All categories

2 years ago

Explain the difference between an oxygen atom and a carbon atom

Physics

1 answer:

Yuliya22 [10]2 years ago

5 0

Answer:

Carbon atoms are made up of 6 protons, 6 electrons, and 6 neutrons, whereas Oxygen atoms are made of 8 protons, 8 neutrons, and 8 electrons.

Explanation:

This may not sound like a big difference at the atomic level, but it is responsible for all of the differences between Oxygen and Carbon.

You might be interested in

If an object force of 50 N is used to move an object a distance of 20 m, what distance must the object be moved if the input for

steposvetlana [31]

Answer:

$\ d_{out} = 100 \ m.$

Explanation:

Given data:

$F_{in} = 50 \ \rm N$

$F_{out} = 10 \ \rm N$

$d_{in} = 20 \ m$

Let the distance traveled by the object in the second case be $d_{out}.$

In the given problem, work done by the forces are same in both the cases.

Thus,

$W_{in} = W_{out}$

$F_{in}.d_{in} = F_{out}.d_{out}$

$\Rightarrow \ d_{out} = \frac{F_{in}.d_{in}}{F_{out}}$

$\ d_{out} = \frac{50 \times 20}{10}$

$\ d_{out} = 100 \ m.$

5 0

3 years ago

Give an example of hypothesis for an experiment and then identify its dependent and independent variables. Write all the steps o

e-lub [12.9K]

An example of a hypothesis for an experiment might be: “A basketball will bounce higher if there is more air it”

Step one would be to make an observation... “hey, my b-ball doesn’t have much air in it, and it isn’t bouncing ver high”

Step two is to form your hypothesis: “A basketball will bounce higher if there is more air it”

Step three is to test your hypothesis: maybe you want to drop the ball from a certain height, deflate it by some amount and then drop it from that same height again, and record how high the ball bounced each time.

Here the independent variable is how much air is in the basketball (what you want to change) and the dependent variable is how high the b-ball will bounce (what will change as a result of the independent variable)

Step four is to record all of your results and step five is to analyze that data. Does your data support your hypothesis? Why or why not?

You should only test one variable at a time because it is easier to tell why the results are how they are; you only have one cause.

Hope this helps!

6 0

3 years ago

A circuit consists of a 12 V battery connected across a single resistor. If the current in the circuit is

finlep [7]

Answer:

4 Ohms

Explanation:

Apply the formula:

Voltage = I (current) . Resistance

You can change it the way you want to use for your purpose.

In this case...

R = V/I

R = 12/3

R = 4 Ohms (Ohm is the unit of measurement of eletrical resistance)

7 0

3 years ago

1. A small light bulb is shining light on a basketball (diameter is 23 cm or 9 inches). The light bulb is 3 m from the closest s

siniylev [52]

Answer:

The size (diameter) of the basketball's shadow on the wall is approximately 53.38 cm

Explanation:

The given parameters of the basketball are;

The diameter of the basketball = 23 cm (9 inches)

The distance of the light bulb from the closest side of the basketball = 3 m

The distance from the ball to the wall = 4 m

The distance from the light source to the center of the ball, d = 3 m + 0.23/2 m = 3.115 m

The angle the light ray makes with the edge of the ball, θ = arctan(0.115/3.115)

Therefore, the ratio of the shadow width divided by 2 to the distance from the light from the wall = 0.115/3.115

The distance from the light from the wall = 3 m + 4 m + 0.23 m = 7.23 m

Therefore;

((The width of the shadow)/2)/(The distance from the light from the wall) = 0.115/3.115

∴ ((The width of the shadow)/2)/(7.23 m) = 0.115/3.115

((The width of the shadow)/2) = 7.23 m × 0.115/3.115 = 16629/62300 m ≈ 0.2669 m = 26.69 cm

The width (diameter) of the shadow on the wall = 2 × 16629/62300 m ≈ 0.5338 m = 53.38 cm

The size (diameter) of the basketball's shadow on the wall ≈ 53.38 cm

4 0

3 years ago

A parallel plate capacitor with plates of area A and plate separation d is charged so that the potential difference between its

faust18 [17]

Answer:

Explanation:

Given a parallel plate capacitor of

Area=A

Distance apart =d

Potential difference, =V

If the distance is reduce to d/2

What is p.d

We know that

Q=CV

Then,

V=Q/C

Then this shows that the voltage is inversely proportional to the capacitance

Therefore,

V∝1/C

So, VC=K

Now, the capacitance of a parallel plate capacitor is given as

C= εA/d

When the distance apart is d

Then,

C1=εA/d

When the distance is half d/2

C2= εA/(d/2)

C2= 2εA/d

Then, applying

VC=K

V1 is voltage of the full capacitor V1=V

V2 is the required voltage let say V'

Then,

V1C1=V2C2

V × εA/d=V' × 2εA/d

VεA/d = 2V'εA/d

Then the εA/d cancels on both sides and remains

V=2V'

Then, V'=V/2

The potential difference is half when the distance between the parallel plate capacitor was reduce to d/2

6 0

4 years ago

Read 2 more answers