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2 years ago

In which phase is the substance when its temperature is 250 degrees Celsius?

Physics

1 answer:

tatuchka [14]2 years ago

3 0

Answer:

o liquid

Explanation:

<h3>liquid fom because it's is liuid</h3>

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The two parents shown in (Figure 1) exert upward forces

timama [110]

Answer:

W = 83 [N] (weight)

Explanation:

We must clarify that the system is in static equilibrium, in such a way that we must perform a sum of forces on the Y-axis equal to zero, in order to determine the downward force corresponding to the weight of the child.

∑Fy = 0

$F_{1}+F_{2}-m*g=0\\18+65=m*g\\m*g = 83 [N]$

And the mass can be determined as follows:

$m*g=83\\m = 83/9.81\\m=8.46[kg]$

5 0

2 years ago

Six moles of an ideal gas are in a cylinder fitted at one end with a movable piston. The initial temperature of the gas is 28.0

mel-nik [20]

Answer:

63.5 °C

Explanation:

The expression for the calculation of work done is shown below as:

$w=P\times \Delta V$

Where, P is the pressure

$\Delta V$ is the change in volume

Also,

Considering the ideal gas equation as:-

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 8.314 J/ K mol

So,

$V=\frac{nRT}{P}$

Also, for change in volume at constant pressure, the above equation can be written as;-

$\Delta V=\frac{nR\times \Delta T}{P}$

So, putting in the expression of the work done, we get that:-

$w=P\times \frac{nR\times \Delta T}{P}=nR\times \Delta T$

Given, initial temperature = 28.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (28.0 + 273.15) K = 301.15 K

W=1770 J

n = 6 moles

So,

$1770\ J=6 moles\times 8.314\ J/ Kmol \times (T_2-301.15\ K)$

Thus,

$T_2=301.15\ K+\frac{1770}{6\times 8.314}\ K$

$T_2=336.63\ K$

The temperature in Celsius = 336.63-273.15 °C = 63.5 °C

<u>The final temperature is:- 63.5 °C</u>

7 0

3 years ago

A condition often referred to as being double jointed is an example of

Makovka662 [10]

It is an example of Hypermobility.

4 0

2 years ago

15 points! An atomic nucleus initially moving at 420 m/s emits an alpha particle in the direction of its velocity, and the remai

alexandr1967 [171]

The alpha particle is emitted at 4235 m/s

Explanation:

We can use the law of conservation of momentum to solve the problem: the total momentum of the original nucleus must be equal to the total momentum after the alpha particle has been emitted. Therefore:

$p_i = p_f\\ Mu=m_1 v_1 + m_2 v_2 =$