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2 years ago

6

Un cubo de hielo de masa 500g a temperatura de -20°C, se introduce en un recipiente en el cual se mantiene la presión constante

y se le suministra calor hasta que se transforme en vapor a 115°C. Determinar la cantidad de calor que se le suministra durante el proceso.

1 answer:

I am Lyosha [343]2 years ago

7 0

Answer:

An ice cube of mass 500g at a temperature of -20 ° C is placed in a container in which the pressure is kept constant and heat is supplied until it transforms into steam at 115 ° C. Determine the amount of heat that is supplied to you during the process.

Explanation:

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A cylindrical metal can is to have no lid. It is to have a volume of 125π in3. What height minimizes the amount of metal used?

Alex Ar [27]

Answer:

Minimum height of metal = 5 inches

Explanation:

Volume of the cylindrical metal = πR²H = 125π

cancelling out π on both sides

R²H = 125

Hence it can be deduced that R² = 25 and H = 5

Hence minimum height of metal = 5 inches

6 0

2 years ago

Gas and dust are the primary components of the interstellar medium. They differ in structural form. True or False

Sedbober [7]

Answer:

True

Explanation:

4 0

2 years ago

a tire is rolling along a road, without slipping with a velocity v. a piece of tape is attached to the tire. When the tape is op

Kryger [21]

Answer:

The right solution will be the "2v".

Explanation:

For something like an object underneath pure rolling the speed at any point is calculated by:

⇒ $v_{rolling}=v_{translational}+v_{rotational}$

Although the angular velocity was indeed closely linked to either the transnational velocity throughout particular instance of pure rolling as:

⇒ $\omega=\frac{v_{translational}}{r}$

Significant meaning is obtained, as speeds are in the same direction. Therefore the speed of rotation becomes supplied by:

⇒ $v_{rotational}=\omega \times r$

On substituting the estimated values, we get

⇒ $=\frac{v_{translational}}{r} \times r$

⇒ $=v_{translational}$

So that the velocity will be:

⇒ $v_{rolling}=v+v$

⇒ $=2v$

4 0

3 years ago

The abbreviation ucs stands for

LenaWriter [7]

It stands for Unified Computing System

I hope I helped :3

3 0

3 years ago

Read 2 more answers

Two charges, X and Y, are placed along the x-axis. Charge X is +18 nC and is placed at x = 0. Charge Y is placed at a location o

Helen [10]

Answer:

Charge Z can be placed at <em>x</em> = -2.7 m or at <em>x</em> = 0.27 m.

Explanation:

The Coulomb force between two charges, $Q_1$ and $Q_2$ , separated by a distance, $d$ , is given

$F = k\dfrac{Q_1Q_2}{r^2}$

<em>k</em> is a constant.

For the charge Z to be at equilibrium, the force exerted on it by charge X must be equal and opposite to the force exerted on it by charge Y.

It is to be placed along the <em>x</em>-axis. Hence, it is on the same line as charges X and Y.

Let the charge on Z be <em>Q</em>. It is positive.

Let the distance from charge X be <em>x m.</em> Then the distance from charge Y will be (0.60 - <em>x</em>) m.

Force due to charge X

$F_X = k\dfrac{18Q}{x^2}$

Force due to charge Y

$F_Y = k\dfrac{-27Q}{(0.60-x)^2}$

Since both forces are equal and opposite,

$F_X = -F_Y$

$k\dfrac{18Q}{x^2} = -k\dfrac{-27Q}{(0.60-x)^2}$

$\dfrac{2}{x^2} = \dfrac{3}{(0.60-x)^2}$

$2(0.60-x)^2 = 3x^2$

$2(0.36-1.20x+x^2) = 3x^2$

$0.72-2.40x+2x^2 = 3x^2$

$x^2+2.40x-0.72 = 0$

Applying the quadratic formula,

$x = \dfrac{-2.40\pm\sqrt{2.40^2 - (4)(1)(-0.72)}}{2} = \dfrac{-2.40\pm\sqrt{8.64}}{2}$

$x = -2.7$ or $x = 0.27$

Charge Z can be placed at <em>x</em> = -2.7 m or at <em>x</em> = 0.27 m

3 0

3 years ago