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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 4.8 mF capacitor in series with a 500 Ω resistor is connected, by a switch, to a 12 V battery. The current through the resisto

ad-work [718]

Answer:

Current will be 81.7 mA

Which is not given in bellow option

Explanation:

We have given capacitance $C=4.8mF=4.8\times 10^{-3}F$

Resistance R = 500 ohm

Voltage V = 12 volt

We know that time constant of RC circuit of RC circuit is given by

$\tau =RC=500\times 4.8\times 10^{-3}=2.4sec$

Time is given as t = 1 sec

We know that current in RC circuit is given by

$i=\frac{v}{R}(1-e^{\frac{-t}{\tau }})$

So current $i=\frac{12}{500}(1-e^{\frac{-1}{2.4 }})=0.00817A=81.7mA$

Which is not given in the following option

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3 years ago

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A sinusoidal transverse wave of amplitude ym = 8.4 cm and wavelength = 5.3 cm travels on a stretched cord. Find the ratio of the

Scilla [17]

Answer:

The ratio is 9.95

Solution:

As per the question:

Amplitude, $y_{m} = 8.4\ cm$

Wavelength, $\lambda = 5.3\ cm$

Now,

To calculate the ratio of the maximum particle speed to the speed of the wave:

For the maximum speed of the particle:

$v_{m} = y_{m}\times \omega$

where

$\omega = 2\pi f$ = angular speed of the particle

Thus

$v_{m} = 2\pi fy_{m}$

Now,

The wave speed is given by:

$v = f\lambda$

Now,

The ratio is given by:

$\frac{v_{m}}{v} = \frac{2\pi fy_{m}}{f\lambda}$

$\frac{v_{m}}{v} = \frac{2\pi \times 8.4}{5.3} = 9.95$

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4 years ago

Se o Universo tem uma idade definida, de aproximadamente 13,7 bilhões de anos, o que existiu antes do big-bang?

UkoKoshka [18]

Ninguém sabe exatamente

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3 years ago

An asteroid is moving along a straight line. A force acts along the displacement of the asteroid and slows it down. The asteroid

Setler79 [48]

Answer:

$- 7.04\times10^{11} J$

$5.03\times10^{5} N$

Explanation:

$m$ = mass of the asteroid = 43000 kg

$v_{o}$ = initial speed of asteroid = 7600 m/s

$v$ = final speed of asteroid = 5000 m/s

$W$ = Work done by the force on asteroid

Using work-change in kinetic energy theorem

$W = (0.5) m (v^{2} - v_{o}^{2})\\W = (0.5) (43000) (5000^{2} - 7600^{2})\\W = - 7.04\times10^{11} J$

$F$ = magnitude of force on asteroid

$d$ = distance traveled by asteroid while it slows down = 1.4 x 10⁶ m

Work done by the force on the asteroid to slow it down is given as

$W = - F d\\- 7.04\times10^{11} = - F (1.4\times10^{6})\\F = 5.03\times10^{5} N$

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3 years ago

Maria makes a graphic organizer to compare open, closed, and short circuits. She adds the labels shown.

Lina20 [59]

The Answer is Label 1:Y Label 2: W

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3 years ago

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