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Stells [14]
4 years ago
14

If a balloon filled with air and sealed is kept in a deep freezer for a period of time, it will shrink in size Determine the the

best reasoning for this outcome.
Physics
2 answers:
Bess [88]4 years ago
6 0
They lose kentic energy and slow down, like the other person said.
Alexandra [31]4 years ago
3 0
As the air molecules cool down they lose kinetic energy. This means they're moving slower and so cannot keep the walls of the rubber balloon expanded by their collisions with it
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What are the 3 laws of Newton
daser333 [38]

Answer:

What are Newton's 1st 2nd and 3rd laws of motion?

They describe the relationship between a body and the forces acting upon it, and its motion in response to those forces. More precisely, the first law defines the force qualitatively, the second law offers a quantitative measure of the force, and the third asserts that a single isolated force doesn't exist.

Explanation:

They describe the relationship between a body and the forces acting upon it, and its motion in response to those forces. More precisely, the first law defines the force qualitatively, the second law offers a quantitative measure of the force, and the third asserts that a single isolated force doesn't exist.

5 0
4 years ago
Radon-222 ( 222/86 Rn) is a radioactive gas with a half-life of 3.82 days. A gas sample contains 4.1 e 8 radon atoms initially.
kow [346]

Answer :

(a) The number of radon atoms will remain after 12 days is, 4.67\times 10^7

(b) The number of radon nuclei have decayed by this time will be, 3.6\times 10^8

Explanation :

<u>For part (a) :</u>

Half-life = 3.82 days

First we have to calculate the rate constant, we use the formula :

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{3.82\text{ days}}

k=1.81\times 10^{-1}\text{ days}^{-1}

Now we have to calculate the number of radon atoms will remain after 12 days.

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 1.81\times 10^{-1}\text{ days}^{-1}

t = time passed by the sample  = 12 days

a = initially number of radon atoms  = 4.1\times 10^8

a - x = number of radon atoms left = ?

Now put all the given values in above equation, we get

12=\frac{2.303}{1.81\times 10^{-1}}\log\frac{4.1\times 10^8}{a-x}

a-x=4.67\times 10^7

Thus, the number of radon atoms will remain after 12 days is, 4.67\times 10^7

<u>For part (b) :</u>

Now we have to calculate the number of radon nuclei will have decayed by this time.

The number of radon nuclei have decayed = Initial number of radon atoms - Number of radon atoms left

The number of radon nuclei have decayed = (4.1\times 10^8)-(4.67\times 10^7)

The number of radon nuclei have decayed = 3.6\times 10^8

Thus, the number of radon nuclei have decayed by this time will be, 3.6\times 10^8

5 0
3 years ago
A boy on a 1.9 kg skateboard initially at rest tosses a(n) 7.8 kg jug of water in the forward direction. if the jug has a speed
Tresset [83]
For this case we first think that the skateboard and the child are one body.
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 1 = jug
 2 = skateboard + boy
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 M1V1i + M2V2i = M1V1f + M2V2f
 Initial rest:
 v1i = v2i = 0
 0 = M1V1f + M2V2f
 Substituting values
 0 = (7.8) (3.2) + (M2) (- 0.65)
 0 = 24.96 + M2 (-0.65)
 -24.96 = (-0.65) M2
 M2 = (-24.96) / (- 0.65) = 38.4 kg
 Then, the child's mass is:
 M2 = Mskateboard + Mb
 Clearing:
 Mb = M2-Mskateboard
 Mb = 38.4 - 1.9
 Mb = 36.5 Kg
 answer:
 the boy's mass is 36.5 Kg
4 0
3 years ago
Why exercise is wise?
Inessa05 [86]

Answer:

Exercise helps people lose weight and lower the risk of some diseases. Exercising regularly lowers a person's risk of developing some diseases, including obesity, type 2 diabetes, and high blood pressure. Exercise also can help keep your body at a healthy weight. Exercise can help a person age well.

8 0
4 years ago
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Quasars are thought to be the nuclei of active galaxies in the early stages of their formation. A typical quasar radiates energy
uranmaximum [27]

Answer:

1.7531 smu/yn

Explanation:

check the attached file below for answer and explanation.

6 0
4 years ago
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