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3 years ago

Definicion del concepto de numero

Mathematics

2 answers:

NNADVOKAT [17]3 years ago

8 0

Answer:

Mathematical concept that expresses an amount in relation to the computer unit; results from counting the elements that make up a set.

"Cardinal number"

A graphic sign or set of graphic signs that expresses or represents that amount.

Step-by-step explanation:

Gre4nikov [31]3 years ago

4 0

Answer:

En esta investigación, los conceptos numéricos cubren las habilidades de contar y comparar. Ambas habilidades son matemáticas básicas que los niños deben dominar antes de que puedan seguir aprendiendo matemáticas avanzadas. Las operaciones numéricas incluyen las habilidades de suma y resta.

Espero que esto ayude. Todavía estoy trabajando en mi español, así que lo siento si escribí algo mal.

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A worker drags a box of mass 50 kg across a horizontal floor. The worker attaches a rope to the box and pulls on the rope so tha

Arlecino [84]

Answer:

A) $F=\displaystyle{\frac{m(a+ \mu g)}{\cos(\alpha) -\mu \sin(\alpha)} }$ with $a = \frac{2\Delta x}{t^2}$

B) The magnitude of the pulling force is 24 % of the magnitude of the gravitational force acting on the box

Step-by-step explanation:

A) The forces acting on the box are the pulling force that the worker exerts on the rope, the friction, the normal force, and the gravitational force. See the attached diagram. Since the pulling force on the rope makes an angle with the horizontal, this will have components in the x and y-axis (see diagram).

In the y-axis, the box does not move, therefore:

$\sum_{y} F = 0\\F\sin(\alpha) + N - F_g = 0\\N = F_g - F\sin(\alpha)\\N = mg - F\sin(\alpha)$ (1)

where $\alpha$ is the given angle of the rope with the horizontal.

In the x-axis, the box moves with an acceleration $a$ that can be calculated as a function of the given displacement and time interval as:

$a = \frac{2\Delta x}{t^2}$ and the force equation is:

$\sum_{x} F = ma\\F\cos(\alpha) -f =ma\\F\cos(\alpha) - \mu N = ma$

now substituting $N$ from equation (1):

$F\cos(\alpha) - \mu N = ma\\F\cos(\alpha) - \mu (mg - F\sin(\alpha))=ma\\F\cos(\alpha) - \mu mg -\mu F\sin(\alpha)=ma\\F\cos(\alpha) -\mu F\sin(\alpha)=ma+ \mu mg\\F(\cos(\alpha) -\mu \sin(\alpha))=ma+ \mu mg\\\boxed{F=\frac{m(a+ \mu g)}{\cos(\alpha) -\mu \sin(\alpha)}}$