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Black_prince [1.1K]
3 years ago
10

When the tube is filled with mercury vapor, as in this case, a sharp drop in the collected current is observed when the accelera

ting potential reaches 4.9 VV. The current drops suddenly to a negligible value and slowly increases again when the accelerating potential is above 4.9 VV. What is the energy EEE absorbed by the atomic electrons in the mercury atom when the accelerating potential is 4.9 VV
Physics
1 answer:
adell [148]3 years ago
3 0

Answer:

4.9 eV .

Explanation:

It is the case of discharge through mercury tube light . In it , mercury  atoms are exited due to which electrons are sent to higher energy  level . Here current drops to zero because electrons are excited to higher level . Energy are absorbed in quantised manner . Energy absorbed by electrons will be 4.9 eV . That means , difference in energy  between two energy level is  4.9 eV .

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Answer:

78 percent

Explanation:

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A triangular plate with height 6 ft and a base of 7 ft is submerged vertically in water so that the top is 2 ft below the surfac
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Answer:

62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56)  } \, dy = 7875 lb

Explanation:

For this problem to be easier to calculate, we can represent the triangle as a right triangle whose right angle is located at the origin of a coordinate system. (See picture attached).

With this disposition of the triangle, we can start finding our integral. The hydrostatic force can be set as an integral with the following shape:

\int\limits^a_bγhxdy

we know that γ=62.5 lb/ft^{3}

from the drawing, we can determine the height (or depth under the water) of each differential area is given by:

h=8-y

x can be found by getting the equation of the line, which we'll get by finding the slope of the line and using one of the points to complete the equation:

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once we got this slope, we can substitute it in the point-slope form of the equation:

y_{2}-y_{1}=m(x_{2}-x_{1})

which yields:

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\int\limits^6_0{(62.5)(8-y)(-\frac{7}{6}y+7)}\,dy

Now that it's all written in terms of y we can now simplify it, so we get:

62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56)}dy

we can now proceed and evaluate it.

When using the power rule on each of the terms, we get the integral to be:

62.5[\frac{7}{18}y^{3}-\frac{49}{6}y^{2}+56y]^{6}_{0}

By using the fundamental theorem of calculus we get:

62.5[(\frac{7}{18}(6)^{3}-\frac{49}{6}(6)^{2}+56(6))-(\frac{7}{18}(0)^{3}-\frac{49}{6}(0)^{2}+56(0))]

When solving we get:

62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56)  } \, dy = 7875 lb

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Explanation:

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