Which model is most likely used to predict a future event?

Fudgin [204]

Answer:

Speed of the light in water= 225,000 km/s

Explanation:

At the speed with which light propagates through a homogeneous and transparent medium, it is a constant characteristic of that medium, and therefore, it changes from one medium to another.

Due to its enormous magnitude, the measurement of the speed of light has required the invention of ingenious procedures that will overcome the inconvenience of short land distances in relation to such extraordinary speed.

Astronomical methods and terrestrial methods have been giving ever closer results. At present, the value c = 299,792,458 km / s is accepted for the speed of light in a vacuum. In any transparent material medium the light propagates with a speed that is always lower than c. Thus, for example, in water it does so at around 75% of the speed of light in a vacuum: about 225,000 km / s.

8 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

Ethane and ethanol are isomers.

alisha [4.7K]

No. The isomers have same molecular formula (same elements in the same number each).

Ethane does not contain oxygen and ethanol does.

3 0

3 years ago

HELPPPPPPPPPPPPPPPPPPPPPPPPPP

aev [14]

Answer:

C

Explanation:

through the desk....here desk is the student's medium to hear the sound. its oblivious because when he lifts his head away from the desk he hears nothing else

6 0

3 years ago

Luis wants to compare the density of three solid objects that are different shapes and sizes. What information does he need to m

tatyana61 [14]

Answer:

u need to make sure that comparison is = to shapes and then find the shapes sizes and add them

7 0

3 years ago