All categories

3 years ago

¿Qué trabajo realiza una carreta que contiene una de masa 800 lb de cemento si se desplaza 100 m por un suelo nivelado?

Physics

1 answer:

Vikentia [17]3 years ago

6 0

Answer:

Ver respuesta en el procedimiento

Explanation:

En este caso, necesitamos la expresión que nos permite hallar el Trabajo realizado por un objeto a través de una distancia dada. Esa expresión es:

W = F * x (1)

Donde:

W: Trabajo en Joules (J)

F: Fuerza en Newton (N)

x: Distancia o radio en metros (m)

En este caso, solo tenemos una masa de 800 lb, y la distancia que se esta recorriendo. No tenemos ningún dato adicional que nos diga si por ejemplo, es empujado por una persona o parecido, solo los datos de la masa que lleva la carreta. Por lo tanto, la única fuerza que podemos asumir se está ejerciendo, sería la fuerza del peso del objeto que se calcula así:

P = mg (2)

Donde:

g: aceleración de gravedad (9,8 m/s²)

Ahora bien si reemplazamos (2) en (1) nos queda:

W = mgx (3)

Y antes de realizar el cálculo debemos convertir la unidad de masa en libra a kilogramo, usando la siguiente conversión: 1 lb -----> 0,454 kg por tanto:

m = 800 * 0,454 = 363.2 kg

Ahora podemos usar (3) y calcular el trabajo:

W = 363,2 * 9,8 * 100

W = 355936 J o simplemente 355,936 kJ

Espero te sirva.

PD: También puedes usar el valor de gravedad como 10 m/s² y tendrás un resultado de 363200 J o 363,2 kJ.

You might be interested in

A pure substance has a/an

Evgesh-ka [11]

Answer:

Element or a compound.

Explanation:

6 0

4 years ago

In an oscillating series RLCcircuit, with a 6.13 Ω resistor and a 15.2 H inductor, find the time required for the maximum energy

ivanzaharov [21]

Answer:

time for maximum energy is 1.718 sec

Explanation:

Maximum energy on the capacitor is given as $=\frac{q_{max}^2}{ 2C}$

Maximum energy $= \frac{(1/6) Q_2}{2C}$

The maximum charge is given by

$q_{max} =Qe^{-Rt/2L}$

Or $\frac{q_{max}}{Q} = e^{-Rt/2L}$

Or $ln\frac{q_{max}}{Q} = \frac{-Rt}{2L}$

solving for t

$t = \frac{2L}{R}\frac{1}{2} ln( 2)$

Putting all value to get desired value

$t = ln(2)\times \frac{L}{R}$

$t = 0.693\times \frac{(15.2}{6.13} = 1.71 sec$

8 0

3 years ago

A long, thin straight wire with linear charge density ? runs down the center of a thin, hollow metal cylinder of radius R. The c

MaRussiya [10]

Answer:

a. E = $\frac{\lambda }{(2\epsilon * r) }$

b. $E = \frac{3(lambda)}{2\pi*\epsilon(outside) }$

Explanation:

The important thing to remember is to use Gauss Law. This is a relation that describes the distribution of electric charge to the resultant electric field.

Linear charge density means charge per unit length of material.

Data:

The metal cylinder is hollow.

The unit length is L.

a.The expression will be as follows:

for charge inside the cylinder, where r < R, the expression is:

E = $\frac{\lambda }{(2\epsilon * r) }$

b. Let's assume that the cylinder is a coaxial cylinder with a radius r > R, then the electrical field strength is given as:

$E = \frac{Q ( enclosed)}{A*\epsilon }$

E = $\frac{\lambda*L+2(\lambda)*L }{(A)*\epsilon(outside) }$

E = $\frac{3(\lambda)L }{(2\pi*R*L*\epsilon(outside) }$

This gives:

$E = \frac{3(lambda)}{2\pi*\epsilon(outside) }$

The solution informs us that there is a surface change taking place on the cylinder. Therefore, there will not be a magnetic field across it.

3 0

4 years ago

A 2 N and an 8 N force pull on an object to the right and a 4 N force pulls on the same object to the left. If the object has a

Elden [556K]

Answer:

a = 12 [m/s²]

Explanation:

To solve this problem we must use Newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

ΣF = m*a

where:

ΣF = sum of forces acting on a body [N] (units of Newtons)

m = mass = 0.5 [kg]

a = acceleration [m/s²]

Let's take the direction of positive forces to the right and negative forces directed to the left

2 + 8 - 4 = 0.5*a

6 = 0.5*a

a = 12 [m/s²]

7 0

3 years ago

A 15.0 kilogram cart initially traveling at 2.0 meters per second east accelerates uniformly at 0.75 meter per second squared ea

Montano1993 [528]

Answer:

Explanation:

Since this is not parabolic motion, it is one-dimensional motion. Very simple. What we are given is

mass: 15.0 kg

initial velocity: 2.0 m/s

acceleration: .75 m/s/s

time: 6.0 seconds

Since we are looking for final velocity, the equation we need for this is

v = v0 + at that says final velocity is equal to the initial velocity plus the acceleration of the object times how long it travels. We don't have a need for the mass here at all.

$v=2.0\frac{m}{s}+.75\frac{m}{s^2}(6.0s)$

Notice that one of the seconds labels to the right of the plus sign cancel out, leaving us with like units...which we HAVE to have if we want to add.

Simplifying a bit gives us

v = 2.0 m/s + 4.5 m/s so

v = 6.5 m/s

4 0

3 years ago