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3 years ago

Which element has the strongest attraction for electrons

2 answers:

6 0

If one atom is overwhelmingly more electronegative than the other atom, the electrons will not be shared and an ionic bond will result. The periodic table below shows the Pauling electronegativity scale. A value of 4.0 is assigned to FLORINE, the most electronegative element.

ITS FLORINE

I BELIVE

Harrizon [31]3 years ago

6 0

I believe the answer is Florine

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Which of the following terms best describes why a skier sliding down a hill eventually comes to a stop?

natta225 [31]

Acceleration means you go faster
inertia means it has a tendency to do nothing or remain unchanged
gravity is what pulls things down to earth
<span>Friction is the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other.
so i would say friction</span>

6 0

3 years ago

Read 2 more answers

If the acceleration of a moving object is zero, which of the following best describes its motion

podryga [215]

If the acceleration of a moving object is zero, the object remains at a constant speed (without friction)

7 0

3 years ago

Read 2 more answers

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:

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

6 0

3 years ago

The acceleration due to gravity is lower on the Moon than on Earth. Which of the following is true about the mass and weight of

Naya [18.7K]

Mass is the same, weight is less

<h3>What is the Weight and mass on Moon ?</h3>

As we know that the mass of the object is the measurement of the quantity of the matter that is present in it

So here we can say that if the mass of the object is m then its total quantity of the matter that is present in it is given as

mass = (density) × (volume)

Now for the weight of the object is defined as the force of gravity due to planet

Fg = mg

so the weight of the object is depending on the acceleration due to gravity of the planet

As we know that the gravity of moon is smaller than the gravity of the earth so here weight on the moon will be smaller than the weight on the Earth

Learn more about Weight on Moon here:

brainly.com/question/4080619

#SPJ4

3 0

2 years ago

What do all waves, including sound and light, have in common

Schach [20]

Answer:

They all have frequency, wavelength, amplitude, speed and also all transfer energy.

6 0

3 years ago