BIOLOGY CLASS LIVE<br><br>meet.google.com/cyh-kzdn-hns

Kelly sits on a rock. Her weight is an action force. What is the reaction force?

Dmitrij [34]

The force ffrom the ground that holds the rock up and prevents her from falling through the gound, unless the ground breaks.

4 0

4 years ago

A ray of light crosses a boundary between two transparent materials. The medium the ray enters has a larger optical density. Whi

Tpy6a [65]

Answer:

The wavelength of the light decreases as it enters into the medium with the greater optical density.

The frequency of the light remains constant as it transitions between materials.

Explanation:

- When a ray of light crosses a boundary between two different materials, it undergoes refraction: the ray changes direction, and it also changes speed, according to the relationship:

$v=\frac{c}{n}$

where v is the speed of the ray of light in the material, c is the speed of light in a vacuum, n is the index of refraction of the material, which is larger for a medium with larger optical density. So, from the equation, we see that the larger the optical density, the smaller the speed of the wave.

- The frequency of the light does not depend on the properties of the medium, so it remains unchanged: therefore the statement

The frequency of the light remains constant as it transitions between materials.

is correct.

- Moreover, the wavelength of the ray of light is related to the speed and the frequency by the equation

$\lambda=\frac{v}{f}$

where v is the speed and f the frequency. Since we have seen that v decreases and f remains constant, this means that the wavelength decreases as well, so the statement

The wavelength of the light decreases as it enters into the medium with the greater optical density.

is also correct.

5 0

3 years ago

Suppose that on a routine trip around the galaxy, your dog puts his head out of the spacecraft window (as dogs often do). The so

neonofarm [45]

Answer: Yes, the dog's head will burn

Explanation:Solar winds are strongly destructive magnetically charged high energy winds. Satellite communication on earth and the occasional geomagnetic storm near the poles are the major disruptive effects caused by solar winds on earth.

Since solar winds are streams of charged particles released from the upperparts of the Sun, called the "corona". This plasma mainly consists of electrons, protons and alpha particles with kinetic energy between 0.5 and 10 keV. Instilled within the solar-wind plasma is the interplanetary magnetic field.

If the dog holds it's breath the heat wave would burn it's head but not outrightly because the breath has ceased for some seconds and it can't absorb the much heat of the solar wind into it's nostrils. The effect of holding its breath would still be disastrous but not as the first case.

7 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).

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