I punch myself really hard. Does that make me strong or weak?

What does soil structure refer to?

Zigmanuir [339]

A this refers to the make up of the soil

6 0

3 years ago

Read 2 more answers

Blank is the change in position of an object

mamaluj [8]

Displacement is your answer :)

6 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

A long, thin solenoid has 390 turns per meter and a radius of 1.20 cm. The current in the solenoid is increasing at a uniform ra

gtnhenbr [62]

To solve this problem it is necessary to apply the concepts related to Faraday's law and the induced emf.

By definition the induced electromotive force is defined as

$\int E dl = -\frac{d\phi}{dt}$

$\int E dl = -(\frac{dB}{dt})A$

Where,

$\phi =$ Electric field

B = Magnetic Field

A = Area

At the theory the magnetic field is defined as,

$B = \mu_0 NI$

Where,

N = Number of loops

I = current

$\mu_0 =$ Permeability constant

We know also that the cross sectional area, is the area from a circle, and the length is equal to the perimeter then

A = \pi r^2

l = 2\pi r

Replacing at the previous equation we have that

$E (2\pi r) = \mu_0 n (\frac{di}{dt})(\pi R^2)$

Where,

R = Radius of the solenoid

r = The distance from the axis

Re-arrange to find the current in function of time,

$\frac{di}{dt} = \frac{Er}{\mu_0 NR^2}$

Replacing our values we have

$\frac{di}{dt} = \frac{(8.00*10^{-6})(0.0348)}{(4\pi*10^{-7})(390)(1.2*10^-2)^2}$

$\frac{di}{dt} = 3.94487A/s$

8 0

3 years ago

Como anticipan la caída del proyectil, con movimiento parabólico para que el misil no haga daño

Elena L [17]

Answer:

Conociendo la velocidad inicial del proyectil y el angulo de lanzamiento con respecto ala horizontal.

Explanation:

Para poder anticipar la caída del proyectil es importante conocer la velocidad inicial del proyectil y el angulo de disparo del proyectil con respecto a la horizontal.

A continuación se presenta un diagrama o esquema donde se pueden ver estas variables y se explicaran a la brevedad:

Para poder encontrar el rango que es la máxima distancia horizontal recorrida por el proyectil debemos utilizar la siguiente ecuación:

$x=(v_{o})_{x} *t\\where:\\(v_{o})_{x} = velocidad inicial x-component [m/s]\\t= time [s]$

Para poder encontrar el tiempo debemos utilizar la siguiente ecuación:

$y=(v_{y} )_{o}*t-0.5*g*t^{2} \\donde:\\(v_{y} )_{o}= velocidad inicial componente y [m/s]\\g = gravity = 9.81 [m/s^2]\\t = time [s]$

En la anterior ecuación, igualamos y = 0, ya que cuando el proyectil cae al suelo la distancia vertical es cero. De esta manera podemos encontrar el tiempo t, ya que conocemos la velocidad inicial del proyectil en la componente y.

Seguidamente reemplazamos t en la primera ecuacion y encontramos la distancia x o el rango.

8 0

3 years ago