The only type of investigation that involves a control group is

Calculate the critical angle for light going from Glycerine to air.

Georgia [21]

The refractive index for glycerine is $n_g=1.473$ , while for air it is $n_a = 1.00$ .

When the light travels from a medium with greater refractive index to a medium with lower refractive index, there is a critical angle over which there is no refraction, but all the light is reflected. This critical angle is given by:
$\theta_c = \arcsin ( \frac{n_2}{n_1} )$
where n1 and n2 are the refractive indices of the two mediums. If we susbtitute the refractive index of glycerine and air in the formula, we find the critical angle for this case:
$\theta_c = \arcsin ( \frac{1.00}{1.473} )=42.8^{\circ}$

6 0

3 years ago

Which of the following changes would cause the fusion rate in the Sun’s core to increase? Check all that apply. View Available H

fredd [130]

Answer:

1. An increase in the core temperature

2. A decrease in the core radius.

Explanation:

The sun is a Main Sequence star. A Main Sequence star is powered by fusing hydrogen into Helium within its core.

For this fusion to take place, a temperature of at least 10 million Kelvin is required, beyond this point, the fusion rate is directly related to the core temperature. If the temperature increases, the fusion rate will greatly increase.

Something similar happens if the core reduces its radius. This can happen at the end of the star's lifetime, shortly before it becomes a red giant. Once the hydrogen is depleted, the core will start to shrink because the force of gravity, and as it gets smaller, gets more compressed, and its temperature increases. The outer layers of remaining Hydrogen that were outside the core now begin to heat up, and as the core continues to shrink, the star gets hot enough to begin the fusion process again, and the fusion rate can even be higher than it was during the first phase of the star, as the star becomes a Red Giant.

7 0

3 years ago

PLEASE HEP FAST PLEASE!!!!! IM SO SCARED!!!! An elevator has the mass of 3 tons of power needed to raise the elevator 50m in 15

gayaneshka [121]

Answer:

Mass=3,height=50 and time=15,g=10,P=?

P=mgh/t....which is 3×10×50=1500/15=100Watts

6 0

3 years ago

Read 2 more answers

A circular loop of flexible iron wire has an initial circumference of 167 cm, but its circumference is decreasing at a constant

Kisachek [45]

Answer:

Part a)

$EMF = 5.6 \times 10^{-3} V$

Part b)

Since the radius is decreasing so induced current will increase the flux through the coil

So it would be clockwise in direction

Explanation:

As we know that magnetic flux linked with the coil is given as

$\phi = \pi r^2 B$

now the rate of change in flux is given as

$\frac{d\phi}{dt} = 2\pi r \frac{dr}{dt} B$

now we know that circumference is decreasing at rate of 15 cm/s

so here we know the length of circumference as

$C = 2\pi r$

So rate of change in circumference is

$\frac{dC}{dt} = 2\pi \frac{dr}{dt}$

$\frac{1}{2\pi}(15 cm) = \frac{dr}{dt}$

final length of circumference at t = 8 s

$C = 167 - (15)(8) = 47$

Part a)

Now the induced EMF is given as

$EMF = (2\pi r)(\frac{1}{2\pi})(0.15)(0.5)$

$EMF = (0.47)(\frac{1}{2\pi})(0.15)(0.5)$

$EMF = 5.6 \times 10^{-3} V$

Part b)

Since the radius is decreasing so induced current will increase the flux through the coil

So it would be clockwise in direction

5 0

3 years ago

Through what potential difference must electrons be accelerated so that they will exhibit wave nature in passing through a pinho

Mila [183]

Answer:

0.15 mV

Explanation:

In order to exhibit wave nature, the de Broglie wavelength of the electron must be of the same size of the diameter of the pinhole, therefore:

$\lambda=0.10 \mu m = 1.0\cdot 10^{-7} m$

The de Broglie wavelength of an electron is

$\lambda = \frac{h}{mv}$

where

$h=6.63\cdot 10^{-34} Js$ is the Planck constant

$m=9.11\cdot 10^{-31} kg$ is the mass of the electron

v is the electron's speed

Therefore, the electron's speed must be

$v=\frac{h}{m\lambda}=\frac{6.63\cdot 10^{-34}}{(9.11\cdot 10^{-31})(1.0\cdot 10^{-7})}=7278 m/s$

When accelerated through a potential difference $\Delta V$ , the kinetic energy gained by the electron is equal to the change in electric potential energy, therefore

$e\Delta V = \frac{1}{2}mv^2$

where

$e=1.6\cdot 10^{-19}$ is the magnitude of the charge of the electron

So, we can find the potential difference needed:

$\Delta V=\frac{mv^2}{2e}=\frac{(9.11\cdot 10^{-31})(7278)^2}{2(1.6\cdot 10^{-19})}=1.5\cdot 10^{-4}V = 0.15 mV$

5 0

3 years ago