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Elena-2011 [213]

3 years ago

9

2.when sugar or another substance is dissolved in water, it disappears from view and forms a homogeneous mixture with the water,

also called a solution. if you can’t see the sugar, how can you tell that it is there?

1 answer:

bekas [8.4K]3 years ago

3 0

You cant really tell when it is dissolved

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When an atom that has no charge looses two electrons it becomes a

alex41 [277]

Answer: it becomes a positive ion

Explanation:

So, when an atom loses 2 electrons there will be no change in the number of neutrons. Therefore, an isotope will not form. Thus, it is concluded that when an atom with no charge loses two electrons, it becomes a positive ion.

8 0

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Suppose a spring has a relaxed length of 28.3 cm. The simulation refers to this as the natural length. This is the length of the

den301095 [7]

Answer:

Explanation:

Normal length of spring = 28.3 cm

stretched length of spring = 38.2 cm

length of extension = 38.2 - 28.3 = 9.9 cm

= 9.9 x 10⁻² m

force applied to stretch = .55 x 9.8 ( mg )

= 5.39 N

Force constant = force applied / extension

= 5.39 / 9.9 x 10⁻²

= .5444 x 10² N /m

= 54.44 N/m

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3 years ago

Two horizontal forces,225N and 165N,are exerted on s canoe in the same direction. Find the net horizontal foce

Readme [11.4K]

Since they are in the same direction, you would add them together. Let’s also assume said direction is positive. 225 N + 165 N = 390 N

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3 years ago

A dog of mass 18 kg runs at a speed of 4 m/s. What is the momentum of the

Andrej [43]

Answer:

A, 72 kg•m/s

Explanation:

p=mv

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3 years ago

If an atomic nucleus were the size of a dime how far away might one of its electrons be?

mihalych1998 [28]

The radius of a nucleus of hydrogen is approximately $r_{n1}=1\cdot 10^{-15}m$ , while we can use the Borh radius as the distance of an electron from the nucleus in a hydrogen atom: $r_{e1}=5.3 \cdot 10^{-11}m$

The radius of a dime is approximately $r_{n2} = 9\cdot 10^{-3}m$ : if we assume that the radius of the nucleus is exactly this value, then we can find how far is the electron by using the proportion
$r_{n1}:r_{e1}=r_{n2}:r_{e2}$
from which we find
$r_{e2}= \frac{r_{e1} r_{n2}}{r_{n1}}= \frac{(5.3 \cdot 10^{-11}m)(9\cdot 10^{-3}m)}{1 \cdot 10^{-15}m}=477 m$

So, if the nucleus had the size of a dime, we would find the electron approximately 500 meters away.

6 0

3 years ago