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sesenic [268]
3 years ago
10

N4M.6 A board has one end wedged under a rock having a mass of 380 kg and is supported by another rock that touches the bottom s

ide of the board at a point 85 cm from the end under the rock. The board is 4.5 m long, has a mass of about 22 kg, and projects essentially horizontally out over a river. Is it safe for an adult with a mass of 62 kg to stand at the unsupported end of the board
Physics
1 answer:
Hitman42 [59]3 years ago
5 0

Answer:

it is safe to stand at the end of the table

Explanation:

For this exercise we use the rotational equilibrium condition

         Στ = 0

         W x₁ - w x₂ - w_table x₃ = 0

         M x₁ - m x₂ - m_table x₃ = 0

where the mass of the large rock is M = 380 kg and its distance to the pivot point x₁ = 850 cm = 0.85m

the mass of the man is 62 kg and the distance

            x₂ = 4.5 - 0.85

            x₂ = 3.65 m

the mass of the table (m_table = 22 kg) is at its geometric center

            x_{cm} = L/2 = 2.25 m

            x₃ = 2.25 -0.85

            x₃ = 1.4 m

let's look for the maximum mass of man

            m_{maximum} = \frac{ M x_1 -m_{table} x_3}{ x_2}

let's calculate

             m_{maximum} = \frac{ 380 \ 0.85 - 22 \ 1.4}{3.65}(380 0.85 - 22 1.4) / 3.65

             m_{maximum} = 80 kg

we can see that the maximum mass that the board supports without turning is greater than the mass of man

             m_{maximum}> m

consequently it is safe to stand at the end of the table

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Answer: G00gle got you bro

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3 years ago
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}}

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\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
2 years ago
A stationary 6-kg shell explodes into three pieces. One 4.0 kg piece moves horizontally along the negative x-axis. The other two
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Answer:

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Based on the above information, the velocity = v is

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Answer:

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Note: All speeds are in the same direction and have the same sign

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Convert the volume 8.06 in.3 to m3, recalling that1in. =2.54cmand100cm=1m. Answer in units of m3.
galina1969 [7]
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