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7nadin3 [17]
4 years ago
9

What are the four most common gases in dry air?

Physics
1 answer:
dimaraw [331]4 years ago
7 0
Nitrogen
oxygen
argon
carbon dioxide
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What type of contraction requires the greatest amount of tension?
mars1129 [50]

Answer:

Isotonic eccentric

Explanation:

First of all, it should be noted that the term contraction has the meaning of joining or shortening. In the field of bodybuilding and workouts with loads, we can define muscle contraction as what happens whenever muscle fibers generate a tension in themselves.

This tension situation occurs in several situations, including when the muscle is shortened, elongated, moving, maintaining the same length or static.

There are different types of muscle contraction, among them we can highlight isotonic contraction, which is divided in turn into concentric and eccentric, isometric, auxotonic and isokinetic contraction.

 Isotonic contraction is the most common type of contraction that occurs in most sports or physical activities that we perform in our day to day. Normally the muscular tensions that we exert are usually accompanied by a shortening and lengthening of the muscle fibers of a muscle. In turn, the isotonic contraction is divided into two, concentric and eccentric.

Concentric concentration: is that which happens when a muscle makes a tension capable of overcoming a resistance, producing a shortening and subsequent mobilization of a part of the body overcoming said resistance. For example, when we take a spoon and take it to our mouths to eat, a concentric shortening occurs. Putting examples in the gym on the bench press the movement of raising the bar, is equivalent to the concentric phase.

Eccentric concentration: we can say that it is one in which, given a resistance, we exert a greater tension with the muscle, so that said muscle is lengthened. In the case of the bench press, the eccentric phase is when we lower the bar to the chest.

8 0
3 years ago
Two point charges exert a 7.35 N force on each other. What will the force become if the distance between them is increased by a
I am Lyosha [343]

Answer :

New force becomes, F' = 1.83 N

Explanation:

Let two point charges exert a force of 7.35 N force on each other. The electric force between two charges is given by :

F=\dfrac{kq_1q_2}{r^2}

q_1\ and\ q_2 are charges

r is the distance between charges if the distance between them is increased by a factor of 2, r' = 2r

New force is given by :

F'=\dfrac{kq^2}{r'^2}

F'=\dfrac{kq^2}{(2r)^2}

F'=\dfrac{1}{4}\dfrac{kq^2}{r^2}

F'=\dfrac{1}{4}\times 7.35

F' = 1.83 N

So, the new force between charges will be 1.83 N. Therefore, this is the required solution.          

3 0
3 years ago
Helppppppppppppppppppppppppppppp
Lady_Fox [76]

Answer:

<em>Option #3</em> is your answer.

Explanation:

8 0
3 years ago
What type of landform is depicted here? a. a mountain b. a depression c. a valley​
Harrizon [31]

Answer:

b.mountain the correct answer

Explanation:

what type of ladform ids depicted here? a.a b. mountain b. a depression c. avaley

6 0
3 years ago
A mass moves back and forth in simple harmonic motion with amplitude A and period T.
Sever21 [200]

a. 0.5 T

- The amplitude A of a simple harmonic motion is the maximum displacement of the system with respect to the equilibrium position

- The period T is the time the system takes to complete one oscillation

During a full time period T, the mass on the spring oscillates back and forth, returning to its original position. This means that the total distance covered by the mass during a period T is 4 times the amplitude (4A), because the amplitude is just half the distance between the maximum and the minimum position, and during a time period the mass goes from the maximum to the minimum, and then back to the maximum.

So, the time t that the mass takes to move through a distance of 2 A can be found by using the proportion

1 T : 4 A = t : 2 A

and solving for t we find

t=\frac{(1T)(2 A)}{4A}=0.5 T

b. 1.25T

Now we want to know the time t that the mass takes to move through a total distance of 5 A. SInce we know that

- the mass takes a time of 1 T to cover a distance of 4A

we can set the following proportion:

1 T : 4 A = t : 5 A

And by solving for t, we find

t=\frac{(1T)(5 A)}{4A}=\frac{5}{4} T=1.25 T

6 0
3 years ago
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