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olya-2409
2 months ago
8

A child runs at speed vc while an adult runs at speed va. if the adult gives the child a head start of distance d, how long will

it take for the adult to catch the child? express your answer in terms of va and vc and
d.
Physics
1 answer:
ValentinkaMS [3.4K]2 months ago
4 0
The child runs at speed vc.
The adult runs at speed va.
<span>The adult allows the child a lead of distance d. 
Speed is calculated by distance divided by time: speed = distance/time.
To find the time it takes for the adult to catch the child, use time = d/v.
</span>Let tc be the child's time and ta the adult's time.
tc = D/vc, with D = D + d, so D + d = tc × vc, therefore D = tc × vc - d.
ta = D/va, where D = ta × va.
Equating both: tc × vc - d = ta × va
Thus, ta = (tc × vc - d) / va
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On the assumption that copper– nickel alloys are random mixtures of copper and nickel atoms, calculate the mass of copper which,
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Response:

65.14 g

Clarification:

The entropy change resulting from mixing two metals is

\delta S= R[(n_{cu}+n_{Ni})ln(n_{cu}+n_{Ni})-n_{cu}lnn_{cu}-n_{Ni}lnn_{Ni}]

Here, R is the real gas constant, n_{cu} denotes the moles of copper, while n_{Ni} signifies the moles of nickel.

Therefore, determining the moles of Nickel yields

n_{Ni}=\frac{m_{Ni}}{M_{Ni}}

n_{Ni}=\frac{100}{58.69} =1.7 moles

Then substituting n_{Ni} = 1.7 into the equation produces ΔS= 15,

with R= 8.314, and solving gives

15= R[(n_{cu}+1.7)ln(n_{cu}+1.7)-n_{cu}lnn_{cu}-1.7ln1.7]

Upon resolving, the quantity of moles of copper in the mixture equals

n_cu= 1.025

Thus, the copper mass is calculated as n_cu×M_cu = 1.025×63.55 = 65.14 g

Consequently, the required mass is found to be = 65.14 g

3 0
1 month ago
The relatively high resistivity of dry skin, about 1×106Ω⋅m, can safely limit the flow of current into deeper tissues of the bod
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Answer:

The resistance of the skin is 98 kΩ

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With assumed radius for the worker's palm,

r = 7 \times 10^{-2} m

Area of the worker's palm,

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R = \frac{10^{6} \times 1.5 \times 10^{-3} }{1.53 \times 10^{-2} }

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kicyunya [3294]

Answer:

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Partial pressure of ozone=\frac{441\times atmospheric\;pressure}{10^9} atm

Inserting the given atmospheric pressure

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Consequently, the partial pressure of ozone is=2.95\times 10^{-7}atm

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