is given 6.5J of energy, determine the electric potential difference.

Gwar [14]

Answer:

The electric potential difference is 32500 volt.

Explanation:

Given that,

Charge $q=2\times10^{-4}C$

Energy = 6.5 J

We need to calculate the electric potential difference

Potential difference :

Potential difference is equal to the energy divide by charge.

Using formula of potential difference

$V=\dfrac{E}{Q}$

$V=\dfrac{6.5}{2\times10^{-4}}$

$V=32500\ volt$

Hence, The electric potential difference is 32500 volt.

8 0

3 years ago