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2 years ago

What is happening when an oxygen molecule is formed from two separate oxygen atoms?

1 answer:

6 0

Two from each oxygen atom. For examples of covalent bonding are water (H2O), and carbon dioxide (CO2). Covalent compounds are mostly gases or liquids with low melting points.

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An object on the surface of the earth has a force of gravity of 682 N. What is the objects mass?

valina [46]

The object's mass is around 69.592 kg

F=mg

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3 years ago

Starting from a pillar, you run 200 m east (the + x-direction) at an average speed of 5.0 m/s and then run 280 m west at an aver

zalisa [80]

Answer:

Total time taken=110 seconds

Total distance traveled=480m

Explanation:

First of all, we find the total time taken:

For that, we use the formula : Distance/Speed= Time

Time for part 1 : 200/5=40 seconds

Time for part 2 : 280/4=70seconds

Total time taken=110 seconds

Total distance traveled=480m

Average Speed= 480/110=4.36 m/s

Total displacement=200-280=-80m (Since this is displacement, we need to find the distance between the initial and final point. Also, I've taken east direction as positive and west as negative)

Average Velocity=-80/110=-0.72 m/s

OR 0.72m/s towards west.

3 0

2 years ago

Write the products for the following neutralization reactions

madam [21]

Answer:

answer 3

Explanation:

5 0

2 years ago

Which of the following is an example of kinetic energy being converted to potential energy?

horsena [70]

Of the following...?? Is there more to this question? :)

4 0

2 years ago

Read 2 more answers

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant

$m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{4}}$ and $C$ is a constant

$\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0$ because $v_{(x)}$ is a constant in this derivation respect to $t$

$m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{2}}$ and $C$ is a constant

6 0

3 years ago