PLEASE HELP ME plsss answer this

1. Una carga Q1 = + 12 μC se coloca a una distancia r = 0.024 m desde una carga Q2 = + 16 μC. a) Determina la magnitud de la fue

lyudmila [28]

Answer:

1. a. 3,000 N

b. Repulsión

2. 46.875 × 10⁶ N/C

3. a. 81,000 J

b. 6.75 × 10⁹ V

Explanation:

1. Los parámetros dados son;

Q₁ = +12 μC, Q₂ = +16 μC

La distancia entre las cargas, r = 0.024

La magnitud de la fuerza electrostática, F, entre cargas se da como sigue;

$F = k \times \dfrac{Q_1 \cdot Q_2}{r^2}$

Donde, k = constante de Coulomb = 9.0 × 10⁹ N · m² / C²

Por lo tanto, obtenemos;

F = 9.0 × 10⁹ × 12 × 10⁻⁶ × 16 × 10⁻⁶ / 0.024² = 3.000

La magnitud de la fuerza electrostática, entre las cargas, F = 3000 N

(b) Dado que tanto Q₁ como Q₂ son cargas positivas, y las cargas iguales se repelen entre sí, la fuerza es la repulsión.

2) La intensidad de un campo eléctrico, E, se da como sigue;

$E = \dfrac{k \cdot Q}{r^2}$

La magnitud de la carga, Q = 24 μC

La distancia donde se mide el campo, r = 48 mm = 0.048 m

Por lo tanto, E = 9.0 × 10⁹ × 12 × 10⁻⁶ / 0.048² = 46,875,000 N / C

La intensidad de un campo eléctrico, E = 46,875,000 N / C = 46.875 × 10⁶ N / C

3. La magnitud de las cargas son;

Q₁ = 24 mC

Q₂ = -12 μC

La distancia entre las cargas, r = 0.032 m

un. El potencial eléctrico de una carga, $U_E$ , se da de la siguiente manera;

$U_E = k \times \dfrac{Q_1 \cdot Q_2}{r}$

Por lo tanto;

$U_E$ = 9.0×10⁹ × 24 × 10⁻³ × (-12) × 10⁻⁶ /0.032 = -81,000

La energía potencial eléctrica entre la carga, Q₁ y Q₂= -81,000 J

b. El potencial eléctrico de Q₁ en Q₂, V₁ = $k \times \dfrac{Q_1 }{r}$

Por lo tanto, V₁ = 9.0×10⁹ × 24 × 10⁻³/0.032 = 6.75 × 10⁹

El potencial eléctrico de Q₁ en Q₂, V₁ = 6.75 × 10⁹ V

3 0

3 years ago

Relative motion can best be defined as

creativ13 [48]

Relative motion is the motion of an object relative to another object,
or as it would appear to an observer stuck to the other object and
riding on it.

6 0

3 years ago

With light microscopy, if the objective lens (lens closest to the specimen) magnifies 40-fold, and the eyepiece lens magnifies 1

Fed [463]

The final magnification will be 400-fold or 400 times the original size of the object.

For magnifying smaller objects, a compound microscope is used.

A compound microscope consists of an objective and an eyepiece, whose diagram is shown in the adjoining image.

The lens near the object is called an objective and the other one is the eyepiece.

Let the magnification of the objective be m1

Let the magnification of the eyepiece be m2

The final magnification by the microscope, M, will be

M = m1 x m2

Putting the values in the above equation

M = 40 x 10

M= 400

Thus, the final magnification will be 400-fold or 400 times the original size of the object.

To know more about "optical instruments", refer to the link given below:

brainly.com/question/13276240?referrer=searchResults\

#SPJ4

6 0

2 years ago

A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altidude

Serjik [45]

Answer:

a) 252 ft/s

b) 1076 ft

Explanation:

The equation for motion for uniform acceleration (we can use it because the rocket is affected only by gravity) is as follows:

Y(t) = Y0 + V0 * t + 1/2 * a * t^2

Where

Y(t): altitude at a given time

Y0: initial altitude

V0: initial speed

a: acceleration, in this case -32.2 ft/s^2 (negative because gravity points down)

We set a 1 dimensional coordinate system with Y pointing up and the origin of coordinates at ground level.

We consider t=0 as the moment where powered flight ended (motor ran oou of fuel), at this moment the altitude was

Y(0) = 89.6 ft

Therefore:

Y0 = 89.6 ft

We also know that the rocket fell to ground 16 seconds later, therefore

Y(16) = 0 ft

So we can write

Y(t=16) = Y0 + V0 * t + 1/2 * a * t^2

V0 * t = Y(t=16) - Y0 - 1/2 * a * t^2

V0 =( Y(t=16) - Y0 - 1/2 * a * t^2 )/t

V0 =( 0 - 89.6 - 1/2 * (-32.2) * 16^2 )/16 = 252 ft/s

In the highest point of flight the rocket will have a speed = 0

The first derivative of the equation of motion is the equation of speed:

V(t) = V0 + a * t

If we equate this to zero we eill find the time at which the rocket achieved it's highest altitude.

0 = V0 + a * t

a * t = 0 - V0

t = -V0/a

t = -252/(-32.2) = 7.83 s

Now, we can take this time value andd plug it back into the position equation

Y(7.83) = 89.6 + 252 * 7.83 + 1/2 * (-32.2) * 7.83^2 = 1076 ft

6 0

3 years ago

True or false: Everything in the universe works with harmony because of balanced energy

Eduardwww [97]

Answer:

Explanation:

True

7 0

3 years ago

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