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kirill [66]
3 years ago
12

Complete the following sentences. a. The motion of the particles is a model of__________. . b. The movement of particles from on

e chamber to another is a model of________. . c. When the average motion of the particles is the same in both chambers, this is a model of _______.
Physics
1 answer:
Leno4ka [110]3 years ago
8 0
Effectiv was the morning time to go home
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An iron ball and an aluminum ball of mass 100 g each are heated to the same temperature and then cooled to a temperature of 20°
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Becaused it the thing was cod that why
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Superman is flying 54.5 m/s when he sees
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348.34 m/s. When Superman reaches the train, his final velocity will be 348.34 m/s.

To solve this problem, we are going to use the kinematics equations for constant aceleration. The key for this problem are the equations d=v_{0} t+\frac{at^{2} }{2} and v_{f} =v_{0} +at where d is distance, v_{0} is the initial velocity, v_{f} is the final velocity, t is time, and a is aceleration.

Superman's initial velocity is v_{0}=54.5\frac{m}{s}, and he will have to cover a distance d = 850m in a time t = 4.22s. Since we know d, v_{0} and t, we have to find the aceleration a in order to find v_{f}.

From the equation d=v_{0} t+\frac{at^{2} }{2} we have to clear a, getting the equation as follows: a=\frac{2(d-v_{0}t) }{t^{2} }.

Substituting the values:

a=\frac{2(850m-54.5\frac{m}{s}.4.22s) }{(4.22s)^{2}}=69.63\frac{m}{s^{2}}

To find v_{f} we use the equation v_{f} =v_{0} +at.

Substituting the values:

v_{f} =54.5\frac{m}{s} +(69.63\frac{m}{s^{2}}.4.22s)=348.34\frac{m}{s}

5 0
3 years ago
A concave lens can only form a. A. real image. B. reversed image. C. virtual image. D. magnified image.
Umnica [9.8K]
A concave lens can only form a virtual image. The correct option among all the options that are given in the question is the third option or option "C". Concave lenses are mostly thinner in the middle compared to its edges. I hope that this answer has come to your help.
4 0
3 years ago
Read 2 more answers
What is amperage?
Shtirlitz [24]

question one b

question 2 i think a

3  d

4 c

5 not sure but wanting to say d

6  letter b

7 not sure

8 idk

9 i have no idea

3 0
3 years ago
Calculate the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition from =7 to =1.
Harrizon [31]

1.549×10-19lJ is the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition from =7 to =1.

The equation E= hcE =hc, where h is Planck's constant and c is the speed of light, describes the inverse relationship between a photon's energy (E) and the wavelength of light ().

The Rydberg formula is used to determine the energy change.

Rydberg's original formula used wavelengths, but we may rewrite it using units of energy instead. The result is the following.

aaΔE=R(1n2f−1n2i) aa

were

2.17810-18lJ is the Rydberg constant.

The initial and ultimate energy levels are ni and nf.

As a change of pace from

n=5 to n=3 gives us

ΔE

=2.178×10-18lJ (132−152)

=2.178×10-18lJ (19−125)

=2.178×10-18lJ×25 - 9/25×9

=2.178×10-18lJ×16/225

=1.549×10-19lJ

Learn more about Rydberg formula here-

brainly.com/question/13185515

#SPJ4

8 0
2 years ago
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