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zepelin [54]
3 years ago
11

Why does the density of a substance remain the same for different amount of the substance

Physics
1 answer:
tresset_1 [31]3 years ago
6 0

Think of it this way: 
-- Any time you have something that means (some number) PER UNIT,
it doesn't matter how many units there are on the table or in the bucket,
because that amount doesn't change the (number) PER UNIT.

-- If oranges cost $1 PER POUND, it doesn't matter how many pounds
you buy, the whole bagful is still $1 PER POUND.

-- If a certain salad dressing has 40 calories PER Tablespoon, it doesn't
matter whether you eat a drop of it or drink the whole jar.  You still get
40 calories PER Tablespoon.

-- Density means '(mass) PER unit of volume'.  Whether you have a tiny
chip of the substance or a whole truckload of it, there's still the same
amount of mass IN EACH unit of volume.

You might be interested in
Is cracking the eggs a physical change or a chemical change and why
matrenka [14]

Answer:

Cracking of an egg is a physical change since the egg and the stuff inside does not change but the shape or appearance of the shell changes.

Explanation:

Hope it helps

3 0
2 years ago
If i want an thrilling job witch should i chose im stuck beetween,
MA_775_DIABLO [31]

Answer:

1. horse rider

Explanation:

5 0
3 years ago
A 4.0-cm tall light bulb is placed at distance of 8.3 cm from a concave mirror having a focal length of 15.2 cm. Determine the i
scoundrel [369]

Answer: the image distance is -18, 28 cm this means behind of the concave mirror. The image size is 2.2 higher that the original  so it has 8.8 cm with the same orientation as original  and it is a virtual imagen.

Explanation: In order to sove the imagen formation for a concave mirror we have to use the following equation:

1/p+1/q=1/f  where p and q represents the distance to the mirror  for the object and imagen, respectively. f is the focal length for the concave mirror.

replacing the values we obtain:

1/8.3+1/q=1/15.2

so 1/q=(1/15.2)-(1/8.3)=-54.7*10^-3

then q=-18.28 cm

The magnification is given by M=-q/p=-(-18,28)/8.3= 2.2

We also add a picture to see the imagen formation for this case.

6 0
3 years ago
Who's really good at physical science?
34kurt
Definitely not me!!!! (unless it's magnetism or circuits)
3 0
3 years ago
A 12.0-g sample of carbon from living matter decays at the rate of 162.5 decays/minute due to the radioactive 14C in it. What wi
Ghella [55]

Answer:

a)143.8 decays/minute

b)0.41 decays/minute

Explanation:

From;

0.693/t1/2 = 2.303/t log (Ao/A)

Where;

t1/2=half-life of C-14= 5670 years

t= time taken to decay

Ao= activity of a living sample

A= activity of the sample under study

a)

0.693/5670 = 2.303/1000 log(162.5/A)

1.22×10^-4 = 2.303×10^-3 log(162.5/A)

1.22×10^-4/2.303×10^-3 = log(162.5/A)

0.53 × 10^-1 = log(162.5/A)

5.3 × 10^-2 = log(162.5/A)

162.5/A = Antilog (5.3 × 10^-2 )

A= 162.5/1.13

A= 143.8 decays/minute

b)

0.693/5670 = 2.303/50000 log(162.5/A)

1.22×10^-4 = 4.61×10^-5 log(162.5/A)

1.22×10^-4/4.61×10^-5 = log(162.5/A)

0.26 × 10^1 = log(162.5/A)

2.6= log(162.5/A)

162.5/A = Antilog (2.6 )

A= 162.5/398.1

A= 0.41 decays/minute

4 0
3 years ago
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