Balloons are often filled with helium gas because it weighs only about one-seventh of what air weighs under identical conditions. The buoyancy force which can be expressed as F b = ρ air gV balloon , will push the balloon upward. If the balloon has a diameter of 12 m and carries two people 85 kg each determine the acceleration of the balloon when it is first released. Assume the density of air is ρ = 1 16 kg/m 3 , and neglect the weight of the ropes and the cage. Answer : 22.4 m/s 2

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Answer 1

Textbook Question

Chapter 1.11, Problem 101RP

Balloons are often filled with helium gas because it weighs only about one-seventh of what air weighs under identical conditions. The buoyancy force which can be expressed as F_b = ρ_airgV_balloon, will push the balloon upward. If the balloon has a diameter of 12 m and carries two people 85 kg each determine the acceleration of the balloon when it is first released. Assume the density of air is ρ = 1 16 kg/m³, and neglect the weight of the ropes and the cage. Answer: 22.4 m/s²

Chapter 1.11, Problem 101RP, Balloons are often filled with helium gas because it weighs only about one-seventh of what air

Expert Solution & Answer

To determine

The acceleration of the balloon when it is first released.

Answer to Problem 101RP

The acceleration of the balloon when it is first released is $22.4 m/s2_$ .

Explanation of Solution

Show the free body diagram of the balloon.

THERMODYNAMICS CONNECT CARD >I<, Chapter 1.11, Problem 101RP

Write the expression of volume of sphere of a balloon.

$Vballoon=4πr3/3$ (I)

Write the expression of the buoyancy force acting on the balloon.

$FB=ρairgVballoon$ (II)

Here, the density of air is $ρH2O$ , the volume of balloon is $Vballoon$ , and the acceleration of gravity is $g$ .

Write the expression of the mass of helium.

$mHe=ρHeVballoon$ (III)

Here, the density of the helium is $ρHe$ and volume of the balloon is $Vballoon$ .

Write the expression of the total mass can carried by balloon.

$mtotal=mHe+mpeople$ (IV)

Here, the total mass of a people is $mpeople$ and the mass of helium is $mHe$ .

Write the expression of the total weight.

$W=mtotalg$ (V)

Write the expression of net force acting on the balloon.

$Fnet=FB−W$ (VI)

Write the expression of acceleration.

$a=Fnetmtotal$ (VII)

Conclusion:

Substitute $6 m$ for $r$ in Equation (I).

$Vballoon=4π(6 m)3/3=904.8 m3$ .

Substitute $1.16 kg/m3$ for $ρair$ , $9.81 m/s2$ for $g$ , and $904.8 m3$ for $Vballoon$ in Equation (II).

$FB=(1.16 kg/m3)(9.81 m/s2)(904.8 m3)=10296 kg⋅m/s2×(1 N1 kg⋅m/s2)=10296 N$

Substitute $1.16/7 kg/m3$ for $ρHe$ and $904.8 m3$ for $Vballoon$ in Equation (III).

$mHe=(1.167 kg/m3)(904.8 m3)=149.9 kg$ .

Substitute $2×85 kg$ for $mpeople$ and $149.9 kg$ for $mHe$ in Equation (V)

$mtotal=(149.9 kg)+(2×85 kg)=319.9 kg$

Substitute $319.9 kg$ for $mtotal$ and $9.81 m/s2$ for $g$ in Equation (VI).

$W=(319.9 kg)(9.81 m/s2)=3138 kg⋅m/s2×(1 N1 kg⋅m/s2)=3138 N$

Substitute $10296 N$ for $FB$ and $3138 N$ for $W$ in Equation (VII).

$Fnet=10296 N−3138 N=7157 N$ .

Substitute $7157 N$ for $Fnet$ and $319.9 kg$ for $mtotal$ in Equation (VIII).

$a=(7157 N)(319.9 kg)=(7157 N)(319.9 kg)×(1 kg⋅m/s21 N)=22.4 m/s2$

Thus, the acceleration of the balloon when it is first released is $22.4 m/s2_$ .

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Answer 2

Textbook Question

Chapter 1.11, Problem 101RP

Balloons are often filled with helium gas because it weighs only about one-seventh of what air weighs under identical conditions. The buoyancy force which can be expressed as F_b = ρ_airgV_balloon, will push the balloon upward. If the balloon has a diameter of 12 m and carries two people 85 kg each determine the acceleration of the balloon when it is first released. Assume the density of air is ρ = 1 16 kg/m³, and neglect the weight of the ropes and the cage. Answer: 22.4 m/s²

Chapter 1.11, Problem 101RP, Balloons are often filled with helium gas because it weighs only about one-seventh of what air

Expert Solution & Answer

To determine

The acceleration of the balloon when it is first released.

Answer to Problem 101RP

The acceleration of the balloon when it is first released is $22.4 m/s2_$ .

Explanation of Solution

Show the free body diagram of the balloon.

THERMODYNAMICS CONNECT CARD >I<, Chapter 1.11, Problem 101RP

Write the expression of volume of sphere of a balloon.

$Vballoon=4πr3/3$ (I)

Write the expression of the buoyancy force acting on the balloon.

$FB=ρairgVballoon$ (II)

Here, the density of air is $ρH2O$ , the volume of balloon is $Vballoon$ , and the acceleration of gravity is $g$ .

Write the expression of the mass of helium.

$mHe=ρHeVballoon$ (III)

Here, the density of the helium is $ρHe$ and volume of the balloon is $Vballoon$ .

Write the expression of the total mass can carried by balloon.

$mtotal=mHe+mpeople$ (IV)

Here, the total mass of a people is $mpeople$ and the mass of helium is $mHe$ .

Write the expression of the total weight.

$W=mtotalg$ (V)

Write the expression of net force acting on the balloon.

$Fnet=FB−W$ (VI)

Write the expression of acceleration.

$a=Fnetmtotal$ (VII)

Conclusion:

Substitute $6 m$ for $r$ in Equation (I).

$Vballoon=4π(6 m)3/3=904.8 m3$ .

Substitute $1.16 kg/m3$ for $ρair$ , $9.81 m/s2$ for $g$ , and $904.8 m3$ for $Vballoon$ in Equation (II).

$FB=(1.16 kg/m3)(9.81 m/s2)(904.8 m3)=10296 kg⋅m/s2×(1 N1 kg⋅m/s2)=10296 N$

Substitute $1.16/7 kg/m3$ for $ρHe$ and $904.8 m3$ for $Vballoon$ in Equation (III).

$mHe=(1.167 kg/m3)(904.8 m3)=149.9 kg$ .

Substitute $2×85 kg$ for $mpeople$ and $149.9 kg$ for $mHe$ in Equation (V)

$mtotal=(149.9 kg)+(2×85 kg)=319.9 kg$

Substitute $319.9 kg$ for $mtotal$ and $9.81 m/s2$ for $g$ in Equation (VI).

$W=(319.9 kg)(9.81 m/s2)=3138 kg⋅m/s2×(1 N1 kg⋅m/s2)=3138 N$

Substitute $10296 N$ for $FB$ and $3138 N$ for $W$ in Equation (VII).

$Fnet=10296 N−3138 N=7157 N$ .

Substitute $7157 N$ for $Fnet$ and $319.9 kg$ for $mtotal$ in Equation (VIII).

$a=(7157 N)(319.9 kg)=(7157 N)(319.9 kg)×(1 kg⋅m/s21 N)=22.4 m/s2$

Thus, the acceleration of the balloon when it is first released is $22.4 m/s2_$ .

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Answer 3

Textbook Question

Chapter 1.11, Problem 101RP

Balloons are often filled with helium gas because it weighs only about one-seventh of what air weighs under identical conditions. The buoyancy force which can be expressed as F_b = ρ_airgV_balloon, will push the balloon upward. If the balloon has a diameter of 12 m and carries two people 85 kg each determine the acceleration of the balloon when it is first released. Assume the density of air is ρ = 1 16 kg/m³, and neglect the weight of the ropes and the cage. Answer: 22.4 m/s²

Chapter 1.11, Problem 101RP, Balloons are often filled with helium gas because it weighs only about one-seventh of what air

Expert Solution & Answer

To determine

The acceleration of the balloon when it is first released.

Answer to Problem 101RP

The acceleration of the balloon when it is first released is $22.4 m/s2_$ .

Explanation of Solution

Show the free body diagram of the balloon.

THERMODYNAMICS CONNECT CARD >I<, Chapter 1.11, Problem 101RP

Write the expression of volume of sphere of a balloon.

$Vballoon=4πr3/3$ (I)

Write the expression of the buoyancy force acting on the balloon.

$FB=ρairgVballoon$ (II)

Here, the density of air is $ρH2O$ , the volume of balloon is $Vballoon$ , and the acceleration of gravity is $g$ .

Write the expression of the mass of helium.

$mHe=ρHeVballoon$ (III)

Here, the density of the helium is $ρHe$ and volume of the balloon is $Vballoon$ .

Write the expression of the total mass can carried by balloon.

$mtotal=mHe+mpeople$ (IV)

Here, the total mass of a people is $mpeople$ and the mass of helium is $mHe$ .

Write the expression of the total weight.

$W=mtotalg$ (V)

Write the expression of net force acting on the balloon.

$Fnet=FB−W$ (VI)

Write the expression of acceleration.

$a=Fnetmtotal$ (VII)

Conclusion:

Substitute $6 m$ for $r$ in Equation (I).

$Vballoon=4π(6 m)3/3=904.8 m3$ .

Substitute $1.16 kg/m3$ for $ρair$ , $9.81 m/s2$ for $g$ , and $904.8 m3$ for $Vballoon$ in Equation (II).

$FB=(1.16 kg/m3)(9.81 m/s2)(904.8 m3)=10296 kg⋅m/s2×(1 N1 kg⋅m/s2)=10296 N$

Substitute $1.16/7 kg/m3$ for $ρHe$ and $904.8 m3$ for $Vballoon$ in Equation (III).

$mHe=(1.167 kg/m3)(904.8 m3)=149.9 kg$ .

Substitute $2×85 kg$ for $mpeople$ and $149.9 kg$ for $mHe$ in Equation (V)

$mtotal=(149.9 kg)+(2×85 kg)=319.9 kg$

Substitute $319.9 kg$ for $mtotal$ and $9.81 m/s2$ for $g$ in Equation (VI).

$W=(319.9 kg)(9.81 m/s2)=3138 kg⋅m/s2×(1 N1 kg⋅m/s2)=3138 N$

Substitute $10296 N$ for $FB$ and $3138 N$ for $W$ in Equation (VII).

$Fnet=10296 N−3138 N=7157 N$ .

Substitute $7157 N$ for $Fnet$ and $319.9 kg$ for $mtotal$ in Equation (VIII).

$a=(7157 N)(319.9 kg)=(7157 N)(319.9 kg)×(1 kg⋅m/s21 N)=22.4 m/s2$

Thus, the acceleration of the balloon when it is first released is $22.4 m/s2_$ .

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Answer 4

Textbook Question

Chapter 1.11, Problem 101RP

Balloons are often filled with helium gas because it weighs only about one-seventh of what air weighs under identical conditions. The buoyancy force which can be expressed as F_b = ρ_airgV_balloon, will push the balloon upward. If the balloon has a diameter of 12 m and carries two people 85 kg each determine the acceleration of the balloon when it is first released. Assume the density of air is ρ = 1 16 kg/m³, and neglect the weight of the ropes and the cage. Answer: 22.4 m/s²

Chapter 1.11, Problem 101RP, Balloons are often filled with helium gas because it weighs only about one-seventh of what air

Expert Solution & Answer

To determine

The acceleration of the balloon when it is first released.

Answer to Problem 101RP

The acceleration of the balloon when it is first released is $22.4 m/s2_$ .

Explanation of Solution

Show the free body diagram of the balloon.

THERMODYNAMICS CONNECT CARD >I<, Chapter 1.11, Problem 101RP

Write the expression of volume of sphere of a balloon.

$Vballoon=4πr3/3$ (I)

Write the expression of the buoyancy force acting on the balloon.

$FB=ρairgVballoon$ (II)

Here, the density of air is $ρH2O$ , the volume of balloon is $Vballoon$ , and the acceleration of gravity is $g$ .

Write the expression of the mass of helium.

$mHe=ρHeVballoon$ (III)

Here, the density of the helium is $ρHe$ and volume of the balloon is $Vballoon$ .

Write the expression of the total mass can carried by balloon.

$mtotal=mHe+mpeople$ (IV)

Here, the total mass of a people is $mpeople$ and the mass of helium is $mHe$ .

Write the expression of the total weight.

$W=mtotalg$ (V)

Write the expression of net force acting on the balloon.

$Fnet=FB−W$ (VI)

Write the expression of acceleration.

$a=Fnetmtotal$ (VII)

Conclusion:

Substitute $6 m$ for $r$ in Equation (I).

$Vballoon=4π(6 m)3/3=904.8 m3$ .

Substitute $1.16 kg/m3$ for $ρair$ , $9.81 m/s2$ for $g$ , and $904.8 m3$ for $Vballoon$ in Equation (II).

$FB=(1.16 kg/m3)(9.81 m/s2)(904.8 m3)=10296 kg⋅m/s2×(1 N1 kg⋅m/s2)=10296 N$

Substitute $1.16/7 kg/m3$ for $ρHe$ and $904.8 m3$ for $Vballoon$ in Equation (III).

$mHe=(1.167 kg/m3)(904.8 m3)=149.9 kg$ .

Substitute $2×85 kg$ for $mpeople$ and $149.9 kg$ for $mHe$ in Equation (V)

$mtotal=(149.9 kg)+(2×85 kg)=319.9 kg$

Substitute $319.9 kg$ for $mtotal$ and $9.81 m/s2$ for $g$ in Equation (VI).

$W=(319.9 kg)(9.81 m/s2)=3138 kg⋅m/s2×(1 N1 kg⋅m/s2)=3138 N$

Substitute $10296 N$ for $FB$ and $3138 N$ for $W$ in Equation (VII).

$Fnet=10296 N−3138 N=7157 N$ .

Substitute $7157 N$ for $Fnet$ and $319.9 kg$ for $mtotal$ in Equation (VIII).

$a=(7157 N)(319.9 kg)=(7157 N)(319.9 kg)×(1 kg⋅m/s21 N)=22.4 m/s2$

Thus, the acceleration of the balloon when it is first released is $22.4 m/s2_$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

The acceleration of the balloon when it is first released.

Answer to Problem 101RP

The acceleration of the balloon when it is first released is $22.4 m/s2_$ .

Explanation of Solution

Show the free body diagram of the balloon.

THERMODYNAMICS CONNECT CARD >I<, Chapter 1.11, Problem 101RP

Write the expression of volume of sphere of a balloon.

$Vballoon=4πr3/3$ (I)

Write the expression of the buoyancy force acting on the balloon.

$FB=ρairgVballoon$ (II)

Here, the density of air is $ρH2O$ , the volume of balloon is $Vballoon$ , and the acceleration of gravity is $g$ .

Write the expression of the mass of helium.

$mHe=ρHeVballoon$ (III)

Here, the density of the helium is $ρHe$ and volume of the balloon is $Vballoon$ .

Write the expression of the total mass can carried by balloon.

$mtotal=mHe+mpeople$ (IV)

Here, the total mass of a people is $mpeople$ and the mass of helium is $mHe$ .

Write the expression of the total weight.

$W=mtotalg$ (V)

Write the expression of net force acting on the balloon.

$Fnet=FB−W$ (VI)

Write the expression of acceleration.

$a=Fnetmtotal$ (VII)

Conclusion:

Substitute $6 m$ for $r$ in Equation (I).

$Vballoon=4π(6 m)3/3=904.8 m3$ .

Substitute $1.16 kg/m3$ for $ρair$ , $9.81 m/s2$ for $g$ , and $904.8 m3$ for $Vballoon$ in Equation (II).

$FB=(1.16 kg/m3)(9.81 m/s2)(904.8 m3)=10296 kg⋅m/s2×(1 N1 kg⋅m/s2)=10296 N$

Substitute $1.16/7 kg/m3$ for $ρHe$ and $904.8 m3$ for $Vballoon$ in Equation (III).

$mHe=(1.167 kg/m3)(904.8 m3)=149.9 kg$ .

Substitute $2×85 kg$ for $mpeople$ and $149.9 kg$ for $mHe$ in Equation (V)

$mtotal=(149.9 kg)+(2×85 kg)=319.9 kg$

Substitute $319.9 kg$ for $mtotal$ and $9.81 m/s2$ for $g$ in Equation (VI).

$W=(319.9 kg)(9.81 m/s2)=3138 kg⋅m/s2×(1 N1 kg⋅m/s2)=3138 N$

Substitute $10296 N$ for $FB$ and $3138 N$ for $W$ in Equation (VII).

$Fnet=10296 N−3138 N=7157 N$ .

Substitute $7157 N$ for $Fnet$ and $319.9 kg$ for $mtotal$ in Equation (VIII).

$a=(7157 N)(319.9 kg)=(7157 N)(319.9 kg)×(1 kg⋅m/s21 N)=22.4 m/s2$

Thus, the acceleration of the balloon when it is first released is $22.4 m/s2_$ .

Answer 8

Expert Solution & Answer

To determine

The acceleration of the balloon when it is first released.

Answer to Problem 101RP

The acceleration of the balloon when it is first released is $22.4 m/s2_$ .

Explanation of Solution

Show the free body diagram of the balloon.

THERMODYNAMICS CONNECT CARD >I<, Chapter 1.11, Problem 101RP

Write the expression of volume of sphere of a balloon.

$Vballoon=4πr3/3$ (I)

Write the expression of the buoyancy force acting on the balloon.

$FB=ρairgVballoon$ (II)

Here, the density of air is $ρH2O$ , the volume of balloon is $Vballoon$ , and the acceleration of gravity is $g$ .

Write the expression of the mass of helium.

$mHe=ρHeVballoon$ (III)

Here, the density of the helium is $ρHe$ and volume of the balloon is $Vballoon$ .

Write the expression of the total mass can carried by balloon.

$mtotal=mHe+mpeople$ (IV)

Here, the total mass of a people is $mpeople$ and the mass of helium is $mHe$ .

Write the expression of the total weight.

$W=mtotalg$ (V)

Write the expression of net force acting on the balloon.

$Fnet=FB−W$ (VI)

Write the expression of acceleration.

$a=Fnetmtotal$ (VII)

Conclusion:

Substitute $6 m$ for $r$ in Equation (I).

$Vballoon=4π(6 m)3/3=904.8 m3$ .

Substitute $1.16 kg/m3$ for $ρair$ , $9.81 m/s2$ for $g$ , and $904.8 m3$ for $Vballoon$ in Equation (II).

$FB=(1.16 kg/m3)(9.81 m/s2)(904.8 m3)=10296 kg⋅m/s2×(1 N1 kg⋅m/s2)=10296 N$

Substitute $1.16/7 kg/m3$ for $ρHe$ and $904.8 m3$ for $Vballoon$ in Equation (III).

$mHe=(1.167 kg/m3)(904.8 m3)=149.9 kg$ .

Substitute $2×85 kg$ for $mpeople$ and $149.9 kg$ for $mHe$ in Equation (V)

$mtotal=(149.9 kg)+(2×85 kg)=319.9 kg$

Substitute $319.9 kg$ for $mtotal$ and $9.81 m/s2$ for $g$ in Equation (VI).

$W=(319.9 kg)(9.81 m/s2)=3138 kg⋅m/s2×(1 N1 kg⋅m/s2)=3138 N$

Substitute $10296 N$ for $FB$ and $3138 N$ for $W$ in Equation (VII).

$Fnet=10296 N−3138 N=7157 N$ .

Substitute $7157 N$ for $Fnet$ and $319.9 kg$ for $mtotal$ in Equation (VIII).

$a=(7157 N)(319.9 kg)=(7157 N)(319.9 kg)×(1 kg⋅m/s21 N)=22.4 m/s2$

Thus, the acceleration of the balloon when it is first released is $22.4 m/s2_$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

The acceleration of the balloon when it is first released.

Answer 12

Answer to Problem 101RP

The acceleration of the balloon when it is first released is $22.4 m/s2_$ .

Answer 13

Explanation of Solution

Show the free body diagram of the balloon.

THERMODYNAMICS CONNECT CARD >I<, Chapter 1.11, Problem 101RP

Write the expression of volume of sphere of a balloon.

$Vballoon=4πr3/3$ (I)

Write the expression of the buoyancy force acting on the balloon.

$FB=ρairgVballoon$ (II)

Here, the density of air is $ρH2O$ , the volume of balloon is $Vballoon$ , and the acceleration of gravity is $g$ .

Write the expression of the mass of helium.

$mHe=ρHeVballoon$ (III)

Here, the density of the helium is $ρHe$ and volume of the balloon is $Vballoon$ .

Write the expression of the total mass can carried by balloon.

$mtotal=mHe+mpeople$ (IV)

Here, the total mass of a people is $mpeople$ and the mass of helium is $mHe$ .

Write the expression of the total weight.

$W=mtotalg$ (V)

Write the expression of net force acting on the balloon.

$Fnet=FB−W$ (VI)

Write the expression of acceleration.

$a=Fnetmtotal$ (VII)

Conclusion:

Substitute $6 m$ for $r$ in Equation (I).

$Vballoon=4π(6 m)3/3=904.8 m3$ .

Substitute $1.16 kg/m3$ for $ρair$ , $9.81 m/s2$ for $g$ , and $904.8 m3$ for $Vballoon$ in Equation (II).

$FB=(1.16 kg/m3)(9.81 m/s2)(904.8 m3)=10296 kg⋅m/s2×(1 N1 kg⋅m/s2)=10296 N$

Substitute $1.16/7 kg/m3$ for $ρHe$ and $904.8 m3$ for $Vballoon$ in Equation (III).

$mHe=(1.167 kg/m3)(904.8 m3)=149.9 kg$ .

Substitute $2×85 kg$ for $mpeople$ and $149.9 kg$ for $mHe$ in Equation (V)

$mtotal=(149.9 kg)+(2×85 kg)=319.9 kg$

Substitute $319.9 kg$ for $mtotal$ and $9.81 m/s2$ for $g$ in Equation (VI).

$W=(319.9 kg)(9.81 m/s2)=3138 kg⋅m/s2×(1 N1 kg⋅m/s2)=3138 N$

Substitute $10296 N$ for $FB$ and $3138 N$ for $W$ in Equation (VII).

$Fnet=10296 N−3138 N=7157 N$ .

Substitute $7157 N$ for $Fnet$ and $319.9 kg$ for $mtotal$ in Equation (VIII).

$a=(7157 N)(319.9 kg)=(7157 N)(319.9 kg)×(1 kg⋅m/s21 N)=22.4 m/s2$

Thus, the acceleration of the balloon when it is first released is $22.4 m/s2_$ .

Answer 14

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Answer to Problem 101RP

Explanation of Solution

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