Heat energy is:

Which is an advantage of storing data digitally?

ra1l [238]

Answer:

D

Explanation:

Because I said so

5 0

2 years ago

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In a thin film experiment, a wedge of air is used between two glass plates. If the wavelength of the incident light in air is 48

cluponka [151]

Answer:

The thickness is $\Delta y = 2.4 *10^{-6} \ m$

Explanation:

From the question we are told that

The wavelength is $\lambda = 480 \ nm = 480*10^{-9} \ m$

The first order of the dark fringe is $m_1 = 16$

The second order of dark fringe considered is $m_2 = 6$

Generally the condition for destructive interference is mathematically represented as

$y = \frac{m \lambda}{2}$

Here y is the path difference between the central maxima(i.e the origin) and any dark fringe

So the path difference between the 16th dark fringe and the 6th dark fringe is mathematically represented as

$y_1 - y_2 = \Delta y = \frac{m_1 \lambda}{2} - \frac{m_2 \lambda}{2}$

=> $y_1 - y_2 = \Delta y = \frac{16 *480*10^{-9}}{2} - \frac{6 *480*10^{-9}}{2}$

=> $y_1 - y_2 = \Delta y = 5 (480*10^{-9})$

=> $\Delta y = 2.4 *10^{-6} \ m$

8 0

3 years ago

1. Which of the following states would likely get hit by both hurricanes and tornadoes? (Points : 1)

bogdanovich [222]

I would have to say Texas because, obviously, its on the coast, and because I know for a fact Oklahoma is VERY prone to Tornadoes and I also know Dallas (and surrounding areas) has a few tornadoes a year:)

I hope I helped:)

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5 0

2 years ago

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The wasted energy in a charger is:

kirza4 [7]

Thermal is the wasted energy in a charger

4 0

3 years ago

On your first day at work as an electrical technician, you are asked to determine the resistance per meter of a long piece of wi

Lostsunrise [7]

Answer:

$0.06\Omega/m$

Explanation:

Firstly, when you measure the voltage across the battery, you get the emf,

E = 13.0 V

In order to proceed we have to assume that the voltmeter offers no loading effect, which is a valid assumption since it has a very high resistance.

Secondly, the wires must be uniform. So the resistance per unit length is constant (say z). Now, even though the ammeter has very little resistance it cannot be ignored as it must be of comparable value/magnitude when compared to the wires. This is can seen in the two cases when currents were measured. Following Ohm's law and the resistance of a length of wire being proportional to it's length, we should have gotten half the current when measuring with the 40 m wire with respect to the 20 m wire ( $I=\frac{V}{R}$ ). But this is not the case.