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3 years ago

If asked to describe a wave, you could say waves are the

2 answers:

8 0

The answer would be : B) continuous transmission of energy from one location to the next.

- A disturbance that travels from one location to another location.

-Hope this helps.

Minchanka [31]3 years ago

6 0

B) continuous transmission of energy from one location to the next.

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Make the following conversion. 34.9 cL = _____ hL. 0.349 0.0349 0.00349 349,000

AnnyKZ [126]

Hey there!

Here is your answer:

The proper answer to this question is option C "0.00349".

Reason:

1 L = 100 cL. Or 1 cL = 0.01 L

34.9 cL = 34.9 / 100 L = 0.349 L

 1 hL = 100 L. 0.349 L = 0.349 / 100 hL = 0.00349 hL

Therefore the answer is option C!

If you need anymore help feel free to ask me!

Hope this helps!

~Nonportrit

4 0

3 years ago

What happens to the speed of the particles if the size of the particle is increased

bearhunter [10]

The particle slows down.

6 0

4 years ago

A child pushes a 75 N toy car across the floor. What is the mass of the car?

GaryK [48]

Answer:

7.6 kg

Explanation:

w=75N

w=mg

m=w÷g

m=75÷9.8

m=7.6kg

8 0

3 years ago

A rough estimate of the radius of a nucleus is provided by the formula r 5 kA1/3, where k is approximately 1.3 × 10213 cm and A

Sphinxa [80]

Answer:

Density of 127 I = $\rm 1.79\times 10^{14}\ g/cm^3.$

Also, $\rm Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.$

Explanation:

Given, the radius of a nucleus is given as

$\rm r=kA^{1/3}$ .

where,

$\rm k = 1.3\times 10^{-13} cm.$

A is the mass number of the nucleus.

The density of the nucleus is defined as the mass of the nucleus M per unit volume V.

$\rm \rho = \dfrac{M}{V}=\dfrac{M}{\dfrac 43 \pi r^3}=\dfrac{M}{\dfrac 43 \pi (kA^{1/3})^3}=\dfrac{M}{\dfrac 43 \pi k^3A}.$

For the nucleus 127 I,

Mass, M = $\rm 2.1\times 10^{-22}\ g.$

Mass number, A = 127.

Therefore, the density of the 127 I nucleus is given by

$\rm \rho = \dfrac{2.1\times 10^{-22}\ g}{\dfrac 43 \times \pi \times (1.3\times 10^{-13})^3\times 127}=1.79\times 10^{14}\ g/cm^3.$

On comparing with the density of the solid iodine,

$\rm \dfrac{Density\ of\ ^{127}I}{Density\ of\ the\ solid\ iodine}=\dfrac{1.79\times 10^{14}\ g/cm^3}{4.93\ g/cm^3}=3.63\times 10^{13}.\\\\\Rightarrow Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.$