When electrons in an atom in an excited state fall to lower energy levels, energy is?

If the Earth and distant stars were stationary (motionless) in space, what would we observe about the wavelength from these star

torisob [31]

There's no such thing as "stationary in space". But if the distance
between the Earth and some stars is not changing, then (A) wavelengths
measured here would match the actual wavelengths emitted from these
stars. 

If a star is moving toward us in space, then (A) Wavelengths measured
would be shorter than the actual wavelengths emitted from that star.

In order to decide what's actually happening, and how that star is moving,
the trick is: How do we know the actual wavelengths the star emitted ?

7 0

3 years ago

You look down an old well, cannot see the bottom, and mutter to yourself "Oh well!". In order to estimate the depth of the well,

telo118 [61]

Answer:

The best estimate of the depth of the well is 2.3 sec.

Explanation:

Given that,

Record time,

$t_{1}=2.19\ sec$

$t_{2}=2.30\ sec$

$t_{3}=2.26\ sec$

$t_{4}=2.29\ sec$

$t_{5}=2.27\ sec$

We need to find the best estimate of the depth of the well

According to record time,

We can write of the record time

$t_{1}=2.19\approx 2.2\ sec$

$t_{2}=2.30\approx 2.3\ sec$

$t_{3}=2.26\approx 2.3\ sec$

$t_{4}=2.29\approx 2.3\ sec$

$t_{5}=2.27\approx 2.3\ sec$

Here, all time is nearest 2.3 sec.

So, we can say that the best estimate of the depth of the well is 2.3 sec.

Hence, The best estimate of the depth of the well is 2.3 sec.

6 0

3 years ago

The speed of sound in air is 345 m/s. A tuning fork vibrates above the open end of a sound resonance tube. If sound waves have w

Delvig [45]

To develop this problem it is necessary to apply the concept of Frequency based on speed and wavelength.

According to the definition the frequency can be expressed as

$f = \frac{v}{\lambda}$

Where,

v = Velocity

$\lambda =$ Wavelength

Our value are given by,

v = 345m/s

$\lambda = 63cm$

Replacing

$f = \frac{345}{0.63}$

$f = 547.61Hz$

Therefore the frequency of the tuning fork is 547.61Hz

6 0

2 years ago

The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio

Sveta_85 [38]

Answer:

temperature on left side is 1.48 times the temperature on right

Explanation:

GIVEN DATA:

$\gamma = 5/3$

T1 = 525 K

T2 = 275 K

We know that

$P_1 = \frac{nRT_1}{v}$

$P_2 = \frac{nrT_2}{v}$

n and v remain same at both side. so we have

$\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}$

$P_1 = \frac{21}{11} P_2$ ..............1

let final pressure is P and temp $T_1 {f} and T_2 {f}$

$P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}$

$P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3}$ ..................2

similarly

$P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3}$ .............3

divide 2 equation by 3rd equation

$\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}$

$T_1 {f} = 1.48 T_2 {f}$

thus, temperature on left side is 1.48 times the temperature on right

6 0

2 years ago

Pressure in a liquid increases with its ..........

Novay_Z [31]

Answer: Pressure increases as the depth increases

Explanation: The pressure in a liquid is due to the weight of the column of water above. Since the particles in a liquid are tightly packed, this pressure acts in all directions.

6 0

2 years ago

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