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3 years ago

The five factors of production include land, _, capital, entrepreneurship, and _.

1 answer:

6 0

<span>The five factors of production include land, labor, capital, entrepreneurship, and knowledge. Land is considered as the primary because these are the pre-existent resources readily available from nature however, nowadays, these are already bought and passed down from generations. Capital is the money used for the production which needs to be returned when one invests on something. Labor is the person’s inputs which is very valuable. Entrepreneurship is how the person transact according to how the people negotiate. And knowledge is one of human’s invaluable efforts of giving the ideas.</span>

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A 90kg woman and a 60kg boy are standing at rest on a frictionless frozen lake. The boy pushes the woman with a 40N horizontal f

uranmaximum [27]

Answer:

The acceleration of the woman is 0.44 m/s²

Explanation:

Given;

mass of the woman, m₁ = 90 kg

mass of the boy, m₂ = 60 kg

The force applied by the boy, f₂ = 40 N

The net horizontal force on the woman = 40 N

Apply Newton's second law of motion to determine the acceleration of the woman;

f = ma

a = f / m

a = 40 / 90

a = 0.44 m/s²

Therefore, the acceleration of the woman is 0.44 m/s²

6 0

3 years ago

A potter spins his wheel at 0.98 rev/s. The wheel has a mass of 4.2 kg and a radius of 0.35 m. He drops a chunk of clay of 2.9 k

Bad White [126]

Answer:

$v_{f,w} = 1.791\,\frac{m}{s}$ , $v_{f,c} = 0.972\,\frac{m}{s}$

Explanation:

The situation can be modelled by applying the Principle of Angular Momentum Conservation:

$I_{w} \cdot \omega_{o} = (I_{w} + I_{c})\cdot \omega_{f}$

The final angular speed is:

$\omega_{f} = \frac{I_{w}}{I_{w}+I_{c}}\cdot \omega_{o}$

$\omega_{f} = \left(\frac{\frac{1}{2}\cdot (4.2\,kg)\cdot (0.35\,m)^{2} }{\frac{1}{2}\cdot (4.2\,kg)\cdot (0.35\,m)^{2} + \frac{1}{2}\cdot (2.9\,kg)\cdot (0.19\,m)^{2}}\right)\cdot (0.98\,\frac{rev}{s} )\cdot \left(\frac{2\pi\,rad}{1\,rev} \right)$

$\omega_{f} \approx 5.116\,\frac{rad}{s}$

The tangential velocities of the wheel and the clay are, respectively:

$v_{f, w} = (0.35\,m)\cdot (5.116\,\frac{rad}{s} )$

$v_{f,w} = 1.791\,\frac{m}{s}$

$v_{f, c} = (0.19\,m) \cdot (5.116\,\frac{rad}{s} )$

$v_{f,c} = 0.972\,\frac{m}{s}$

5 0

3 years ago

An astronaut circling the earth at an altitude of 400 km is horrified to discover that a cloud of space debris is moving in the

elena-14-01-66 [18.8K]

One of the essential concepts to solve this problem is the utilization of the equations of centripetal and gravitational force.

From them it will be possible to find the speed of the body with which the estimated time can be calculated through the kinematic equations of motion. At the same time for the calculation of this speed it is necessary to clarify that this will remain twice the ship, because as we know by relativity, when moving in the same magnitude but in the opposite direction, with respect to the ship the debris will be double speed.

By equilibrium the centrifugal force and the gravitational force are equal therefore

$F_c = F_g$

$\frac{mv^2_{orbit}}{r} = \frac{GMm}{r^2}$

Where

m = mass spacecraft

v = velocity

G = Gravitational Universal Constant

M = Mass of earth

$r \rightarrow R+h \Rightarrow$ Radius of earth and orbit

Re-arrange to find the velocity

$\frac{mv^2_{orbit}}{r} = \frac{GMm}{r^2}$

$\frac{v^2_{orbit}}{r} = \frac{GM}{r^2}$

$v^2_{orbit}=\frac{GM}{r}$

$v_{orbit} = \sqrt{\frac{GM}{r}}$

$v_{orbit} = \sqrt{\frac{GM}{R+h}}$

Replacing with our values we have

$v_{orbit} = \sqrt{\frac{(6.67*10^{-11})(5.98*10^{24})}{6.37*10^6+0.4*10^6}}$

$v_{orbit} = 7676m/s$

From the cinematic equations of motion we have to

$t = \frac{d}{2v_{orbit}} \rightarrow$ Remember that the speed is double for the counter-direction of the trajectories.

Replacing

$t = \frac{29000m}{7676m/s}$

$t = 3.778s$

Therefore the time required is 3.778s

4 0

3 years ago

Tap on the photo. For each diagram, explain why the light behaves in the way that it does.

dem82 [27]

Answer:

Diagram 1, 3 and 4 can be explained with the phenomenon of refraction.

Refraction occurs when a ray of light crosses the interface between two mediums with different optical density: when this occurs, the ray of light is bent and its speed changes, according to Snell's law

$n_1 sin \theta_1 = n_2 sin \theta_2$

where $n_1,n_2$ are the refractive index of the 1st and 2nd medium

$\theta_1, \theta_2$ are the angle that the incident ray and the refracted ray makes with the normal to the interface

In diagram, 1, the ray of light arrives perpendicularly to the interface, so it is refracted through the medium but it doesn't change its direction (only its speed).

In diagram 3, the ray of light is refracted twice: at the 1st interface and at the 2nd interface. In the 1st case, it goes from a medium with lower refractive index to a medium with higher refractive index ( $n_1$ ), this means that $\theta_2$ , so the ray bends towards the normal. Vice-versa, in the 2nd case the ray goes from a medium with higher refractive index to a medium with lower refractive index ( $n_1>n_2$ ), so it bends away from the normal ( $\theta_2>\theta_1$ ).

In diagram 4, the ray of light is also refracted twice. The ray of light here acts exactly the same as in diagram 3, h

However, this time the 2nd interface is the opposite direction with respect to diagram 3, so in this case the ray of light at the 2nd interface bends in the opposite direction (still away from the normal).

Diagram 2 instead is an example of reflection, that occurs when a ray of light bounces off the interface between the two mediums, withouth entering the 2nd medium.

According to the law of reflection:

- The incoming ray, the reflected ray and the normal to the boundary are all in the same plane

- The angle of incidence is equal to the angle of reflection (both are measured relative to the normal to the boundary)

Therefore in this diagram, the ray of light hits the boundary at approx. 45 degrees from the normal, and then it is reflected back approximately at 45 degrees on the other side with respect to the normal.

3 0

4 years ago

Based on what you have just read, what are the advantages of storing music digitally compared to using an analog system?

melamori03 [73]

Storing music digitally requires less storage room than analog records or tapes. Digital music is easier to copy and the copies are the same as the original. The quality of the signal does not degrade over long periods of time.

7 0

3 years ago

Read 2 more answers

The five factors of production include land, _______, capital, entrepreneurship, and _______.

The five factors of production include land, _, capital, entrepreneurship, and _.