The units and dimensions of the constant b in the retarding force b υ n if ( a ) n = 1 .

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 5, Problem 27P

(a)

To determine

The units and dimensions of the constant $b$ in the retarding force $bυn$ if $(a)n=1$ .

(a)

Expert Solution

Answer to Problem 27P

Dimension of $b=MT−1$

The unit of $b is kg s-1$

Explanation of Solution

Given Information:

$n=1$

Formula Used:

The drag force equation is $Fd=bvn$

Calculation:

The drag force equation is $Fd=bvn$

Dimension of force $=[MLT−2]$

The dimension of speed $=[LT−1]$

Solving the drag force equation for $b$ with $n=1$

$b=Fdv$

Substitute the dimensions of $Fd$ and $v$ and simplify to obtain the dimension of $b$ as

$[b]=[ML T −2][L T −1][b]=MT−1$

So, the units ofb is $kg s-1$ .

(b)

To determine

Using dimensional analysis, determine the units and dimensions of the constant $b$ in the retarding force $bυn$ if $n=2$ .

(b)

Expert Solution

Answer to Problem 27P

Dimension of $b=ML−1$

The units of $b is kg.m-1$

Explanation of Solution

Given Information:

$n=2$

Calculation:

The drag force equation is $Fd=bvn$

Dimension of force $=[MLT−2]$

The dimension of speed $=[LT−1]$

Solving the drag force equation for $b$ with $n=2$ ,

$b=Fdv2$

Substituting the dimensions of $Fd$ and $v$ and simplify to obtain the dimension of $b$ as

$[b]=[ML T −2] [ L T −1 ]2=ML−1$

So, the units of $b are kg m−1$ .

(c)

To determine

To Show: Air resistance of a falling object with a circular cross section should be approximately $12ρπr2υ2, where ρ=1.20 kg/m3$ , the density of air, is dimensionally consistent.

(c)

Expert Solution

Explanation of Solution

Given Information:

Air resistance of a falling object with a circular cross section should be approximately $12ρπr2υ2, where ρ=1.20 kg/m3$ , the density of air.

Calculation:

According to the expression of drag force,

$Fd=12ρπr2υ2$

So the dimensions of drag force is

$[Fd]=[12ρπr2υ2]or =[ML−3][L]2[LT −1]2or =MLT−2$

The drag force is given as

$Fd=bv2$

The dimension of drag force is

$[Fd]=[bv2]=[ML−1][LT −1]2=MLT−2$

Conclusion:

Hence, the expression for drag force is dimensionallyconsistent .

(d)

To determine

To Find: The terminal speed of the skydiver.

(d)

Expert Solution

Answer to Problem 27P

$56.9 m/s$

Explanation of Solution

Given Information:

Mass of skydiver = $56 kg$

Disk of radius = $0.30 m$

Density of air near the surface of Earth is $120 kg/m3$

Formula Used:

The terminal speed is given by relation

$vt=2mgρπr2$

Calculation:

Mass $m=56 kg$

Acceleration due to gravity $g=9.81 m/s2$

Density $ρ=1.2 kg/m3$

Radius $r=0.3 m$

The terminal speed is given by relation

$vt= 2mg ρπ r 2 = 2( 56 kg ) ( 9.81m/s ) 2 π( 1.2 kg/m 3 ) ( 0.3m ) 2 =56.9 m/s$

(e)

To determine

To Calculate: The terminal velocity at the given height.

(e)

Expert Solution

Answer to Problem 27P

The terminal velocity is $86.9 m/s$

Explanation of Solution

Given Information:

Height = $8 km$

Density of air near the surface of Earth is $0.514kg/m3$

Formula Used:

The terminal speed is given by relation

$vt=2mgρπr2$

Calculation:

Mass, $m=56 kg$

Acceleration due to gravity, $g=9.81 m/s2$

Density, $ρ=0.514 kg/m3$

Radius $r=0.3 m$

Substitute the values:

$vt= 2mg ρπ r 2 = 2( 56 kg ) ( 9.81m/s ) 2 π( 0.514 kg/m 3 ) ( 0.3 m ) 2 =272.12 m/s$

Conclusion:

Thus, the terminal speed at the given height is $272.12 m/s$

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

(λvacuum =640nm) red light (λ vacuum = 640 nm) and green light perpendicularly on a soap film (n=1.31) A mixture of (a vacuum = 512 nm) shines that has air on both side. What is the minimum nonzero thickness of the film, so that destructive interference to look red in reflected light? nm Causes it

Suppose the inteference pattern shown in the figure below is produced by monochromatic light passing through a diffraction grating, that has 260 lines/mm, and onto a screen 1.40m away. What is the wavelength of light if the distance between the dashed lines is 180cm? nm

How many (whole) dark fringes will produced on on an infinitely large screen if red light (2)=700 nm) is incident on two slits that are 20.0 μm apart?

Answer 2

Question

Chapter 5, Problem 27P

(a)

To determine

The units and dimensions of the constant $b$ in the retarding force $bυn$ if $(a)n=1$ .

(a)

Expert Solution

Answer to Problem 27P

Dimension of $b=MT−1$

The unit of $b is kg s-1$

Explanation of Solution

Given Information:

$n=1$

Formula Used:

The drag force equation is $Fd=bvn$

Calculation:

The drag force equation is $Fd=bvn$

Dimension of force $=[MLT−2]$

The dimension of speed $=[LT−1]$

Solving the drag force equation for $b$ with $n=1$

$b=Fdv$

Substitute the dimensions of $Fd$ and $v$ and simplify to obtain the dimension of $b$ as

$[b]=[ML T −2][L T −1][b]=MT−1$

So, the units ofb is $kg s-1$ .

(b)

To determine

Using dimensional analysis, determine the units and dimensions of the constant $b$ in the retarding force $bυn$ if $n=2$ .

(b)

Expert Solution

Answer to Problem 27P

Dimension of $b=ML−1$

The units of $b is kg.m-1$

Explanation of Solution

Given Information:

$n=2$

Calculation:

The drag force equation is $Fd=bvn$

Dimension of force $=[MLT−2]$

The dimension of speed $=[LT−1]$

Solving the drag force equation for $b$ with $n=2$ ,

$b=Fdv2$

Substituting the dimensions of $Fd$ and $v$ and simplify to obtain the dimension of $b$ as

$[b]=[ML T −2] [ L T −1 ]2=ML−1$

So, the units of $b are kg m−1$ .

(c)

To determine

To Show: Air resistance of a falling object with a circular cross section should be approximately $12ρπr2υ2, where ρ=1.20 kg/m3$ , the density of air, is dimensionally consistent.

(c)

Expert Solution

Explanation of Solution

Given Information:

Air resistance of a falling object with a circular cross section should be approximately $12ρπr2υ2, where ρ=1.20 kg/m3$ , the density of air.

Calculation:

According to the expression of drag force,

$Fd=12ρπr2υ2$

So the dimensions of drag force is

$[Fd]=[12ρπr2υ2]or =[ML−3][L]2[LT −1]2or =MLT−2$

The drag force is given as

$Fd=bv2$

The dimension of drag force is

$[Fd]=[bv2]=[ML−1][LT −1]2=MLT−2$

Conclusion:

Hence, the expression for drag force is dimensionallyconsistent .

(d)

To determine

To Find: The terminal speed of the skydiver.

(d)

Expert Solution

Answer to Problem 27P

$56.9 m/s$

Explanation of Solution

Given Information:

Mass of skydiver = $56 kg$

Disk of radius = $0.30 m$

Density of air near the surface of Earth is $120 kg/m3$

Formula Used:

The terminal speed is given by relation

$vt=2mgρπr2$

Calculation:

Mass $m=56 kg$

Acceleration due to gravity $g=9.81 m/s2$

Density $ρ=1.2 kg/m3$

Radius $r=0.3 m$

The terminal speed is given by relation

$vt= 2mg ρπ r 2 = 2( 56 kg ) ( 9.81m/s ) 2 π( 1.2 kg/m 3 ) ( 0.3m ) 2 =56.9 m/s$

(e)

To determine

To Calculate: The terminal velocity at the given height.

(e)

Expert Solution

Answer to Problem 27P

The terminal velocity is $86.9 m/s$

Explanation of Solution

Given Information:

Height = $8 km$

Density of air near the surface of Earth is $0.514kg/m3$

Formula Used:

The terminal speed is given by relation

$vt=2mgρπr2$

Calculation:

Mass, $m=56 kg$

Acceleration due to gravity, $g=9.81 m/s2$

Density, $ρ=0.514 kg/m3$

Radius $r=0.3 m$

Substitute the values:

$vt= 2mg ρπ r 2 = 2( 56 kg ) ( 9.81m/s ) 2 π( 0.514 kg/m 3 ) ( 0.3 m ) 2 =272.12 m/s$

Conclusion:

Thus, the terminal speed at the given height is $272.12 m/s$

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 5, Problem 27P

(a)

To determine

The units and dimensions of the constant $b$ in the retarding force $bυn$ if $(a)n=1$ .

(a)

Expert Solution

Answer to Problem 27P

Dimension of $b=MT−1$

The unit of $b is kg s-1$

Explanation of Solution

Given Information:

$n=1$

Formula Used:

The drag force equation is $Fd=bvn$

Calculation:

The drag force equation is $Fd=bvn$

Dimension of force $=[MLT−2]$

The dimension of speed $=[LT−1]$

Solving the drag force equation for $b$ with $n=1$

$b=Fdv$

Substitute the dimensions of $Fd$ and $v$ and simplify to obtain the dimension of $b$ as

$[b]=[ML T −2][L T −1][b]=MT−1$

So, the units ofb is $kg s-1$ .

(b)

To determine

Using dimensional analysis, determine the units and dimensions of the constant $b$ in the retarding force $bυn$ if $n=2$ .

(b)

Expert Solution

Answer to Problem 27P

Dimension of $b=ML−1$

The units of $b is kg.m-1$

Explanation of Solution

Given Information:

$n=2$

Calculation:

The drag force equation is $Fd=bvn$

Dimension of force $=[MLT−2]$

The dimension of speed $=[LT−1]$

Solving the drag force equation for $b$ with $n=2$ ,

$b=Fdv2$

Substituting the dimensions of $Fd$ and $v$ and simplify to obtain the dimension of $b$ as

$[b]=[ML T −2] [ L T −1 ]2=ML−1$

So, the units of $b are kg m−1$ .

(c)

To determine

To Show: Air resistance of a falling object with a circular cross section should be approximately $12ρπr2υ2, where ρ=1.20 kg/m3$ , the density of air, is dimensionally consistent.

(c)

Expert Solution

Explanation of Solution

Given Information:

Air resistance of a falling object with a circular cross section should be approximately $12ρπr2υ2, where ρ=1.20 kg/m3$ , the density of air.

Calculation:

According to the expression of drag force,

$Fd=12ρπr2υ2$

So the dimensions of drag force is

$[Fd]=[12ρπr2υ2]or =[ML−3][L]2[LT −1]2or =MLT−2$

The drag force is given as

$Fd=bv2$

The dimension of drag force is

$[Fd]=[bv2]=[ML−1][LT −1]2=MLT−2$

Conclusion:

Hence, the expression for drag force is dimensionallyconsistent .

(d)

To determine

To Find: The terminal speed of the skydiver.

(d)

Expert Solution

Answer to Problem 27P

$56.9 m/s$

Explanation of Solution

Given Information:

Mass of skydiver = $56 kg$

Disk of radius = $0.30 m$

Density of air near the surface of Earth is $120 kg/m3$

Formula Used:

The terminal speed is given by relation

$vt=2mgρπr2$

Calculation:

Mass $m=56 kg$

Acceleration due to gravity $g=9.81 m/s2$

Density $ρ=1.2 kg/m3$

Radius $r=0.3 m$

The terminal speed is given by relation

$vt= 2mg ρπ r 2 = 2( 56 kg ) ( 9.81m/s ) 2 π( 1.2 kg/m 3 ) ( 0.3m ) 2 =56.9 m/s$

(e)

To determine

To Calculate: The terminal velocity at the given height.

(e)

Expert Solution

Answer to Problem 27P

The terminal velocity is $86.9 m/s$

Explanation of Solution

Given Information:

Height = $8 km$

Density of air near the surface of Earth is $0.514kg/m3$

Formula Used:

The terminal speed is given by relation

$vt=2mgρπr2$

Calculation:

Mass, $m=56 kg$

Acceleration due to gravity, $g=9.81 m/s2$

Density, $ρ=0.514 kg/m3$

Radius $r=0.3 m$

Substitute the values:

$vt= 2mg ρπ r 2 = 2( 56 kg ) ( 9.81m/s ) 2 π( 0.514 kg/m 3 ) ( 0.3 m ) 2 =272.12 m/s$

Conclusion:

Thus, the terminal speed at the given height is $272.12 m/s$

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 5, Problem 27P

(a)

To determine

The units and dimensions of the constant $b$ in the retarding force $bυn$ if $(a)n=1$ .

(a)

Expert Solution

Answer to Problem 27P

Dimension of $b=MT−1$

The unit of $b is kg s-1$

Explanation of Solution

Given Information:

$n=1$

Formula Used:

The drag force equation is $Fd=bvn$

Calculation:

The drag force equation is $Fd=bvn$

Dimension of force $=[MLT−2]$

The dimension of speed $=[LT−1]$

Solving the drag force equation for $b$ with $n=1$

$b=Fdv$

Substitute the dimensions of $Fd$ and $v$ and simplify to obtain the dimension of $b$ as

$[b]=[ML T −2][L T −1][b]=MT−1$

So, the units ofb is $kg s-1$ .

(b)

To determine

Using dimensional analysis, determine the units and dimensions of the constant $b$ in the retarding force $bυn$ if $n=2$ .

(b)

Expert Solution

Answer to Problem 27P

Dimension of $b=ML−1$

The units of $b is kg.m-1$

Explanation of Solution

Given Information:

$n=2$

Calculation:

The drag force equation is $Fd=bvn$

Dimension of force $=[MLT−2]$

The dimension of speed $=[LT−1]$

Solving the drag force equation for $b$ with $n=2$ ,

$b=Fdv2$

Substituting the dimensions of $Fd$ and $v$ and simplify to obtain the dimension of $b$ as

$[b]=[ML T −2] [ L T −1 ]2=ML−1$

So, the units of $b are kg m−1$ .

(c)

To determine

To Show: Air resistance of a falling object with a circular cross section should be approximately $12ρπr2υ2, where ρ=1.20 kg/m3$ , the density of air, is dimensionally consistent.

(c)

Expert Solution

Explanation of Solution

Given Information:

Air resistance of a falling object with a circular cross section should be approximately $12ρπr2υ2, where ρ=1.20 kg/m3$ , the density of air.

Calculation:

According to the expression of drag force,

$Fd=12ρπr2υ2$

So the dimensions of drag force is

$[Fd]=[12ρπr2υ2]or =[ML−3][L]2[LT −1]2or =MLT−2$

The drag force is given as

$Fd=bv2$

The dimension of drag force is

$[Fd]=[bv2]=[ML−1][LT −1]2=MLT−2$

Conclusion:

Hence, the expression for drag force is dimensionallyconsistent .

(d)

To determine

To Find: The terminal speed of the skydiver.

(d)

Expert Solution

Answer to Problem 27P

$56.9 m/s$

Explanation of Solution

Given Information:

Mass of skydiver = $56 kg$

Disk of radius = $0.30 m$

Density of air near the surface of Earth is $120 kg/m3$

Formula Used:

The terminal speed is given by relation

$vt=2mgρπr2$

Calculation:

Mass $m=56 kg$

Acceleration due to gravity $g=9.81 m/s2$

Density $ρ=1.2 kg/m3$

Radius $r=0.3 m$

The terminal speed is given by relation

$vt= 2mg ρπ r 2 = 2( 56 kg ) ( 9.81m/s ) 2 π( 1.2 kg/m 3 ) ( 0.3m ) 2 =56.9 m/s$

(e)

To determine

To Calculate: The terminal velocity at the given height.

(e)

Expert Solution

Answer to Problem 27P

The terminal velocity is $86.9 m/s$

Explanation of Solution

Given Information:

Height = $8 km$

Density of air near the surface of Earth is $0.514kg/m3$

Formula Used:

The terminal speed is given by relation

$vt=2mgρπr2$

Calculation:

Mass, $m=56 kg$

Acceleration due to gravity, $g=9.81 m/s2$

Density, $ρ=0.514 kg/m3$

Radius $r=0.3 m$

Substitute the values:

$vt= 2mg ρπ r 2 = 2( 56 kg ) ( 9.81m/s ) 2 π( 0.514 kg/m 3 ) ( 0.3 m ) 2 =272.12 m/s$

Conclusion:

Thus, the terminal speed at the given height is $272.12 m/s$

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 5, Problem 27P

(a)

To determine

The units and dimensions of the constant $b$ in the retarding force $bυn$ if $(a)n=1$ .

(a)

Expert Solution

Answer to Problem 27P

Dimension of $b=MT−1$

The unit of $b is kg s-1$

Explanation of Solution

Given Information:

$n=1$

Formula Used:

The drag force equation is $Fd=bvn$

Calculation:

The drag force equation is $Fd=bvn$

Dimension of force $=[MLT−2]$

The dimension of speed $=[LT−1]$

Solving the drag force equation for $b$ with $n=1$

$b=Fdv$

Substitute the dimensions of $Fd$ and $v$ and simplify to obtain the dimension of $b$ as

$[b]=[ML T −2][L T −1][b]=MT−1$

So, the units ofb is $kg s-1$ .

(b)

To determine

Using dimensional analysis, determine the units and dimensions of the constant $b$ in the retarding force $bυn$ if $n=2$ .

(b)

Expert Solution

Answer to Problem 27P

Dimension of $b=ML−1$

The units of $b is kg.m-1$

Explanation of Solution

Given Information:

$n=2$

Calculation:

The drag force equation is $Fd=bvn$

Dimension of force $=[MLT−2]$

The dimension of speed $=[LT−1]$

Solving the drag force equation for $b$ with $n=2$ ,

$b=Fdv2$

Substituting the dimensions of $Fd$ and $v$ and simplify to obtain the dimension of $b$ as

$[b]=[ML T −2] [ L T −1 ]2=ML−1$

So, the units of $b are kg m−1$ .

(c)

To determine

To Show: Air resistance of a falling object with a circular cross section should be approximately $12ρπr2υ2, where ρ=1.20 kg/m3$ , the density of air, is dimensionally consistent.

(c)

Expert Solution

Explanation of Solution

Given Information:

Air resistance of a falling object with a circular cross section should be approximately $12ρπr2υ2, where ρ=1.20 kg/m3$ , the density of air.

Calculation:

According to the expression of drag force,

$Fd=12ρπr2υ2$

So the dimensions of drag force is

$[Fd]=[12ρπr2υ2]or =[ML−3][L]2[LT −1]2or =MLT−2$

The drag force is given as

$Fd=bv2$

The dimension of drag force is

$[Fd]=[bv2]=[ML−1][LT −1]2=MLT−2$

Conclusion:

Hence, the expression for drag force is dimensionallyconsistent .

(d)

To determine

To Find: The terminal speed of the skydiver.

(d)

Expert Solution

Answer to Problem 27P

$56.9 m/s$

Explanation of Solution

Given Information:

Mass of skydiver = $56 kg$

Disk of radius = $0.30 m$

Density of air near the surface of Earth is $120 kg/m3$

Formula Used:

The terminal speed is given by relation

$vt=2mgρπr2$

Calculation:

Mass $m=56 kg$

Acceleration due to gravity $g=9.81 m/s2$

Density $ρ=1.2 kg/m3$

Radius $r=0.3 m$

The terminal speed is given by relation

$vt= 2mg ρπ r 2 = 2( 56 kg ) ( 9.81m/s ) 2 π( 1.2 kg/m 3 ) ( 0.3m ) 2 =56.9 m/s$

(e)

To determine

To Calculate: The terminal velocity at the given height.

(e)

Expert Solution

Answer to Problem 27P

The terminal velocity is $86.9 m/s$

Explanation of Solution

Given Information:

Height = $8 km$

Density of air near the surface of Earth is $0.514kg/m3$

Formula Used:

The terminal speed is given by relation

$vt=2mgρπr2$

Calculation:

Mass, $m=56 kg$

Acceleration due to gravity, $g=9.81 m/s2$

Density, $ρ=0.514 kg/m3$

Radius $r=0.3 m$

Substitute the values:

$vt= 2mg ρπ r 2 = 2( 56 kg ) ( 9.81m/s ) 2 π( 0.514 kg/m 3 ) ( 0.3 m ) 2 =272.12 m/s$

Conclusion:

Thus, the terminal speed at the given height is $272.12 m/s$

Answer 6

Chapter 5, Problem 27P

(a)

To determine

The units and dimensions of the constant $b$ in the retarding force $bυn$ if $(a)n=1$ .

(a)

Expert Solution

Answer to Problem 27P

Dimension of $b=MT−1$

The unit of $b is kg s-1$

Explanation of Solution

Given Information:

$n=1$

Formula Used:

The drag force equation is $Fd=bvn$

Calculation:

The drag force equation is $Fd=bvn$

Dimension of force $=[MLT−2]$

The dimension of speed $=[LT−1]$

Solving the drag force equation for $b$ with $n=1$

$b=Fdv$

Substitute the dimensions of $Fd$ and $v$ and simplify to obtain the dimension of $b$ as

$[b]=[ML T −2][L T −1][b]=MT−1$

So, the units ofb is $kg s-1$ .

Answer 7

Chapter 5, Problem 27P

(a)

To determine

The units and dimensions of the constant $b$ in the retarding force $bυn$ if $(a)n=1$ .

Answer 8

(a)

Expert Solution

Answer to Problem 27P

Dimension of $b=MT−1$

The unit of $b is kg s-1$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given Information:

$n=1$

Formula Used:

The drag force equation is $Fd=bvn$

Calculation:

The drag force equation is $Fd=bvn$

Dimension of force $=[MLT−2]$

The dimension of speed $=[LT−1]$

Solving the drag force equation for $b$ with $n=1$

$b=Fdv$

Substitute the dimensions of $Fd$ and $v$ and simplify to obtain the dimension of $b$ as

$[b]=[ML T −2][L T −1][b]=MT−1$

So, the units ofb is $kg s-1$ .

Answer 13

(b)

To determine

Using dimensional analysis, determine the units and dimensions of the constant $b$ in the retarding force $bυn$ if $n=2$ .

(b)

Expert Solution

Answer to Problem 27P

Dimension of $b=ML−1$

The units of $b is kg.m-1$

Explanation of Solution

Given Information:

$n=2$

Calculation:

The drag force equation is $Fd=bvn$

Dimension of force $=[MLT−2]$

The dimension of speed $=[LT−1]$

Solving the drag force equation for $b$ with $n=2$ ,

$b=Fdv2$

Substituting the dimensions of $Fd$ and $v$ and simplify to obtain the dimension of $b$ as

$[b]=[ML T −2] [ L T −1 ]2=ML−1$

So, the units of $b are kg m−1$ .

Answer 14

(b)

To determine

Using dimensional analysis, determine the units and dimensions of the constant $b$ in the retarding force $bυn$ if $n=2$ .

Answer 15

(b)

Expert Solution

Answer to Problem 27P

Dimension of $b=ML−1$

The units of $b is kg.m-1$

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given Information:

$n=2$

Calculation:

The drag force equation is $Fd=bvn$

Dimension of force $=[MLT−2]$

The dimension of speed $=[LT−1]$

Solving the drag force equation for $b$ with $n=2$ ,

$b=Fdv2$

Substituting the dimensions of $Fd$ and $v$ and simplify to obtain the dimension of $b$ as

$[b]=[ML T −2] [ L T −1 ]2=ML−1$

So, the units of $b are kg m−1$ .

Answer 20

(c)

To determine

To Show: Air resistance of a falling object with a circular cross section should be approximately $12ρπr2υ2, where ρ=1.20 kg/m3$ , the density of air, is dimensionally consistent.

(c)

Expert Solution

Explanation of Solution

Given Information:

Air resistance of a falling object with a circular cross section should be approximately $12ρπr2υ2, where ρ=1.20 kg/m3$ , the density of air.

Calculation:

According to the expression of drag force,

$Fd=12ρπr2υ2$

So the dimensions of drag force is

$[Fd]=[12ρπr2υ2]or =[ML−3][L]2[LT −1]2or =MLT−2$

The drag force is given as

$Fd=bv2$

The dimension of drag force is

$[Fd]=[bv2]=[ML−1][LT −1]2=MLT−2$

Conclusion:

Hence, the expression for drag force is dimensionallyconsistent .

Answer 21

(c)

To determine

To Show: Air resistance of a falling object with a circular cross section should be approximately $12ρπr2υ2, where ρ=1.20 kg/m3$ , the density of air, is dimensionally consistent.

Answer 22

(c)

Expert Solution

Explanation of Solution

Given Information:

Air resistance of a falling object with a circular cross section should be approximately $12ρπr2υ2, where ρ=1.20 kg/m3$ , the density of air.

Calculation:

According to the expression of drag force,

$Fd=12ρπr2υ2$

So the dimensions of drag force is

$[Fd]=[12ρπr2υ2]or =[ML−3][L]2[LT −1]2or =MLT−2$

The drag force is given as

$Fd=bv2$

The dimension of drag force is

$[Fd]=[bv2]=[ML−1][LT −1]2=MLT−2$

Conclusion:

Hence, the expression for drag force is dimensionallyconsistent .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

(d)

To determine

To Find: The terminal speed of the skydiver.

(d)

Expert Solution

Answer to Problem 27P

$56.9 m/s$

Explanation of Solution

Given Information:

Mass of skydiver = $56 kg$

Disk of radius = $0.30 m$

Density of air near the surface of Earth is $120 kg/m3$

Formula Used:

The terminal speed is given by relation

$vt=2mgρπr2$

Calculation:

Mass $m=56 kg$

Acceleration due to gravity $g=9.81 m/s2$

Density $ρ=1.2 kg/m3$

Radius $r=0.3 m$

The terminal speed is given by relation

$vt= 2mg ρπ r 2 = 2( 56 kg ) ( 9.81m/s ) 2 π( 1.2 kg/m 3 ) ( 0.3m ) 2 =56.9 m/s$

Answer 27

(d)

To determine

To Find: The terminal speed of the skydiver.

Answer 28

(d)

Expert Solution

Answer to Problem 27P

$56.9 m/s$

Answer 29

(d)

Expert Solution

Answer 30

(d)

Expert Solution

Answer 31

Expert Solution

Answer 32

Explanation of Solution

Given Information:

Mass of skydiver = $56 kg$

Disk of radius = $0.30 m$

Density of air near the surface of Earth is $120 kg/m3$

Formula Used:

The terminal speed is given by relation

$vt=2mgρπr2$

Calculation:

Mass $m=56 kg$

Acceleration due to gravity $g=9.81 m/s2$

Density $ρ=1.2 kg/m3$

Radius $r=0.3 m$

The terminal speed is given by relation

$vt= 2mg ρπ r 2 = 2( 56 kg ) ( 9.81m/s ) 2 π( 1.2 kg/m 3 ) ( 0.3m ) 2 =56.9 m/s$

Answer 33

(e)

To determine

To Calculate: The terminal velocity at the given height.

(e)

Expert Solution

Answer to Problem 27P

The terminal velocity is $86.9 m/s$

Explanation of Solution

Given Information:

Height = $8 km$

Density of air near the surface of Earth is $0.514kg/m3$

Formula Used:

The terminal speed is given by relation

$vt=2mgρπr2$

Calculation:

Mass, $m=56 kg$

Acceleration due to gravity, $g=9.81 m/s2$

Density, $ρ=0.514 kg/m3$

Radius $r=0.3 m$

Substitute the values:

$vt= 2mg ρπ r 2 = 2( 56 kg ) ( 9.81m/s ) 2 π( 0.514 kg/m 3 ) ( 0.3 m ) 2 =272.12 m/s$

Conclusion:

Thus, the terminal speed at the given height is $272.12 m/s$

Answer 34

(e)

To determine

To Calculate: The terminal velocity at the given height.

Answer 35

(e)

Expert Solution

Answer to Problem 27P

The terminal velocity is $86.9 m/s$

Answer 36

(e)

Expert Solution

Answer 37

(e)

Expert Solution

Answer 38

Expert Solution

Answer 39

Explanation of Solution

Given Information:

Height = $8 km$

Density of air near the surface of Earth is $0.514kg/m3$

Formula Used:

The terminal speed is given by relation

$vt=2mgρπr2$

Calculation:

Mass, $m=56 kg$

Acceleration due to gravity, $g=9.81 m/s2$

Density, $ρ=0.514 kg/m3$

Radius $r=0.3 m$

Substitute the values:

$vt= 2mg ρπ r 2 = 2( 56 kg ) ( 9.81m/s ) 2 π( 0.514 kg/m 3 ) ( 0.3 m ) 2 =272.12 m/s$

Conclusion:

Thus, the terminal speed at the given height is $272.12 m/s$

Answer 40

Ch. 5 - Prob. 1P Ch. 5 - Prob. 2P Ch. 5 - Prob. 3P Ch. 5 - Prob. 4P Ch. 5 - Prob. 5P Ch. 5 - Prob. 6P Ch. 5 - Prob. 7P Ch. 5 - Prob. 8P Ch. 5 - Prob. 9P Ch. 5 - Prob. 10P

The units and dimensions of the constant b in the retarding force b υ n if ( a ) n = 1 .

Concept explainers

The units and dimensions of the constant $b$ in the retarding force $bυn$ if $(a)n=1$ .

Answer to Problem 27P

Explanation of Solution

Using dimensional analysis, determine the units and dimensions of the constant $b$ in the retarding force $bυn$ if $n=2$ .

Answer to Problem 27P

Explanation of Solution

To Show: Air resistance of a falling object with a circular cross section should be approximately $12ρπr2υ2, where ρ=1.20 kg/m3$ , the density of air, is dimensionally consistent.

Explanation of Solution

To Find: The terminal speed of the skydiver.

Answer to Problem 27P

Explanation of Solution

To Calculate: The terminal velocity at the given height.

Answer to Problem 27P

Explanation of Solution

Want to see more full solutions like this?

Chapter 5 Solutions