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irina [24]
3 years ago
12

Quanto vale a resultante de duas forças de mesmo módulo mesma direção e sentidos opostos atuando sobre um corpo de massa igual a

2 kg qual é o valor da aceleração desse corpo nessas condições
Physics
2 answers:
Crank3 years ago
6 0
A resultante das duas forças será zero, já q os sentidos são opostos e sãos iguais em módulo, elas se anulam. Logo, se a força resultante é zero, e F=ma, aceleração também será igual a zero.
Anton [14]3 years ago
6 0

Answer:

This question is in Portuguese, this means:

"How much is the result of two forces of the same module in the same direction and opposite ways acting on a body with a mass equal to 2 kg? what is the value of the acceleration of that body in these conditions?"

Now, i will answer in English.

When we are doing math with forces, we can use "Linear overlay", this means that we can add an subtract the forces between them.

Now, if you apply two forces of the same magnitude and same direction but in opposite ways (one is pointing in x positive and the other in x negative) we can directly add them:

F*(+x) + F(-x) = F(+x) - F(+x) = 0

Where (+x) and (-x) refer to the direction at which the forces are applied,

and it has sense, if you pull with the same force by opposite sides, the total force that the object suffers is zero.

And because the total force that the object suffers is zero, by the second Newton's equation f*m = a, the acceleration that the object suffers is also zero.

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Susan plans to grow cabbages and broccoli in her vegetable garden. Cabbages and broccoli grow best in clay soil. Which of these
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The correct option is B.
As we are told in the question, the cabbage and broccoli grow best in clay soil. From the options we are given, it can be seen that it is only option B that has the highest amount of clay soil. The soil that will be used should be one that has the highest content of clay.
7 0
3 years ago
Of the following transitions in the Bohr hydrogen atom, the ________ transition results in the absorption of the highest-energy
8_murik_8 [283]

Answer:

The <em><u>n = 2 → n = 3</u></em> transition results in the absorption of the highest-energy photon.

Explanation:

E_n=-13.6\times \frac{Z^2}{n^2}ev

Formula used for the radius of the n^{th} orbit will be,

where,

E_n = energy of n^{th} orbit

n = number of orbit

Z = atomic number

Here: Z = 1 (hydrogen atom)

Energy of the first orbit in H atom .

E_1=-13.6\times \frac{Z^2}{1^2} eV=-13.6 eV

Energy of the second orbit in H atom .

E_2=-13.6\times \frac{Z^2}{(2)^2} eV=-3.40 eV

Energy of the third orbit in H atom .

E_3=-13.6\times \frac{Z^2}{(9)^2} eV=-1.51 eV

Energy of the fifth orbit in H atom .

E_5=-13.6\times \frac{Z^2}{(2)^2} eV=-0.544 eV

Energy of the sixth orbit in H atom .

E_6=-13.6\times \frac{Z^2}{(2)^2} eV=-0.378 eV

Energy of the seventh orbit in H atom .

E_7=-13.6\times \frac{Z^2}{(2)^2} eV=-278 eV

During an absorption of energy electron jumps from lower state to higher state.So,  absorption will take place in :

1) n = 2 → n = 3

2) n=  5 → n = 6

Energy absorbed when: n = 2 → n = 3

E=E_3-E_2

E=(-1.51 eV) -(-3.40 eV)=1.89 eV

Energy absorbed when: n = 5 → n = 6

E'=E_6-E_5

E'=(-0.378 eV)-(-0.544 eV) =0.166 eV

1.89 eV > 0.166 eV

E> E'

So,the n = 2 → n = 3 transition results in the absorption of the highest-energy photon.

4 0
3 years ago
How does the structure of mitochondrial on affect its structure
anyanavicka [17]
It is itself. This question does not make sense.
7 0
3 years ago
What does the REVOLUTION of Earth around the Sun bring us and how long does it take?
olasank [31]

Answer:

takes 365 days and it bring us seasons such as, spring,winter and fall.

7 0
3 years ago
An amusement park ride consists of a rotating circular platform 8.26 m in diameter from which 10 kg seats are suspended at the e
VashaNatasha [74]

To solve this problem we will begin by finding the necessary and effective distances that act as components of the centripetal and gravity Forces. Later using the same relationships we will find the speed of the body. The second part of the problem will use the equations previously found to find the tension.

PART A) We will begin by finding the two net distances.

r = \frac{8.26}{2} = 4.13m

And the distance 'd' is

d = lsin\theta

d = 1.14 sin 16.2\°

d = 0.318m

Through the free-body diagram the tension components are given by

Tcos\theta = mg

Tsin\theta = \frac{mv^2}{R}

Here we can watch that,

R = r+d

Dividing both expression we have that,

tan\theta = \frac{v^2}{Rg}

Replacing the values,

tan(16.2) = \frac{v^2}{(4.13+0.318)(9.8)}

v = 4.83371m/s

PART B) Using the vertical component we can find the tension,

Tcos\theta = mg

T = \frac{mg}{cos\theta}

T = \frac{(10+26.2)(9.8)}{cos(16.2)}

T = 369.42N

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3 years ago
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