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aniked [119]
3 years ago
15

The difference in energy () between vibrational energy levels is determined by the nature of the bond. If a photon of light poss

esses exactly this amount of energy, the bond can absorb the photon to promote a excitation. True or false?
Physics
1 answer:
forsale [732]3 years ago
5 0

Answer:

the statement is true

Explanation:

The bonds in the molecules give rise to irrational energy levels, are quantized, that is, they are discrete

In order for a photon to create a transition between the given levels it has to have the same energy as these two levels, it is in this case if the photon has the energy of the bond can be promised to the electron and be absorbed.

In summary the statement is true

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What is the unit for resistance?
AnnyKZ [126]

For electrical resistance, the unit is Ohm. The symbol is capital greek O.

4 0
3 years ago
The drawing shows a large cube (mass = 28.6 kg) being accelerated across a horizontal frictionless surface by a horizontal force
MrRissso [65]

Answer:

P= 454.11 N

Explanation:

Since P is the only horizontal force acting on the system, it can be defined as the product of the acceleration by the total mass of the system (both cubes).

P= (M+m)*a\\a = \frac{P}{28.6 +4.3}\\a = \frac{P}{32.9}

The friction force between both cubes (F) is defined as the normal force acting on the smaller cube multiplied by the coefficient of static friction. Since both cubes are subject to the same acceleration:

F = m * a*\mu \\F= 4.3*0.710*\frac{P}{32.9}\\F=3.053*\frac{P}{32.9}

In order for the small cube to not slide down, the friction force must equal the weight of the small cube:

3.053*\frac{P}{32.9} = 4.3 * g\\\\P = \frac{4.3*9.8*32.9}{3.053} \\P= 454.11 N

The smallest magnitude that P can have in order to keep the small cube from sliding downward is 454.11 N

8 0
3 years ago
In Millikan's experiment, an oil drop of radius 1.362 μm and density 0.888 g/cm3 is suspended in chamber C when a downward-point
Misha Larkins [42]

Answer:

The charge on the oil drop is 3e

Explanation:

F = qE

Where;

F is the applied force in Newton

E is the electric field potential N/C

q is charge in C

Given;

Radius, r = 1.362 μm = 1.362 X 10⁻⁶ m

density, ρ = 0.888 g/cm³ = 0.888 X 10³ kg/m³

Electric field potential = 1.92 ✕ 10⁵ N/C

F =mg

mass of the oil drop = density, ρ  X volume of the oil drop

volume of the oil drop (spherical) =  (4/3)πr³ = 1.3333π(1.362 X 10⁻⁶)³

⇒ volume of the oil drop = 10.584 X 10⁻¹⁸ m³

mass of the oil drop = 0.888 X 10³ (kg/m³) X 10.584 X 10⁻¹⁸ (m³)

⇒ mass of the oil drop = 9.399 X 10⁻¹⁵ kg

⇒ F =mg = 9.399 X 10⁻¹⁵ kg X 9.8 = 9.21 X10⁻¹⁴ N

F = qE

q = F/E

q = (9.21 X10⁻¹⁴)/(1.92 ✕ 10⁵) = 4.797 X 10⁻¹⁹ C

In terms of e

1e = 1.6 X10⁻¹⁹ C

=  (4.797 X 10⁻¹⁹ C)/(1.6 X10⁻¹⁹ C) = 3e

Therefore, the charge on the oil drop is 3e

7 0
3 years ago
What are the constituent metals of stainless steel?
bogdanovich [222]

<span>Stainless steel is a metal alloy that made up mainly of carbon and chromium. In combination with low carbon contents, chromium is highly reactive element that imparts remarkable resistance to corrosion and heat.</span>

Moreover, stainless steel is mixed up with sufficient nickel, which is an essential allying element in the series of stainless steel grades. Other components are manganese, molybdenum, silicon, titanium, aluminum, niobium, copper, nitrogen, and sulfur.

 

6 0
3 years ago
150J of heat energy
arsen [322]

Btu/(lb-°F) J/(g-°C i mean this is the correct answer

6 0
2 years ago
Read 2 more answers
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