Important minerals such as copper are found in what?

Find the equivalent resistance between points A and B in the drawing

densk [106]

20.o hope this helps if not I am sorry

4 0

3 years ago

A sphere is originally at a temperature of 500°c. The sphere is melted and recast, without loss of mass, into a cube with the sa

aleksandr82 [10.1K]

Answer: The value of the celsius temperature of the cube is 472.2°c.

Explanation:

The expression for the power radiated is as follows;

$P=A\epsilon\sigma T^{4}$

Here, A is the area, $\sigma$ is the stefan's constant, $\epsilon$ is the emissivity and T is the temperature.

It is given in the problem that A sphere is originally at a temperature of 500°c. The sphere is melted and recast, without loss of mass, into a cube with the same emissivity as the sphere.

Then the expression for the radiated power for the cube and the sphere can be expressed as;

$A_{1}\epsilon \e\sigma T_{1}^{4}=A_{2}\epsilon \e\sigma T_{2}^{4}$

Here, $A_{1}$ is the area of the sphere, $A_{2}$ is the area of the cube, $T_{1}$ is the temperature of the sphere and $T_{2}$ is the temperature of the cube.

The radiated powers and emissivity of the cube and the sphere are same.

$A_{1}T_{1}^{4}=A_{2}T_{2}^{4}$

The area of the sphere is $A_{1}=4\pi \times r^{2}$ .

Here, r is the radius of the sphere.

The area of the cube is $A_{2}=6\times a^{2}$ .

Here, a is the edge of the cube.

Put $A_{1}=4\pi \times r^{2}$ and $A_{2}=6\times a^{2}$ .

$T_{2}=T_{1}(\frac{2\pi }{3}\times (\frac{r}{a})^{2})^{\frac{1}{4}}$ ....(1)

The masses and the densities of the sphere and the cube are same. Then the volumes are also same.

$V_{1}=V_{2}$

Here, $V_{1},V_{1}$ are the volumes of the sphere and the cube.

$\frac{4}{3}\pi r^{3}=a^{3}$

$\frac{r}{a}=(\frac{3}{4\pi })^{\frac{1}{3}}$

Put this value in the equation (1).

$T_{2}=T_{1}(\frac{2\pi }{3}\times (\frac{r}{a})^{2})^{\frac{1}{4}}$ $T_{2}=T_{1}(\frac{2\pi }{3}\times ((\frac{3}{4\pi })^{\frac{1}{3}})^{2})^{\frac{1}{4}}$

Put T_{1}=500°c.

$T_{2}=(500)(\frac{2\pi }{3}\times (\frac{3}{4\pi })^{\frac{2}{3}})^{\frac{1}{4}}$

$T_{2}=472.2^{\circ}c$

Therefore, the value of the celsius temperature of the cube is 472.7°c.

5 0

3 years ago

What is the density of 53.4 wt queous naoh solution if 16.7 ml of the solution diluted to 2.00 l gives 0.169 m naoh?

vampirchik [111]

The density of 53.4 wt aqueous NaOH solution is 0.809 g/ml

Given data:

The mass percent of NaOH is 53.4.
Volume of NaOH diluted is 16.7 ml.
The volume of diluted solution is 2.00 L =2000 ml.
Concentration of diluted solution is 0.169 M.

First, we find the initial concentration of NaOH by using the following formulae,

M₁V₁ = M₂V₂

Where,

M₁ is the initial molarity of NaOH

M₂ is the molarity after dilution

V₁ is the initial volume of NaOH

V₂ is the final volume after dilution.

Substituting the values,

M₁ × 16.7 ml = 0.169 M × 2000 ml,

M₁ = $\frac{0.169 M *2000 ml}{16.7 ml}$

M₁ = 20.2 M.

Thus, the initial concentration of NaOH is 20.2 M.

we know, 1 M solution contains 1 mol of substance present in 1 L solution,

Thus, 20.2 M solution will have 20.2 mols of NaOH.

Now, we can find the mass of NaOH by using the number of moles and molar mass.

molar mass of NaOH is 40 g/mol.

Mass = no. of moles × molar mass

= 20.2 mol × 40 g/mol

= 808 g.

Thus, the mass of NaOH is 808g.

53.4 wt of NaOH means 53.4 g of NaOH in a 100 g solution,

Thus, 808 g of NaOH will be present in ,

⇒ $\frac{53.4 g NaOH}{100 g solution} = \frac{808 g NaOH}{x g solution}$

⇒ 1513.1 g

Now, Convert the grams of NaOH to milliliters, using the density of NaOH at room temperature.

The density of NaOH at room temperature is 1.515 g/ml,

Density = $\frac{mass}{volume}$

⇒ 1.515 g/mol = $\frac{1513.1 g}{volume}$

⇒ volume = $\frac{1513.1 g}{1.515 g/mol}$

⇒ volume = 998.7 ml.

Thus, the volume of NaOH is 998.7 ml.

Hence, we know,

the mass of NaOH is 808 g
the volume of NaOH is 998.7 ml

Substituting the values,

Density = 808 g / 998.7 g/ml

⇒ Density = 0.809 g/ml

Thus, the density of 53.4 wt aqueous NaOH is 0.809 g/ml.

To learn more about Density here

brainly.com/question/15164682

#SPJ4

3 0

2 years ago

A wire delivers 12.0 C of charge in 4.0 s. What is the current in the wire? 3.0 A 8.0 A 16 A 48 A

const2013 [10]

3.0 A i got it off Quizlet and there usually always right lol can't submit tho my answers to short.... Dot dot dot

5 0

4 years ago

Read 2 more answers

Please need help fast

iVinArrow [24]

(a) See graph in attachment

The appropriate graph to draw in this part is a graph of velocity vs time.

In this problem, we have a horse that accelerates from 0 m/s to 15 m/s in 10 s.

Assuming the acceleration of the horse is uniform, it means that the velocity (y-coordinate of the graph) must increase linearly with the time: therefore, the velocity-time graph will appear as a straight line, having the final point at

t = 10 s

v = 15 m/s

(b) $1.5 m/s^2$