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8_murik_8 [283]
3 years ago
10

Which of the following has least mass

Physics
1 answer:
makvit [3.9K]3 years ago
4 0
Answer is 1 molecule of S
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PLZ HELP, BRAINLIEST !!
Tems11 [23]
Density = mass divided by volume so if you use that formula Liquid A turns out to be a higher density so it flows slower.
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A bowl-shaped depression formed by a mountain glacier is termed a(n) ____________.
Andrej [43]
A bowl-shaped depression formed by a mountain glacier is termed a cirque.
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Which are ways to improve the design of this experiment? Check all that apply.
Arte-miy333 [17]

Answer:

* Experiment with a higher range of materials

* Use a galvanometer.

* Vary in number of coils of the electromagnet

Explanation:

This is an experiment of electricity and magnetism, in general the best way to improve the results are:

* Experiment with a higher range of materials

   allowing to know the scope of the initial assumptions

* Use a galvanometer.

  The more accurate the readings the error of the derived quantities is the less which will improve the precision of the experiment.

* Vary in number of coils of the electromagnet

  Since it allows to have greater magnetic fields and therefore expand the range of measurements

3 0
3 years ago
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If the density of an object is 5.2 g/cm3, and volume is 3.7 cm3, what is its mass
netineya [11]
Here's the equation you use: Density = mass/volume

1) 5.2g/cm^3 = m/3.7cm^3

2) m = 5.2g/cm^3 x 3.7cm^3

3) m = 19.24g

You can check the answer by plugging it in

19.24g/3.7cm^3
= 5.2g/cm^3
6 0
3 years ago
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A tree falls in a forest. How many years must pass before the 14C activity in 1.03 g of the tree's carbon drops to 1.02 decay pe
Illusion [34]

Answer:

t = 5.59x10⁴ y

Explanation:

To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:

A_{t} = A_{0}\cdot e^{- \lambda t}    (1)

<em>where A_{t}: is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.</em>

To find A₀ we can use the following equation:  

A_{0} = N_{0} \lambda   (2)

<em>where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon </em>

From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:        

N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}}

<em>where m_{T}: is the tree's carbon mass, N_{A}: is the Avogadro's number and m_{^{12}C}: is the ¹²C mass.  </em>

N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C    

Similarly, from equation (2) λ is:

\lambda = \frac{Ln(2)}{t_{1/2}}

<em>where t 1/2: is the half-life of ¹⁴C= 5700 years </em>

\lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1}

So, the initial activity A₀ is:  

A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y    

Finally, we can calculate the time from equation (1):

t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y              

I hope it helps you!

4 0
3 years ago
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