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Andreas93 [3]
3 years ago
8

Looking at systems is important in science because

Physics
1 answer:
Vikentia [17]3 years ago
7 0

Answer:

systems help identify parts and simplify solutions.

Explanation:

Systems are very important in science because they are made up of many parts and can help simplify solutions better.

A system is made up of several parts that are interacting together as a whole. They are very important in solving scientific problems and looking introspectively into a problem.

The parts that makes up the system are equally important as the whole. Systems are typified by what transpires at the level of their parts. Viewing a problem from a system wide angle helps scientists have better knowledge and can be vital in solving scientific problems.

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Amy gets up from her seat on the train to move closer to the front of the train. Just as she begins to walk toward the front, th
Misha Larkins [42]
The answer is B this can be easily shown if you have ever gotten up on a bus its the same effect but if you have trouble  just elimination works A and D are outrageous and wrong C i can see the confusion but a train or a bus making you go in motion but your body wants to keep moving so it keeps in motion 
5 0
3 years ago
El extremo de una cuerda se mueve de tal manera que las partículas de la cuerda emplean 0.25 segundos para producir una vibració
insens350 [35]

Responder:

T = 0.25 segundos

f = 4 Hz

Explicación:

Paso uno:

datos dados

se nos dice que las partículas tardan 0,25 segundos en completar una vibración

por lo tanto, el período es de 0,25 segundos

¿Qué es un período de vibración?

Un período T es el tiempo necesario para que un ciclo completo de vibración pase por un punto determinado.

la expresión para el período se da a continuación.

T = 1 / f

¿Cuál es la frecuencia de vibración?

Básicamente, se refiere a la frecuencia con la que ocurren las vibraciones.

la frecuencia es una tasa, se expresa como la inversa del período

f = 1 / T

t = 1 / 0.25

f = 4 Hz

6 0
3 years ago
The density for gold is 19.3 g/cm3. What would be the mass of a 45 cm3 piece of gold?
lyudmila [28]

Answer:

868.5 g

Explanation:

Mass= Density x Volume

Mass= 19.3 x 45

=868.5

6 0
3 years ago
Read 2 more answers
The four major Earth systems are the atmosphere, lithosphere, hydrosphere, and geosphere. true and false
love history [14]
False its atmosphere, lithosphere, hydrosphere, and boisphere
5 0
3 years ago
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Find the area of the surface.the part of the plane x 2y 3z = 1 that lies inside the cylinder x2 y2 = 6
AlladinOne [14]

I assume you mean the plane x+2y+3z=1. Its area over the region

R = \left\{(x,y) ~:~ x^2 + y^2 \le 6\right\}

is given by the integral

\displaystyle \iint_R dA = \iint_R \sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2} \, dx \, dy

where z=f(x,y) = \frac{1 - x - 2y}3.

We have

\dfrac{\partial f}{\partial x} = -\dfrac13

\dfrac{\partial f}{\partial y} = -\dfrac23

so that the area element is

dA = \sqrt{1 + \left(-\dfrac13\right)^2 + \left(-\dfrac23\right)^2} \,dx\,dy = \dfrac{\sqrt{14}}3\,dx\,dy

Then we have

\displaystyle \iint_R dA = \frac{\sqrt{14}}3 \iint_R dx \, dy

and the remaining integral is exactly the area of the disk x^2+y^2\le6. Its radius is √6, so its area is π (√6)² = 6π. So the area of the surface is

\displaystyle \iint_R dA = \frac{\sqrt{14}}3 \cdot 6\pi = \boxed{2\sqrt{14}\pi}

8 0
2 years ago
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