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3 years ago

Properties such as luster conductivity and flexibility are used to describe what type of elements

1 answer:

4 0

Answer: metals.

Justification:

There are 118 elements which you can find in the periodic table ordered by atomic number. Those elements my be classified into metals, non-metals and metalloids.

The metals are placed on the left side of the periodic table. The metals share the properties of luster, conductivity and flexibility.

The properties of non-metals (which are on the right side of the periodic table) are opposite to those of metals: opaque, low conductivity and brittle.

Metalloids have in between properties.

Copper, for example is a metal: it has luster, is flexible and is highly conductive of the electricity (and temperature).

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If all of the forces acting on an object are balanced, then:

kondaur [170]

A. the direction the object is moving in will not change.

B. the acceleration of the object will be 0 m/s2

Explanation:

We can answer this problem by using Newton's second law of motion:

$\sum F = ma$

where

$\sum F$ is the net force acting on an object

m is the mass of the object

a is its acceleration

In this problem, all the forces acting on the object are balanced, therefore the net force is zero:

$\sum F=0$

which means that also the acceleration is zero:

$a=0$

Acceleration is equal to the rate of change of velocity: therefore, zero acceleration means that the velocity of the object does not change. We can now analyze the given statements:

A. the direction the object is moving in will not change. --> TRUE, because the velocity is not changing.

B. the acceleration of the object will be 0 m/s2 --> TRUE, as we stated above

c. the object will not be in motion. --> FALSE: we just know that its velocity is constant, but it can be different from zero

D. the velocity of the object will be 0 m/s. --> FALSE, for the same reason stated in C

Learn more about Newton's second law of motion:

brainly.com/question/3820012

#LearnwithBrainly

3 0

3 years ago

Sound is a type of _____ wave.

saul85 [17]

Sound is an example of a mechanical wave. Mechanical waves are the kinds of waves that cannot be propagated without a medium. As such, these waves cannot travel through a vacuum, just like how sound cannot travel through space, since space is a vacuum.

8 0

3 years ago

How would doubling the mass of Earth affect the gravity we experience?

N76 [4]

The strength of the gravitational force between two objects depends on two factors, mass and distance. the force of gravity the masses exert on each other. If one of the masses is doubled, the force of gravity between the objects is doubled. increases, the force of gravity decreases.

5 0

3 years ago

Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force

Margaret [11]

Answer:

Speed of the spacecraft right before the collision: $\displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}$ .

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

the kinetic energy of this spacecraft, and
the (gravitational) potential energy of this spacecraft.

Let $m$ denote the mass of this spacecraft. At a distance of $R$ from the center of the earth (with mass $M_\text{e}$ ), the gravitational potential energy ( $\mathrm{GPE}$ ) of this spacecraft would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}$ .

Initially, $R$ (the denominator of this fraction) is infinitely large. Therefore, the initial value of $\mathrm{GPE}$ will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy ( $\rm KE$ ) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance $R$ between the spacecraft and the center of the earth would be approximately equal to $R_\text{e}$ , the radius of the earth.

The $\mathrm{GPE}$ of the spacecraft at that moment would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ .

Subtract this value from zero to find the loss in the $\rm GPE$ of this spacecraft:

$\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}$

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the $\rm GPE$ of this spacecraft would be equal to the size of the gain in its $\rm KE$ .

Therefore, right before collision, the $\rm KE$ of this spacecraft would be:

$\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}$ .

On the other hand, let $v$ denote the speed of this spacecraft. The following equation that relates $v\!$ and $m$ to $\rm KE$ :

$\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2$ .

Rearrange this equation to find an equation for $v$ :

$\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}$ .

It is already found that right before the collision, $\displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ . Make use of this equation to find $v$ at that moment:

$\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}$ .