91影视

According to Newton's Second Law of Motion, the weight of a body is equal to the product of the body's mass and acceleration due to gravity. It can be expressed in terms of Newtons.

The radius of the spherical balloon is r = 7.15m.

The mass of the skin and structure of the balloon is m_balloon = 930kg.

Now, understand this question with the help of a diagram, as given below.

So, as you can see, four forces are acting on the spherical balloon.

Force in an upward direction tends to lift the balloon. It can be represented by,

F = 蟻 _air.V_{air .}g

Here, g is the acceleration due to gravity, 蟻_air is the density of air which can be taken as 1.225 kg/m³ (standard value), V_air and is the volume of the air displaced by the balloon, and it is equal to the volume of the balloon, i.e.,

Substitute the value of r_balloon in the above formula.

\begin{aligned}{V_{{\rm{balloon}}}}=\frac{4}{3}\pi{\left( {7.15}\right)^3}\;{{\rm{m}}^3}\\{V_{{\rm{balloon}}}}=1531.11\;{{\rm{m}}^3}\end{aligned}

Substitute the values of 蟻 _air, V_{air ,}g in the formula for F.

The balloon experiences force due to the weight of the skin and structure of the balloon. It can be represented by W_balloon.

W_balloon= m_balloon . g

Substitute the values in the above formula.

Force experienced by the balloon due to the weight of the helium gas filled in the balloon. It can be represented by W_He.

W_He = m_{He . g}

W_He = 蟻_He.V_{He . g}

贬别谤别,蚁_He is the density of the Helium gas, which can be taken as 0.1785kg/m³ (standard value), and V_He is the volume of Helium gas present inside the spherical balloon, equal to the balloon's volume, i.e., 1531.1m³.

Substitute the values of 蟻_He, V_{He, g} in the formula for W_He.

\begin{aligned}{W_{{\rm{He}}}}=0.1785{{{\rm{kg}}}\mathord{\left/{\vphantom{{{\rm{kg}}}{{{\rm{m}}^3}}}}\right\{{{\rm{m}}^3}}}\times1531.11\;{{\rm{m}}^3}\times9.81\;{{\rm{m}}\mathord{\left/{\vphantom{{\rm{m}}{{{\rm{s}}^2}}}}\right\{{{\rm{s}}^2}}}\\{W_{{\rm{He}}}}=2681.1\;{{{\rm{kg}}\cdot{\rm{m}}}\mathord{\left/{\vphantom{{{\rm{kg}}\cdot{\rm{m}}}{{{\rm{s}}^2}}}}\right\{{{\rm{s}}^2}}}\\{W_{{\rm{He}}}}=2681.1\;{\rm{N}}\end{aligned}

Force experienced by the balloon due to the weight of the cargo attached to the balloon. It can be represented by W_He.

W_cargo = m_cargo . g

Now, considering the spherical balloon in its equilibrium condition, we have,

\begin{aligned}F-{W_{{\rm{balloon}}}}-{W_{{\rm{He}}}}-{W_{{\rm{cargo}}}}=0\\{m_{{\rm{cargo}}}}\cdot g=F-{W_{{\rm{balloon}}}}-{W_{{\rm{He}}}}\end{aligned}

Substitute the values of all forces in the above formula.

\begin{aligned}{m_{cargo}}\cdot g=18399.73-9123.3-2681.1\;{\rm{N}}\\{m_{cargo}}=\frac{{6595.33\;{\rm{N}}}}{g}\\{m_{cargo}}=\frac{{6595.33\;{{{\rm{kg}}\cdot{\rm{m}}}\mathord{\left/{\vphantom{{{\rm{kg}}\cdot{\rm{m}}}{{{\rm{s}}^2}}}}\right\{{{\rm{s}}^2}}}}}{{9.81\;{{\rm{m}}\mathord{\left/{\vphantom{{\rm{m}}{{{\rm{s}}^2}}}}\right\{{{\rm{s}}^2}}}}}\\{m_{carg o}}=672.31\;{\rm{kg}}\end{aligned}

Hence, the mass of the spherical balloon's cargo can lift is 672.31 Kg.