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riadik2000 [5.3K]
3 years ago
14

Could anyone help me with my project? Plz it’s overdue & I’m Don’t understand

Physics
2 answers:
posledela3 years ago
8 0

The example says skateboard. Something to do a vehicle. Motorcycle or bike are both vehicles which uses the following energy.

viktelen [127]3 years ago
5 0
Message me I can help!
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According to Newton's law of universal gravitation, which statements are true?
kogti [31]

Answer:

c.  As we gain mass, the force of gravity on us increases

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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
Multiple-Concept Example 9 reviews the concepts that are important in this problem. A drag racer, starting from rest, speeds up
Mademuasel [1]

Answer:

V = 90.51 m/s

Explanation:

From the given information:

Initial speed (u) = 0

Distance (S) = 391 m

Acceleration (a) = 18.9 m/s²

Using the relation for the equation of motion:

v² - u² = 2as

v² - 0² = 2as

v² = 2as

v = \sqrt{2as}

v = \sqrt{2*18.9*391}

v = 121.57 m/s

After the parachute opens:

The initial velocity = 121.57 m/ss

Distance S' = 332 m

Acceleration = -9.92 m/s²

How fast is the racer can be determined by using the relation:

V=  \sqrt{v^2 + 2aS'}

V = \sqrt{121.57^2+ 2 (-9.92)(332)}

V = 90.51 m/s

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Can i have the physical science grade 12 june 2021 paper (common paper)​
Veseljchak [2.6K]

Answer:

u have to purchase it via online e-commerce platforms

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Is a table made from minerals?
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A table can be made from minerals but other kinds of solids as well 
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2 years ago
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