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3 years ago

In the periodic table of elements, Au stands for the element_____________, after the the Latin word "aurum."

Physics

1 answer:

Effectus [21]3 years ago

6 0

Answer:

gold

Explanation:

au stands for gold in the periodic table

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How much momentum does a 10 kg object have when is moving 30 m/s

nikklg [1K]

P = m x v

P = 30 x 10
=300

3 0

3 years ago

Why are the very first stars thought to have been much more massive than the sun? (a the clouds that made them were much more ma

pashok25 [27]

I think the correct answer would be that the temperature of the clouds that made the very first stars where thought to be higher since the clouds are made up of hydrogen and helium. Hope this answers the question. Have a nice day.

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3 years ago

Read 2 more answers

How much heat is needed to bring 12.0 g of water from 28.3 °C to 43.87 °C, if the specific heat capacity of water is 4.184 J/(g•

Vika [28.1K]

Answer:

Heat capacity, Q = 781.74 Joules

Explanation:

Given the following data;

Mass = 12g

Initial temperature = 28.3°C

Final temperature = 43.87°C

Specific heat capacity of water = 4.184J/g°C

To find the quantity of heat needed?

Heat capacity is given by the formula;

$Q = mcdt$

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

dt = T2 - T1

dt = 43.87 - 28.3

dt = 15.57°C

Substituting into the equation, we have;

$Q = 12*4.184*15.57$

Q = 781.74 Joules

7 0

3 years ago

Why is the weight of an object more on Earth than on the Moon?

JulijaS [17]

<span>Why is the weight of an object more on Earth than on the Moon? i would say c</span>

8 0

3 years ago

The sound from a loud speaker has an intensity level of 80 db at a distance of 2.0m. Consider the speaker to be a point source,

Tamiku [17]

Answer:

2.83m

Explanation:

The information that we have is

Intensity at 2.0 m: $I=80dB$ and $r_{1}=2m$

we need an intensity level of: $I_{2}=40dB$

thus, we are looking for the distance $r_{2}$ .

which we can find with the law for intensity and distance:

$(\frac{r_{2}}{r_{1}} )^2=\frac{I_{1}}{I_{2}}$

we solve for $r_{2}$ :

$\frac{r_{2}}{r_{1}}=\sqrt{\frac{I_{1}}{I_{2}} }\\\\r_{2}=r_{1}\sqrt{\frac{I_{1}}{I_{2}} }$

and we substitute the known values:

$r_{2}=(2m)\sqrt{\frac{80dB}{40dB} }\\\\r_{2}=(2m)\sqrt{2}\\ r_{2}=2.83m$

at a distance of 2.83m the intensity level is 40dB

5 0

3 years ago