Plzzzz help me this is due today. only need help with these two questions and the element is oxygen.

¿Cómo se llama el aparato que convierte la energía cinética en energía eléctrica?

Mumz [18]

Answer:

sorry don't understand the language

6 0

3 years ago

1. horizontal row in the periodic table____ 2. vertical column in the periodic table____ 3. A repetition of properties occurs wh

Anna11 [10]

Answer:

1. Periods

2. Groups of families

3. the statement is true about Periodic law

4. metals

5.transition metals

6.atomic radius

7. cations

8. electrons

9. electronegativity of the atom

10. ionization energy

Explanation:

1. horizontal row in the periodic table____are called Periods

2. vertical column in the periodic table____ are called Groups of families

3. A repetition of properties occurs when elements are arranged in order of increasing atomic number.____ True about Periodic Law (most elements that fall within the same families has similar properties)

4. type of element that is a good conductor of heat and electric current____ are metals that are found in group 1 and 2 of the periodic table Boron, tin, silver, Zinc, Alluminium, Germanium, Pollonium, antimony, e.t.c

5. type of element characterized by the presence of electrons in the dorbital____are called transition metals

6. one-half the distance between the nuclei of two atoms when the atoms are joined____ are called atomic radius

7. type of ion formed by Group 2A elements____are Cations

8. subatomic particles that are transferred to form positive and negative ions____ are called electrons

9. ability of an atom to attract electrons when the atom is in a compound____ is called electronegativity of the atom

10. energy required to remove an electron from an atom is called ionization energy

4 0

4 years ago

The hanging mass is released and allowed to fall. When it has traveled a distance h (that is, the moment just before it hits the

abruzzese [7]

Answer:

Gravitational potential energy to kinetic energy

Explanation:

In this case you have a case about conservation of energy.

When the mass is released and allowed to fall, its energy is completely gravitational potential energy with a value of U = mgh. m is the mass, g is the gravitational constant and h is the height to the floor from the mass.

While the mass is falling down part of its potential energy converts to kinetic energy of value K=1/2mv^2, because the mass has been acquiring more and more velocity.

Thus, the kinetic energy is increasing while the potential energy is decreasing.

When the mass is just above the floor (the moment just before the mass hits the floor) all its potential energy has been converted to kinetic energy.

Then, you have that the kinetic energy of the mass when the mass is just above the floor, is equal to the potential energy when the mass is at height of h. That is:

$mgh=\frac{1}{2}mv^2$

This is how the law of conservation of energy is fulfilled.

5 0

3 years ago

Se debe usar una rampa de 5 m para elevar muebles que pesen 490 N hasta una plataforma a una altura de 2 m. Su propuesta de que

AfilCa [17]

Answer:

F = 180 N

Explanation:

To find the required force you first calculate the angle of the ramp, by using the following relation:

$sin\theta=\frac{2m}{5m}=0.4\\\\\theta=sin^{-1}(0.4)=23.57\°$

Next, you use the Newton second law to know what is the x component (in a rotated coordinate system) of the gravitational force:

$F_x=mgsin\theta$

the required force must be, at least, the last force Fx. You know that the weight of the object is 490N = mg. Hence, you have:

$F=F_x=(450N)sin(23.57\°)=180N$

the required force is 180N

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TRANSLATION:

Para encontrar la fuerza requerida, primero calcule el ángulo de la rampa, utilizando la siguiente relación:

Luego, usa la segunda ley de Newton para saber cuál es el componente x (en un sistema de coordenadas girado) de la fuerza gravitacional:

la fuerza requerida debe ser, al menos, la última fuerza Fx. Sabes que el peso del objeto es 490N = mg. Por lo tanto, tienes:

3 0

3 years ago

Find net force acting on O

klemol [59]

The net force acting on the center O is 4.495 X 10⁹ N

<u>Explanation:</u>

Given:

Charge-

q₁ = 4C

q₂ = 3C

q₃ = -5C

q₄ = 2C

q₅ = 1C

Side of the square = 4cm

Net force acting on the center, F = ?

Length of the side of the square is 4cm

The length of the diagonal will be 4√2

The length of half diagonal = 4√2 / 2 = 2√2

Force acting on the center of the square is:

$F = k({\frac{Qq_1}{r^2} + \frac{Qq_2}{r^2} + \frac{Qq_3}{r^2} + \frac{Qq_4}{r^2} )\\\\\\$

$F = \frac{kQ}{r^2} (q_1 + q_2 + q_3 + q_4)$

Here, k is coulomb's constant and the value is

k = 8.99 x 10⁹ N m2 / C2

On substituting the value, we get:

$F = \frac{8.99 X 10^9 X 1}{(2\sqrt{2} )^2} ( 4C + 3C - 5C + 2C)\\\\F = \frac{8.99 X 10^9}{8} (4C)\\\\F = 4.495 X 10^9 N$

Therefore, the net force acting on the center O is 4.495 X 10⁹ N

8 0

3 years ago