There is no way to ascertain the phases present in the container when equilibrium is established because we are not furnished with enough information.

The most important information needed to have a deeper understanding and provide solution is temperature.
Liquid water and ice can both exist at the same temperature.
So we cannot conclude based on the information at hand.

5 0

3 years ago

How much current will pass through a 12.5 ohm resistor when it is connected to ta 115 volt source of electrical potential?

Marrrta [24]

Answer:

9.2 amperes

Explanation:

Ohm's law states that the voltage V across a conductor of resistance R is given by $V = R I$

Here, voltage V is proportional to the current I.

For voltage, unit is volts (V)

For current, unit is amperes (A)

For resistance, unit is Ohms (Ω)

Put R = 12.5 and V = 115 in V=RI

$115=12.5I\\I=\frac{115}{12.5}\\ =9.2\,\,amperes$

8 0

4 years ago

A block of mass m is hung from the ceiling by the system of massless springs consisting of two layers. The upper layer consists

alukav5142 [94]

Answer:

T₀ = 2π $\sqrt{\frac{m}{k} }$ T = $\sqrt{\frac{5}{6} }$ T₀

Explanation:

When the block is oscillating it forms a simple harmonic motion, which in the case of a spring and a mass has an angular velocity

w = $\sqrt{k/m}$

To apply this formula to our case, let's look for the equivalent constant of the springs.

Let's start with the springs in parallels.

* the three springs in the upper part, when stretched, lengthen the same distance, therefore the total force is

F_total = F₁ + F₂ + F₃

the springs fulfill Hooke's law and indicate that the spring constant is the same for all three,

F_total = - k x - k x - kx = -3k x

therefore the equivalent constant for the combination of the springs at the top is

k₁ = 3 k

* the two springs at the bottom

following the same reasoning the force at the bottom is

F_total2 = - 2 k x

the equivalent constant at the bottom is

k₂ = 2 k

now let's work the two springs are equivalent that are in series

the top spring is stretched by an amount x₁ and the bottom spring is stretched x₂

x₂ = x -x₁

x₂ + x₁ = x

if we consider that the springs have no masses we can use Hooke's law

$-\frac{F_{1} }{k_{1} } - \frac{F_{2}}{k_{2} } = \frac{F}{k_{eq} }$

therefore the equivalent constant is the series combination is

$\frac{1}{k_{eq} } = \frac{1}{k_{1} } + \frac{1}{k_{2} }$

we substitute the values

\frac{1}{k_{eq} } = \frac{1}{3k } + \frac{1}{2k }

\frac{1}{k_{eq} } = \frac{5}{6k} }

k_eq = $\frac{6k}{5}$

therefore the angular velocity is

w = $\sqrt{\frac{6k}{5m} }$

angular velocity, frequency, and period are related

w = 2π f = 2π / T

T = 2π / w

T = 2π $\sqrt{\frac{5m}{6k} }$

T₀ = 2π $\sqrt{\frac{m}{k} }$

T = $\sqrt{\frac{5}{6} }$ T₀

8 0

3 years ago