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3 years ago

7

A hot air balloon of total mass M (including passengers and luggage) is moving with a downward acceleration of magnitude a. As i

t approaches a mountain, the captain needs to accelerate upwards. He decides to throw enough ballast over board to achieve an upward acceleration of magnitude a/2. What fraction of the initial mass does he have to drop?

1 answer:

frez [133]3 years ago

5 0

Answer:

When balloon moves in the downward direction two forces acts on it.

i) Force exerted by air in the upward direction

ii) Weight

According to newton’s second law of motion:

Sum of forces = Ma

W – F = Ma

Mg – F = Ma …….. (i)

when some of the mass m is dropped and balloon is moving in upward direction with acceleration a/2 then,

F – W = (M-m)a/2

F – (M-m)g = (M-m)a/2

F – Mg + mg = Ma/2 – ma/2 ….. (ii)

Adding equation (i) and (ii)

mg = M(3a/2) – ma/2

m(g + a/2) = M(3a/2)

m = M(3a/2)/(g + a/2)

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A 0.4000 kg sample of methanol at 16.0ºC is mixed with 0.4000 kg of water at 85.0ºC. Assuming no heat loss to the surroundings,

AVprozaik [17]

Answer:

T_finalmix = 59.5 [°C].

Explanation:

In order to solve this problem, a thermal balance must be performed, where the heat is transferred from water to methanol, at the end the temperature of the water and methanol must be equal once the thermal balance is achieved.

$Q_{water}=Q_{methanol}$

where:

$Q_{water}=m_{water}*Cp_{water}*(T_{waterinitial}-T_{final})$

mwater = mass of the water = 0.4 [kg]

Cp_water = specific heat of the water = 4180 [J/kg*°C]

T_waterinitial = initial temperature of the water = 85 [°C]

T_finalmix = final temperature of the mix [°C]

$Q_{methanol}=m_{methanol}*Cp_{methanol}*(T_{final}-T_{initialmethanol})$

Now replacing:

$0.4*4180*(85-T_{final})=0.4*2450*(T_{final}-16)\\142120-1672*T_{final}=980*T_{final}-15680\\157800=2652*T_{final}\\T_{final}=59.5[C]$

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3 years ago

A practical rule is that a radioactive nuclide is essentially gone after 10 half-lives. What percentage of the original radioact

ArbitrLikvidat [17]

Answer:

0.09 % of the original radioactive nucllde its left after 10 half-lives
It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

Explanation:

The equation for radioactive decay its:

$N ( t) \ = \ N_0 \ e^{ \ - \frac{t}{\tau}}$ ,

where N(t) its quantity of material at time t, $N_0$ its the initial quantity of material and $\tau$ its the mean lifetime of the radioactive element.

The half-life $t_{\frac{1}{2}}$ its the time at which the quantity of material its the half of the initial value, so, we can find:

$N (t_{\frac{1}{2} }) \ = \ N_0 \ e^{ \ - \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}$

so:

$\ N_0 \ e^{ \ - \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{N_0}{2}$

$e^{ \ - \frac{t_{\frac{1}{2}}}{\tau}} \ = \frac{1}{2}$

$- \frac{t_{\frac{1}{2}}}{\tau}} \ = - \ ln( 2 )$

$t_{\frac{1}{2}}\ = \tau ln( 2 )$

So, after 10 half-lives, we got:

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ - \frac{10 \ t_{\frac{1}{2}}}{\tau}}$

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ - \frac{10 \ \tau \ ln( 2 ) }{\tau}}$

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ e^{ \ - 10 \ \ ln( 2 ) }$

$N ( 10 \ t_{\frac{1}{2}}) \ = \ N_0 \ * \ 9.76 * 10^{-4}$

So, we got that a 0.09 % of the original radioactive nucllde its left.

Putonioum-239 has a half-life of 24,110 years. So, 10 half-life will take to pass

$10 \ * \ 24,110 \ years \ = \ 241,100 \ years$

It will take 241,100 years for 10 half-lives of plutonium-239 to pass.

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4 years ago

Read 2 more answers

Mercury is a liquid metal having a density of 13.6 g/mL. What is the

sergij07 [2.7K]

Answer:

$\boxed {\boxed {\sf v \approx 33.088 \ mL}}$

Explanation:

The formula for density is:

$d= \frac{m}{v}$

where <em>m</em> is the mass and <em>v</em> is the volume.

The mass is 0.45 kilograms and the density is 13.6 grams per milliliter. The density is given in grams, so we must convert the mass.

There are 1000 grams in 1 kilogram or $\frac{1000 \ g} {1 \ kg}$ . We can multiply the mass by this ratio.

$0.45 \ kg * \frac{1000 \ g} {1 \ kg} = 0.45 * 1000 \ g = 450 \ g$

Now we have values for the mass and density:

$13.6 \ g/mL =\frac{450 \ g}{ v}$

Cross multiply.

$\frac {13.6 \ g/mL}{1} =\frac{450 \ g}{ v}$

$13.6 \ g/mL * v=450 \ g * 1 \\13.6 \ g/mL * v=450 \ g$

We are trying to find the volume, so we must isolate that variable.

13.6 and v are being multiplied. The inverse of multiplication is division. Divide both sides by 13.6

$\frac {13.6 \ g/mL*v } { 13.6 \ g/mL} = \frac{ 450 \ g} {13.6 \ g/mL}$

$v= \frac{ 450 \ g} {13.6 \ g/mL}$

The grams will cancel.

$v= 33.0882353 \ mL\\v \approx 33.088 \ mL$

The volume is about 33.088 milliliters.

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3 years ago

Who was elected senator of Illinois in 1858

Tamiku [17]

Stephen A. Douglas was elected senator of Illinois in 1858. He bested Abraham Lincoln in the election for senate.

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4 years ago

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An isolated, irregularly shaped piece of platinum carries -8.89 × 10-9 C of charge and is in electric equilibrium. The size of t

rusak2 [61]

Answer:

c) equals V

Explanation:

This is because, since the isolated, irregularly shaped piece of platinum is in electric equilibrium, the electric potential at all points on its surface is V. So that, the potential difference across any point is zero. This implies that diametrically opposite sides have the same potential and thus, the potential at other points of the surface is V since it is in electric equilibrium.

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