All categories

1 year ago

In what phase change do molecules completely overcome the intermolecular

Physics

1 answer:

Anika [276]1 year ago

8 0

(B.) I know it’s not that hard or easy so don’t be made if wrong

You might be interested in

Gravitational potential energy is due to....

mote1985 [20]

Answer:

Explanation:

I only think its A because of the gravity part...sorry im not good at explaining

5 0

2 years ago

5. What is the density of 4.5 mL of a liquid that has a mass of 1.3 grams?

antoniya [11.8K]

Answer:

A. 0.289g/mL

Explanation:

Using the equation for density which is d = m/v or density = mass/volume, we input 1.3g/4.5mL and get 0.289g/mL.

5 0

2 years ago

Read 2 more answers

Volcanoes change the earth by

SOVA2 [1]

I believe the answer is b) slowly heating the surface

8 0

2 years ago

Read 2 more answers

Does the orbital period of a planet depend on the mass of the planet or on the mass of the star that it orbits?

jasenka [17]

Answer:

The orbital period of a planet depends on the mass of the planet.

Explanation:

A less massive planet will take longer to complete one period than a more massive planet.

8 0

1 year ago

A sound is recorded at 19 decibels. What is the intensity of the sound?

sp2606 [1]

$1 \times 10^{-10.1} \mathrm{Wm}^{-2}$ is the intensity of the sound.

Answer: Option B

<u>Explanation:</u>

The range of sound intensity that people can recognize is so large (including 13 magnitude levels). The intensity of the weakest audible noise is called the hearing threshold. (intensity about $1 \times 10^{-12} \mathrm{Wm}^{-2}$ ). Because it is difficult to imagine numbers in such a large range, it is advisable to use a scale from 0 to 100.

This is the goal of the decibel scale (dB). Because logarithm has the property of recording a large number and returning a small number, the dB scale is based on a logarithmic scale. The scale is defined so that the hearing threshold has intensity level of sound as 0.

$\text { Intensity }(d B)=(10 d B) \times \log _{10}\left(\frac{I}{I_{0}}\right)$

Where,

I = Intensity of the sound produced

$I_{0}$ = Standard Intensity of sound of 60 decibels = $1 \times 10^{-12} \mathrm{Wm}^{-2}$

So for 19 decibels, determine I as follows,

$19 d B=(10 d B) \times \log _{10}\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)$

$\log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=\frac{19}{10}$

$\log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=1.9$

When log goes to other side, express in 10 to the power of that side value,

$\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)=10^{1.9}$

$I=1 \times 10^{-12} \mathrm{Wm}^{-2} \times 10^{1.9}=1 \times 10^{-12-1.9}=1 \times 10^{-10.1} \mathrm{Wm}^{-2}$

5 0

2 years ago