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3 years ago

What is spontaneous emission?

Physics

2 answers:

Oxana [17]3 years ago

5 0

Answer:

Spontaneous emission is a quantum effect, which in a semiclassical picture can be described as an emission which is stimulated by vacuum noise.

Irina-Kira [14]3 years ago

4 0

Spontaneous emission is the process in which a quantum mechanical system transits from an excited energy state to a lower energy state and emits a quantized amount of energy in the form of a photon. <em><u>definition from wikipedia </u></em>

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Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy the eq

gayaneshka [121]

Answer:

The volume is decreasing at 160 cm³/min

Explanation:

Given;

Boyle's law, PV = C

where;

P is pressure of the gas

V is volume of the gas

C is constant

Differentiate this equation using product rule:

$V\frac{dp}{dt} +P\frac{dv}{dt} = \frac{d(C)}{dt}$

Given;

$\frac{dP}{dt}$ (increasing pressure rate of the gas) = 40 kPa/min

V (volume of the gas) = 600 cm³

P (pressure of the gas) = 150 kPa

Substitute in these values in the differential equation above and calculate the rate at which the volume is decreasing ( $\frac{dv}{dt}$ );

(600 x 40) + (150 x $\frac{dv}{dt}$ ) = 0

$\frac{dv}{dt} = -\frac{(600*40)}{150} = -160 \ cm^3/min$

Therefore, the volume is decreasing at 160 cm³/min

3 0

2 years ago

The magnitude of a force is:

lys-0071 [83]

Answer:

Explanation:

force is how hard it is pulled or pushed

3 0

2 years ago

A student measures the time it takes for two reactions to be completed reaction eight is completed and 57 seconds and reaction b

Maslowich

The answer to this problem is 1200

8 0

2 years ago

When the air resistance can be ignored the velocity of an object dropped initially from rest is given by the following equation

ad-work [718]

Answer:

I am confused of your question. Do you want final velocity? To get final velocity, use (initial V)+(Gravity*Time)

Explanation:

8 0

3 years ago

Read 2 more answers

At the same moment, one rock is dropped and one is thrown downward with an initial velocity of 29m/s from the top of a building

Inessa [10]

Answer:

The thrown rock strike 2.42 seconds earlier.

Explanation:

This is an uniformly accelerated motion problem, so in order to find the arrival time we will use the following formula:

$x=vo*t+\frac{1}{2} a*t^2\\where\\x=distance\\vo=initial velocity\\a=acceleration$

So now we have an equation and unkown value.

for the thrown rock

$\frac{1}{2}(9.8)*t^2+29*t-300=0$

for the dropped rock

$\frac{1}{2}(9.8)*t^2+0*t-300=0$

solving both equation with the quadratic formula:

$\frac{-b\±\sqrt{b^2-4*a*c} }{2*a}$

we have:

the thrown rock arrives on t=5.4 sec

the dropped rock arrives on t=7.82 sec

so the thrown rock arrives 2.42 seconds earlier (7.82-5.4=2.42)

6 0

2 years ago