The number of times terminal speed greater than for steel sphere that of the wooden sphere.

Question

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Answer 1

Question

Chapter 3, Problem 92P

(a)

To determine

The number of times terminal speed greater than for steel sphere that of the wooden sphere.

(a)

Expert Solution

Answer to Problem 92P

The steel’s sphere terminal speed is 4 times greater than for the velocity of wooden sphere.

Explanation of Solution

Write the expression for terminal velocity of the object.

$vterm=2mgρA$ (I)

Here, $m$ is the mass, $g$ is the acceleration due to gravity, $ρ$ is the density of air, and $A$ is the cross section area of the object.

The cross sectional area of the sphere is,

$A=πr2=π(d2)2$ (II)

Here, $r$ is the radius of the sphere and $d$ is the diameter of the sphere.

Rewrite the above equation for terminal velocity of wooden sphere.

$vterm(wood)=2mgρAwood$ (III)

Here, $Awood$ is the cross section area of the wooden sphere.

Rewrite the equation (I) for terminal velocity of steel sphere.

$vterm(steel)=2mgρAsteel$ (IV)

Here, $Asteel$ is the cross section area of the steel sphere.

Conclusion:

Solve the equation (III) and (IV).

$vterm(wood)vterm(steel)=2mgρAwood2mgρAsteel=AsteelAwood$

Rewrite the above relation by using equation (2).

$vterm(wood)vterm(steel)=π(dsteel2)2π(dwood2)2=(dsteel)2(dwood)2=dsteeldwood$

Substitute $40 cm$ for $dwood$ and $10 cm$ for $dsteel$ in above relation.

$vterm(wood)vterm(steel)=10 cm40 cmvterm(wood)vterm(steel)=14vterm(wood)=14vterm(steel)$

Therefore, the steel’s sphere terminal speed is 4 times greater than for the velocity of wooden sphere, due to its smaller cross sectional area.

(b)

To determine

The ratio of the terminal speed of the steel sphere to that of wooden sphere.

(b)

Expert Solution

Answer to Problem 92P

The terminal velocity of the steel sphere is twice the velocity of the wooden sphere due to its higher mass.

Explanation of Solution

From part (a),

The ratio of the terminal speed of the steel sphere to that of wooden sphere for different mass of the sphere with same cross sectional area is,

$vterm(wood)vterm(steel)=2mwoodgρA2msteelgρA=mwoodmsteel$

Conclusion:

Substitute $40 kg$ for $msteel$ and $10 kg$ for $mwood$ in above relation.

$vterm(wood)vterm(steel)=10 kg40 kgvterm(wood)vterm(steel)=12vterm(wood)=12vterm(steel)$

Therefore, the terminal velocity of the steel sphere is twice the velocity of the wooden sphere due to its higher mass.

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Answer 2

Question

Chapter 3, Problem 92P

(a)

To determine

The number of times terminal speed greater than for steel sphere that of the wooden sphere.

(a)

Expert Solution

Answer to Problem 92P

The steel’s sphere terminal speed is 4 times greater than for the velocity of wooden sphere.

Explanation of Solution

Write the expression for terminal velocity of the object.

$vterm=2mgρA$ (I)

Here, $m$ is the mass, $g$ is the acceleration due to gravity, $ρ$ is the density of air, and $A$ is the cross section area of the object.

The cross sectional area of the sphere is,

$A=πr2=π(d2)2$ (II)

Here, $r$ is the radius of the sphere and $d$ is the diameter of the sphere.

Rewrite the above equation for terminal velocity of wooden sphere.

$vterm(wood)=2mgρAwood$ (III)

Here, $Awood$ is the cross section area of the wooden sphere.

Rewrite the equation (I) for terminal velocity of steel sphere.

$vterm(steel)=2mgρAsteel$ (IV)

Here, $Asteel$ is the cross section area of the steel sphere.

Conclusion:

Solve the equation (III) and (IV).

$vterm(wood)vterm(steel)=2mgρAwood2mgρAsteel=AsteelAwood$

Rewrite the above relation by using equation (2).

$vterm(wood)vterm(steel)=π(dsteel2)2π(dwood2)2=(dsteel)2(dwood)2=dsteeldwood$

Substitute $40 cm$ for $dwood$ and $10 cm$ for $dsteel$ in above relation.

$vterm(wood)vterm(steel)=10 cm40 cmvterm(wood)vterm(steel)=14vterm(wood)=14vterm(steel)$

Therefore, the steel’s sphere terminal speed is 4 times greater than for the velocity of wooden sphere, due to its smaller cross sectional area.

(b)

To determine

The ratio of the terminal speed of the steel sphere to that of wooden sphere.

(b)

Expert Solution

Answer to Problem 92P

The terminal velocity of the steel sphere is twice the velocity of the wooden sphere due to its higher mass.

Explanation of Solution

From part (a),

The ratio of the terminal speed of the steel sphere to that of wooden sphere for different mass of the sphere with same cross sectional area is,

$vterm(wood)vterm(steel)=2mwoodgρA2msteelgρA=mwoodmsteel$

Conclusion:

Substitute $40 kg$ for $msteel$ and $10 kg$ for $mwood$ in above relation.

$vterm(wood)vterm(steel)=10 kg40 kgvterm(wood)vterm(steel)=12vterm(wood)=12vterm(steel)$

Therefore, the terminal velocity of the steel sphere is twice the velocity of the wooden sphere due to its higher mass.

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Answer 3

Question

Chapter 3, Problem 92P

(a)

To determine

The number of times terminal speed greater than for steel sphere that of the wooden sphere.

(a)

Expert Solution

Answer to Problem 92P

The steel’s sphere terminal speed is 4 times greater than for the velocity of wooden sphere.

Explanation of Solution

Write the expression for terminal velocity of the object.

$vterm=2mgρA$ (I)

Here, $m$ is the mass, $g$ is the acceleration due to gravity, $ρ$ is the density of air, and $A$ is the cross section area of the object.

The cross sectional area of the sphere is,

$A=πr2=π(d2)2$ (II)

Here, $r$ is the radius of the sphere and $d$ is the diameter of the sphere.

Rewrite the above equation for terminal velocity of wooden sphere.

$vterm(wood)=2mgρAwood$ (III)

Here, $Awood$ is the cross section area of the wooden sphere.

Rewrite the equation (I) for terminal velocity of steel sphere.

$vterm(steel)=2mgρAsteel$ (IV)

Here, $Asteel$ is the cross section area of the steel sphere.

Conclusion:

Solve the equation (III) and (IV).

$vterm(wood)vterm(steel)=2mgρAwood2mgρAsteel=AsteelAwood$

Rewrite the above relation by using equation (2).

$vterm(wood)vterm(steel)=π(dsteel2)2π(dwood2)2=(dsteel)2(dwood)2=dsteeldwood$

Substitute $40 cm$ for $dwood$ and $10 cm$ for $dsteel$ in above relation.

$vterm(wood)vterm(steel)=10 cm40 cmvterm(wood)vterm(steel)=14vterm(wood)=14vterm(steel)$

Therefore, the steel’s sphere terminal speed is 4 times greater than for the velocity of wooden sphere, due to its smaller cross sectional area.

(b)

To determine

The ratio of the terminal speed of the steel sphere to that of wooden sphere.

(b)

Expert Solution

Answer to Problem 92P

The terminal velocity of the steel sphere is twice the velocity of the wooden sphere due to its higher mass.

Explanation of Solution

From part (a),

The ratio of the terminal speed of the steel sphere to that of wooden sphere for different mass of the sphere with same cross sectional area is,

$vterm(wood)vterm(steel)=2mwoodgρA2msteelgρA=mwoodmsteel$

Conclusion:

Substitute $40 kg$ for $msteel$ and $10 kg$ for $mwood$ in above relation.

$vterm(wood)vterm(steel)=10 kg40 kgvterm(wood)vterm(steel)=12vterm(wood)=12vterm(steel)$

Therefore, the terminal velocity of the steel sphere is twice the velocity of the wooden sphere due to its higher mass.

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Answer 4

Question

Chapter 3, Problem 92P

(a)

To determine

The number of times terminal speed greater than for steel sphere that of the wooden sphere.

(a)

Expert Solution

Answer to Problem 92P

The steel’s sphere terminal speed is 4 times greater than for the velocity of wooden sphere.

Explanation of Solution

Write the expression for terminal velocity of the object.

$vterm=2mgρA$ (I)

Here, $m$ is the mass, $g$ is the acceleration due to gravity, $ρ$ is the density of air, and $A$ is the cross section area of the object.

The cross sectional area of the sphere is,

$A=πr2=π(d2)2$ (II)

Here, $r$ is the radius of the sphere and $d$ is the diameter of the sphere.

Rewrite the above equation for terminal velocity of wooden sphere.

$vterm(wood)=2mgρAwood$ (III)

Here, $Awood$ is the cross section area of the wooden sphere.

Rewrite the equation (I) for terminal velocity of steel sphere.

$vterm(steel)=2mgρAsteel$ (IV)

Here, $Asteel$ is the cross section area of the steel sphere.

Conclusion:

Solve the equation (III) and (IV).

$vterm(wood)vterm(steel)=2mgρAwood2mgρAsteel=AsteelAwood$

Rewrite the above relation by using equation (2).

$vterm(wood)vterm(steel)=π(dsteel2)2π(dwood2)2=(dsteel)2(dwood)2=dsteeldwood$

Substitute $40 cm$ for $dwood$ and $10 cm$ for $dsteel$ in above relation.

$vterm(wood)vterm(steel)=10 cm40 cmvterm(wood)vterm(steel)=14vterm(wood)=14vterm(steel)$

Therefore, the steel’s sphere terminal speed is 4 times greater than for the velocity of wooden sphere, due to its smaller cross sectional area.

(b)

To determine

The ratio of the terminal speed of the steel sphere to that of wooden sphere.

(b)

Expert Solution

Answer to Problem 92P

The terminal velocity of the steel sphere is twice the velocity of the wooden sphere due to its higher mass.

Explanation of Solution

From part (a),

The ratio of the terminal speed of the steel sphere to that of wooden sphere for different mass of the sphere with same cross sectional area is,

$vterm(wood)vterm(steel)=2mwoodgρA2msteelgρA=mwoodmsteel$

Conclusion:

Substitute $40 kg$ for $msteel$ and $10 kg$ for $mwood$ in above relation.

$vterm(wood)vterm(steel)=10 kg40 kgvterm(wood)vterm(steel)=12vterm(wood)=12vterm(steel)$

Therefore, the terminal velocity of the steel sphere is twice the velocity of the wooden sphere due to its higher mass.

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Answer 5

Chapter 3, Problem 92P

(a)

To determine

The number of times terminal speed greater than for steel sphere that of the wooden sphere.

(a)

Expert Solution

Answer to Problem 92P

The steel’s sphere terminal speed is 4 times greater than for the velocity of wooden sphere.

Explanation of Solution

Write the expression for terminal velocity of the object.

$vterm=2mgρA$ (I)

Here, $m$ is the mass, $g$ is the acceleration due to gravity, $ρ$ is the density of air, and $A$ is the cross section area of the object.

The cross sectional area of the sphere is,

$A=πr2=π(d2)2$ (II)

Here, $r$ is the radius of the sphere and $d$ is the diameter of the sphere.

Rewrite the above equation for terminal velocity of wooden sphere.

$vterm(wood)=2mgρAwood$ (III)

Here, $Awood$ is the cross section area of the wooden sphere.

Rewrite the equation (I) for terminal velocity of steel sphere.

$vterm(steel)=2mgρAsteel$ (IV)

Here, $Asteel$ is the cross section area of the steel sphere.

Conclusion:

Solve the equation (III) and (IV).

$vterm(wood)vterm(steel)=2mgρAwood2mgρAsteel=AsteelAwood$

Rewrite the above relation by using equation (2).

$vterm(wood)vterm(steel)=π(dsteel2)2π(dwood2)2=(dsteel)2(dwood)2=dsteeldwood$

Substitute $40 cm$ for $dwood$ and $10 cm$ for $dsteel$ in above relation.

$vterm(wood)vterm(steel)=10 cm40 cmvterm(wood)vterm(steel)=14vterm(wood)=14vterm(steel)$

Therefore, the steel’s sphere terminal speed is 4 times greater than for the velocity of wooden sphere, due to its smaller cross sectional area.

(b)

To determine

The ratio of the terminal speed of the steel sphere to that of wooden sphere.

(b)

Expert Solution

Answer to Problem 92P

The terminal velocity of the steel sphere is twice the velocity of the wooden sphere due to its higher mass.

Explanation of Solution

From part (a),

The ratio of the terminal speed of the steel sphere to that of wooden sphere for different mass of the sphere with same cross sectional area is,

$vterm(wood)vterm(steel)=2mwoodgρA2msteelgρA=mwoodmsteel$

Conclusion:

Substitute $40 kg$ for $msteel$ and $10 kg$ for $mwood$ in above relation.

$vterm(wood)vterm(steel)=10 kg40 kgvterm(wood)vterm(steel)=12vterm(wood)=12vterm(steel)$

Therefore, the terminal velocity of the steel sphere is twice the velocity of the wooden sphere due to its higher mass.

Answer 6

Chapter 3, Problem 92P

(a)

To determine

The number of times terminal speed greater than for steel sphere that of the wooden sphere.

(a)

Expert Solution

Answer to Problem 92P

The steel’s sphere terminal speed is 4 times greater than for the velocity of wooden sphere.

Explanation of Solution

Write the expression for terminal velocity of the object.

$vterm=2mgρA$ (I)

Here, $m$ is the mass, $g$ is the acceleration due to gravity, $ρ$ is the density of air, and $A$ is the cross section area of the object.

The cross sectional area of the sphere is,

$A=πr2=π(d2)2$ (II)

Here, $r$ is the radius of the sphere and $d$ is the diameter of the sphere.

Rewrite the above equation for terminal velocity of wooden sphere.

$vterm(wood)=2mgρAwood$ (III)

Here, $Awood$ is the cross section area of the wooden sphere.

Rewrite the equation (I) for terminal velocity of steel sphere.

$vterm(steel)=2mgρAsteel$ (IV)

Here, $Asteel$ is the cross section area of the steel sphere.

Conclusion:

Solve the equation (III) and (IV).

$vterm(wood)vterm(steel)=2mgρAwood2mgρAsteel=AsteelAwood$

Rewrite the above relation by using equation (2).

$vterm(wood)vterm(steel)=π(dsteel2)2π(dwood2)2=(dsteel)2(dwood)2=dsteeldwood$

Substitute $40 cm$ for $dwood$ and $10 cm$ for $dsteel$ in above relation.

$vterm(wood)vterm(steel)=10 cm40 cmvterm(wood)vterm(steel)=14vterm(wood)=14vterm(steel)$

Therefore, the steel’s sphere terminal speed is 4 times greater than for the velocity of wooden sphere, due to its smaller cross sectional area.

Answer 7

Chapter 3, Problem 92P

(a)

To determine

The number of times terminal speed greater than for steel sphere that of the wooden sphere.

Answer 8

(a)

Expert Solution

Answer to Problem 92P

The steel’s sphere terminal speed is 4 times greater than for the velocity of wooden sphere.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Write the expression for terminal velocity of the object.

$vterm=2mgρA$ (I)

Here, $m$ is the mass, $g$ is the acceleration due to gravity, $ρ$ is the density of air, and $A$ is the cross section area of the object.

The cross sectional area of the sphere is,

$A=πr2=π(d2)2$ (II)

Here, $r$ is the radius of the sphere and $d$ is the diameter of the sphere.

Rewrite the above equation for terminal velocity of wooden sphere.

$vterm(wood)=2mgρAwood$ (III)

Here, $Awood$ is the cross section area of the wooden sphere.

Rewrite the equation (I) for terminal velocity of steel sphere.

$vterm(steel)=2mgρAsteel$ (IV)

Here, $Asteel$ is the cross section area of the steel sphere.

Conclusion:

Solve the equation (III) and (IV).

$vterm(wood)vterm(steel)=2mgρAwood2mgρAsteel=AsteelAwood$

Rewrite the above relation by using equation (2).

$vterm(wood)vterm(steel)=π(dsteel2)2π(dwood2)2=(dsteel)2(dwood)2=dsteeldwood$

Substitute $40 cm$ for $dwood$ and $10 cm$ for $dsteel$ in above relation.

$vterm(wood)vterm(steel)=10 cm40 cmvterm(wood)vterm(steel)=14vterm(wood)=14vterm(steel)$

Therefore, the steel’s sphere terminal speed is 4 times greater than for the velocity of wooden sphere, due to its smaller cross sectional area.

Answer 13

(b)

To determine

The ratio of the terminal speed of the steel sphere to that of wooden sphere.

(b)

Expert Solution

Answer to Problem 92P

The terminal velocity of the steel sphere is twice the velocity of the wooden sphere due to its higher mass.

Explanation of Solution

From part (a),

The ratio of the terminal speed of the steel sphere to that of wooden sphere for different mass of the sphere with same cross sectional area is,

$vterm(wood)vterm(steel)=2mwoodgρA2msteelgρA=mwoodmsteel$

Conclusion:

Substitute $40 kg$ for $msteel$ and $10 kg$ for $mwood$ in above relation.

$vterm(wood)vterm(steel)=10 kg40 kgvterm(wood)vterm(steel)=12vterm(wood)=12vterm(steel)$

Therefore, the terminal velocity of the steel sphere is twice the velocity of the wooden sphere due to its higher mass.

Answer 14

(b)

To determine

The ratio of the terminal speed of the steel sphere to that of wooden sphere.

Answer 15

(b)

Expert Solution

Answer to Problem 92P

The terminal velocity of the steel sphere is twice the velocity of the wooden sphere due to its higher mass.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

From part (a),

The ratio of the terminal speed of the steel sphere to that of wooden sphere for different mass of the sphere with same cross sectional area is,

$vterm(wood)vterm(steel)=2mwoodgρA2msteelgρA=mwoodmsteel$

Conclusion:

Substitute $40 kg$ for $msteel$ and $10 kg$ for $mwood$ in above relation.

$vterm(wood)vterm(steel)=10 kg40 kgvterm(wood)vterm(steel)=12vterm(wood)=12vterm(steel)$

Therefore, the terminal velocity of the steel sphere is twice the velocity of the wooden sphere due to its higher mass.

Answer 20

Ch. 3.2 - Prob. 3.1CC Ch. 3.3 - Prob. 3.2CC Ch. 3.4 - Prob. 3.3CC Ch. 3.4 - Prob. 3.4CC Ch. 3.5 - Prob. 3.5CC Ch. 3.6 - Prob. 3.6CC Ch. 3.7 - Acceleration of a Skydiver Figure 3.27 shows a...Ch. 3 - Prob. 1Q Ch. 3 - Prob. 2Q Ch. 3 - Prob. 3Q

The number of times terminal speed greater than for steel sphere that of the wooden sphere.

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The number of times terminal speed greater than for steel sphere that of the wooden sphere.

Answer to Problem 92P

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The ratio of the terminal speed of the steel sphere to that of wooden sphere.

Answer to Problem 92P

Explanation of Solution

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