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Drupady [299]
3 years ago
10

3. What can you infer about an ion that has a positive charge?

Physics
1 answer:
IRINA_888 [86]3 years ago
4 0

Answer: This ion is derived from atom who lost one electron.

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If you pour hot soup into a bowl, and the bowl stays cool to the touch, you can assume that...
sveta [45]

Answer:

The soup still is cool, or since its recent it will take a while to get warmer

Explanation:

3 0
3 years ago
Two identical horizontal sheets of glass have a thin film of air of thickness t between them. The glass has refractive index 1.4
Gre4nikov [31]

Answer:

the wavelength λ of the light when it is traveling in air = 560 nm

the smallest thickness t of the air film = 140 nm

Explanation:

From the question; the path difference is Δx = 2t  (since the condition of the phase difference in the maxima and minima gets interchanged)

Now for constructive interference;

Δx= (m+ \frac{1}{2} \lambda)

replacing ;

Δx = 2t   ; we have:

2t = (m+ \frac{1}{2} \lambda)

Given that thickness t = 700 nm

Then

2× 700 = (m+ \frac{1}{2} \lambda)     --- equation (1)

For thickness t = 980 nm that is next to constructive interference

2× 980 = (m+ \frac{1}{2} \lambda)     ----- equation (2)

Equating the difference of equation (2) and equation (1); we have:'

λ = (2 × 980) - ( 2× 700 )

λ = 1960 - 1400

λ = 560 nm

Thus;  the wavelength λ of the light when it is traveling in air = 560 nm

b)  

For the smallest thickness t_{min} ; \ \ \ m =0

∴ 2t_{min} =\frac{\lambda}{2}

t_{min} =\frac{\lambda}{4}

t_{min} =\frac{560}{4}

t_{min} =140 \ \  nm

Thus, the smallest thickness t of the air film = 140 nm

7 0
4 years ago
Read 2 more answers
A photoelectric experiment is performed where green light with a wavelength of 546.1 nm is shined on a metal plate, creating a p
PtichkaEL [24]

Answer:

\phi=1.55 [eV]

Explanation:

We can use the work function equation for a photoelectric experiment:

\phi=\frac{hc}{\lambda}-K_{max}

  • h is the plank constant
  • c is the speed of light
  • λ is the wave length
  • K is the kinetic energy (or K=eΔV)

So we will have:

\phi=\frac{hc}{\lambda}-e\Delta V

\phi=\frac{6.63*10^{-34}*3*10^{8}}{546.1*10^{-9}}-0.728eV    

\phi=3.64*10^{-19}[J]-0.728 [eV]

\phi=(3.64*10^{-19}[J]*\frac{1eV}{1.6*10^{-19}[J]})-0.728 [eV]

\phi=2.28 [eV] - 0.728 [eV]

\phi=1.55 [eV]

I hope it helps you!

8 0
4 years ago
Read 2 more answers
A bumblebee is flying to the right when a breeze causes the bee to slow down with a constant leftward acceleration of magnitude
romanna [79]

Answer:

3.75 m/s

Explanation:

Given that a bumblebee is flying to the right when a breeze causes the bee to slow down with a constant leftward acceleration of magnitude 0.50 m/s^2. After 2.0 s the bee is moving to the right with a speed of 2.75 m/ s

What was the velocity of the bumblebee right before the breeze?

Since the breeze blows with the acceleration of 0.5 m/s^2 for 2 seconds, we can calculate the magnitude of the velocity at which it moves.

Acceleration = velocity/ time

Substitute the acceleration and time into the formula

0.5 = V/2

Cross multiply

V = 2 × 0.5

V = 1 m/s

The bumblebee is travelling right ward while the wind travels leftward.

The relative velocity = 2.75 m/s

Let the bumblebee speed = Vb

While the wind = VW

Vb - Vw = 2.75

Substitute the wind speed into the formula

Vb - 1 = 2.75

Vb = 2.75 + 1

Vb = 3.75 m/s

Therefore, the velocity of the bumblebee right before the breeze is 3.75 m/s

8 0
3 years ago
At the start of an experiment, the potential energy of an object was 326 joules and its kinetic energy was 117 joules. At the en
Oksanka [162]

he mechanical energy of the system decreased due to the transfer of energy on a macroscopic scale to energy on a molecular scale.

7 0
3 years ago
Read 2 more answers
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