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Likurg_2 [28]
3 years ago
10

Boss tweed encourages his associates to work "like a well-oiled machine", thus providing his workers with

Physics
2 answers:
Arisa [49]3 years ago
6 0
<span>Boss tweed encourage his associates to work like a well oiled machine thus providing his workers with benefits that encourages them to work like a well oiled machine. A well oiled machine means that it does not stop, its work is continuous and efficient.</span>
Mademuasel [1]3 years ago
4 0

Answer:

A. an analogy

Explanation:

Page 90 of 150 PF Downloaded Lesson:

Analogies are representations that help in understanding a problem, a

set of circumstances, or even an unfamiliar object. In using an

analogy, you might compare something you’re familiar with to

something you’re less certain about in an attempt to understand the

unfamiliar thing. Or you may form an analogy to make a concept clear

to someone else. For example, to give employees an idea of what’s

expected of them, a supervisor might refer to a good department as “a

hive of busy bees.” To help the people you supervise get an idea of

the mental attitude you would prefer they adopt, you could refer to a

clear, sparkling mountain stream as an analogy of an open, flowing

mind—nothing hidden at the bottom, always moving toward new

destinations and new ideas.

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Speed is velocity in a given direction<br> O True<br> O False
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Answer:

FALSE

Explanation:

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3 years ago
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ANTONII [103]
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What is the meaning of the reference point in electric potential?.
mrs_skeptik [129]

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2 years ago
The Franck-Hertz experiment involved shooting electrons into a low-density gas of mercury atoms and observing discrete amounts o
Anuta_ua [19.1K]

Answer:

the final kinetic energy is 0.9eV

Explanation:

To find the kinetic energy of the electron just after the collision with hydrogen atoms you take into account that the energy of the electron in the hydrogen atoms are given by the expression:

E_n=\frac{-13.6eV}{n^2}

you can assume that the shot electron excites the electron of the hydrogen atom to the first excited state, that is

E_{n_2-n_1}=-13.6eV[\frac{1}{n_2^2}-\frac{1}{n_1^2}]\\\\E_{2-1}=-13.6eV[\frac{1}{2^2}-\frac{1}{1}]=-10.2eV

-10.2eV is the energy that the shot electron losses in the excitation of the electron of the hydrogen atom. Hence, the final kinetic energy of the shot electron after it has given -10.2eV of its energy is:

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Explanation:

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