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3 years ago

9

determine the longest wavelength of light required to remove an electron from a sample of potassium metal, if the binding energy

for an electron in k is 1.76 × 103 kj/mol.

1 answer:

9966 [12]3 years ago

5 0

Answer:

The wavelength of light is 68 nm

Explanation:

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Which of the following processes DOES NOT occur during the convective transfer of heat within a container of air? Group of answe

vovangra [49]

The volume of a warmed part of the air is reduced and its density increases.

Explanation:

In a convective form of heat transfer, the volume of a warmed part of air is not reduced and its density does not increase.

During convection, heat causes the warm part of the air to expand and its volume increases. When volume increases, density is reduced.

Convection is a form of heat transfer that involves the actual movement of particles of the medium.
It usually occurs in fluids i.e gases and liquids.
In convection, the cold part exerts a buoyant force on the warmer air below and causes it to rise.
As the warmer part is rising the cooler part replaces it and a convective cell is formed in the process.

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3 years ago

The vertical height of an ocean wave from a crest to a trough is 10 meters. What is the amplitude of the ocean wave?

natima [27]

The amplitude of the wave would be 5 meters.

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PtichkaEL [24]

15) a
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3 years ago

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if vector u has lenght 70 and direction 40 degrees, and vector v has length 85 and direction 335 degrees what is the length and

Anastaziya [24]

Answer:

Magnitude of resultant = 131.15

Direction of resultant = 3.97°

Explanation:

||u|| = 70

θ = 40°

$\vec{u}_x=||u||cos\theta \\\Rightarrow \vec{u}_x=70cos40=53.62$

$\vec{u}_y=||u||sin\theta \\\Rightarrow \vec{u}_y=70sin40=44.99$

||v|| = 85

θ = 335°

$\vec{v}_x=||v||cos\theta \\\Rightarrow \vec{v}_x=85cos335=77.03$

$\vec{v}_y=||v||sin\theta \\\Rightarrow \vec{v}_y=85sin335=-35.92$

Resultant

$R=\sqrt{R_x^2+R_y^2}\\\Rightarrow R=\sqrt{(\vec{u}_x+\vec{v}_x)^2+(\vec{u}_y+\vec{v}_y)^2}\\\Rightarrow R =\sqrt{(70cos40+85cos335)^2+(70sin40+85sin335)^2}\\\Rightarrow R =131.15$

$\theta=tan^{-1}\frac{R_y}{R_x}\\\Rightarrow \theta=tan^{-1}\frac{70sin40+85sin335}{70cos40+85cos335}\\\Rightarrow \theta=tan^{-1}0.069=3.97^{\circ}$

Magnitude of resultant = 131.15

Direction of resultant = 3.97°

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