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3 years ago

What’s the difference between applied and pure research

Physics

1 answer:

Ket [755]3 years ago

8 0

Answer:

Pure research is conducted without any specific goal in mind. Applied research is conducted in order to solve a specific and practical problem, unlike pure research. Hope this helped.

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A bullet is dropped into a river from a very high bridge. At the same time, another bullet is fired from a gun straight down tow

anastassius [24]

Answer:

Option (c)

Explanation:

Both the bullets have same acceleration because they both falls under the influence of acceleration due to gravity.

The bullet which is fired from the gun has some initial velocity but the bullet which is dropped has zero initial velocity.

the acceleration is acting on both the bullets which is equal to the acceleration due to gravity and they both in motion in the influence of gravity.

8 0

3 years ago

What is the frequency of a photon with an energy of 4. 56 x 10^-19 j

Sauron [17]

The frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.

<h3>What is a frequency?</h3>

The number of waves that travel through a particular point in a given length of time is described by frequency. So, if a wave takes half a second to pass, the frequency is 2 per second.

Given that the energy of the photon is 4.56 x 10⁻¹⁹ J. Therefore, the frequency of the photon can be written as,

$\rm \gamma = \dfrac{E}{h} = \dfrac{4.56x10^{-19} J}{6.626 \times 10^{-34}\ Jsec^{-1}}\\\\\\\gamma = 6.88 \times 10^{14}\ s^{-1}$

Hence, the frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.

Learn more about Frequency:

brainly.com/question/5102661

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5 0

2 years ago

Read 2 more answers

On the Moon, the acceleration due to the effect of gravity is only about 1/6 of that on Earth. An astronaut whose weight on Eart

vazorg [7]

And on the moon, the with is 595/6 N

3 0

3 years ago

What is the magnitude of the magnetic field at a point midway between them if the top one carries a current of 19.5 A and the bo

Phantasy [73]

Answer:

The magnetic field will be $\large{\dfrac{1.4 \times 10^{-4}}{d}} T$ , '2d' being the distance the wires.

Explanation:

From Biot-Savart's law, the magnetic field ( $\large{\overrightarrow{B}}$ ) at a distance ' $r$ ' due to a current carrying conductor carrying current ' $I$ ' is given by

$\large{\overrightarrow{B} = \dfrac{\mu_{0}I}{4 \pi}} \int \dfrac{\overrightarrow{dl} \times \hat{r}}{r^{2}}}$

where ' $\overrightarrow{dl}$ ' is an elemental length along the direction of the current flow through the conductor.

Using this law, the magnetic field due to straight current carrying conductor having current ' $I$ ', at a distance ' $d$ ' is given by

$\large{\overrightarrow{B}} = \dfrac{\mu_{0}I}{2 \pi d}$

According to the figure if ' $I_{t}$ ' be the current carried by the top wire, ' $I_{b}$ ' be the current carried by the bottom wire and ' $2d$ ' be the distance between them, then the direction of the magnetic field at 'P', which is midway between them, will be perpendicular towards the plane of the screen, shown by the $\bigotimes$ symbol and that due to the bottom wire at 'P' will be perpendicular away from the plane of the screen, shown by $\bigodot$ symbol.