What kind of device can you use to to separate visible light into its different colors

Un cuerpo de m=0,5 Kg se desplaza horizontalmente con v=4m/s y luego de un lapso de tiempo se mueve con v=20 m/s. ¿cual ha sido

Lilit [14]

Responder:

<h2>64 Julios </h2>

Explicación:

La energía cinética se expresa mediante la fórmula KE = 1 / 2mv² donde;

m es la masa del cuerpo

v es la velocidad del objeto

Dado que el cuerpo se mueve horizontalmente con v = 4 m / sy después de un período de tiempo se mueve con v = 20 m / s, entonces la variación en la velocidad será de 20 m / s - 4 m / s = 16 m / s.

Parámetros dados

masa del objeto m = 0,5 kg

Variación de velocidad = 16 m / s

Variación de la energía cinética = 1/2 * 0,5 * 16²

Variación de la energía cinética = 1/2 * 0,5 * 256

Variación de la energía cinética = 0,5 * 128

<em>Variación de la energía cinética = 64 Julios</em>

7 0

3 years ago

A force that causes an object to change its motion is called?

GarryVolchara [31]

Unbalanced force.

The definition of an unbalanced force is a force that changes the position, speed or direction of the object to which it is applied.

When an object is still it means that is is balanced force. So you would need to add more force to another side to make it move. Which is unbalanced force.

8 0

3 years ago

A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Co

Tems11 [23]

(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

<h3>Center mass of the airplane</h3>

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>

The wheels under the wind do not pass through the center line.
The position of the front wheel is constant and it is zero mark (origin).
The rear wheels are at 21.7 m mark

Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

<h3>Mass of the plane at the position of the rear wheels</h3>

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

$X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}$

$18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg$

<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

$W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN$

<h3>Mass of the plane at the position of the front wheel</h3>

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

<h3>Force exerted by the ground on the front wheel</h3>

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

Learn more about center mass here: brainly.com/question/13499822

7 0

2 years ago

Which of these is not a part of the<br> cardiovascular system?

lilavasa [31]

Answer:I don't see any

Explanation:

6 0

3 years ago

Read 2 more answers

The pressure on a fluid at rest in a pipe increases by 20 Pa. How does this change in pressure affect the pressure on the fluid

timofeeve [1]

Answer:The change in pressure can affect the pressure on the fluid through the radius and diameter of the pipe.

r^² x Pressure (pa).

Therefore the narrower the other part of the pile, the greater the pressure on the fluid at such part, the wider in other part the lesser the pressure on the fluid at this part.

Explanation:

4 0

3 years ago