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Elis [28]
3 years ago
7

Does the densoty of an object change if you add more of it because i did an expirement that proved that the density does change

so im confused, thx
Physics
2 answers:
Marysya12 [62]3 years ago
8 0
No the density does not change. Density is a ratio D=m/v no matter how much of a substance you have its mass will be proportional.
When doing density labs sometimes you might get different answers due to errors that are unavoidable.
stealth61 [152]3 years ago
4 0

If your experiment somehow 'proved' that the density of a substance changed, then your experiment was faulty and produced an error result.

The density of a SUBSTANCE doesn't depend on how much of the SUBSTANCE there is.  One drop of the substance has the same density as a supertanker full of the SAME substance.

You just have to be careful that when you add more of it, the stuff that you add is PURE, and doesn't have anything else in it except more of the SAME SUBSTANCE.  If it has something else in it, then the density will APPEAR to change.

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Suppose a fireworks shell explodes, breaking into three large pieces for which air resistance is negligible. How is the motion o
denis-greek [22]

The center of mass isn't affected by the explosion.

To find the answer, we need to know about the trajectory of motion at zero external force.

<h3>How is the trajectory of an object changed when the net external force on it is zero?</h3>
  • When there's no net external force acting on an object, its momentum doesn't change with time.
  • As its momentum doesn't change, so it continues with the original trajectory.
<h3>Why doesn't the trajectory of firework change when it's exploded?</h3>
  • When a firework is exploded, its internal forces are changed, but there's no external force.
  • So, although the fragments follow different trajectories, but the trajectory of center of mass remains unchanged.

Thus, we can conclude that the center of mass isn't affected by the explosion.

Learn more about the trajectory of exploded firework here:

brainly.com/question/17151547

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3 0
2 years ago
Which of these is the easiest mineral identification property to use when you want to know whether you have pyrite or actual gol
Makovka662 [10]

Answer:

c. streak

Explanation:

Pyrite is a mineral that looks like gold but actually is iron disulfide.

Pyrite and gold have comparable luster.

Pyrite and gold have different tones of yellow. This can be determined by their streak. Streak is the powdered form of a mineral. A streak of mineral can be found just by rubbing the mineral on a rough surface and comparing the colors.

Pyrite is diamagnetic which is not a strong form of magnetism. Gold is also diamagnetic

5 0
3 years ago
While traveling along a highway a driver slows from 32 m/s and comes to a stop with an acceleration of -6 m/s2. How long did it
diamong [38]

Answer: 6s

Explanation:

Vs=32m/s  speed at beginning of slowing down

Vf=0m/s     stop speed

a= -6 m/s²  acceleration

----------------

Use equation for acceleration :

a=(Vf-Vs)/t

a*t=Vf-Vs

t=(Vf-Vs)/a

t=(0-36)/-6

t=-36/-6

t=6 s

7 0
3 years ago
What is a watt a unit of?<br><br> light<br><br><br> time<br><br><br> distance<br><br><br> power
weeeeeb [17]

Answer:

power...............m....

6 0
3 years ago
Read 2 more answers
Michael Jordan, el célebre basquetbolista, ganó el torneo de clavadas de la NBA en 1988. Para lograr la hazaña saltó 1.35 metros
kozerog [31]

(a) 0.40 s

First of all, let's find the initial speed at which Jordan jumps from the ground.

The maximum height is h = 1.35 m. We can use the following equation:

v^2-u^2=2gh

where

v = 0 is the velocity at the maximum height

u is the initial velocity

g=-9.8 m/s^2 is the acceleration of gravity

Solving for u,

u=\sqrt{-2gh}=\sqrt{-2(-9.8)(1.35)}=5.14 m/s

The time needed to reach the maximum height can now be found by using the equation

v=u+gt

Solving for t,

t=\frac{v-u}{g}=\frac{0-5.14}{-9.8}=0.52s

Now we can find the velocity at which Jordan reaches a point 20 cm below the maximum height, so at a height of

h' = 1.35 - 0.20 = 1.15 m

Using again the equation

v'^2-u^2=2gh'

we find

v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(1.15)}=1.97 m/s

And the corresponding time is

t'=\frac{v'-u}{g}=\frac{1.97-5.14}{-9.8}=0.32s

So the time to go from h' to h is

\Delta t = t-t'=0.52-0.32=0.20 s

And since we have also to take into account the fall down (after Jordan reached the maximum height), which is symmetrical, we have to multiply this time by 2 to get the total time of permanence in the highest 20 cm of motion:

\Delta t=2\cdot 0.20 = 0.40 s

(b) 0.08 s

This part is easier since we need to calculate only the velocity at a height of h' = 0.20 m:

v'^2-u^2=2gh'

v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(0.20)}=4.74 m/s

And the corresponding time is

t'=\frac{v'-u}{g}=\frac{4.74-5.14}{-9.8}=0.04s

So this is the time needed to go from h=0 to h=20 cm; again, we have to take into account the motion downwards, so we have to multiply this by 2:

\Delta t = 2\cdot 0.04 =0.08 s

8 0
4 years ago
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