Owen and Dina are at rest in frame S′, which is moving with a speed of 0.600 c with respect to frame S. They play a game of catch while Ed, at rest in frame S, watches the action (Fig. P26.45). Owen throws the ball to Dina with a speed of 0.800 c (according to Owen) and their separation (measured in S′) is equal to 1.80 × 10 12 m. (a) According to Dina, how fast is the ball moving? (b) According to Dina, what time interval is required for the ball to reach her? According to Ed, (c) how far apart are Owen and Dina, and (d) how fast is the ball moving? Figure. P26.45

Question

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Answer 1

Textbook Question

Chapter 26, Problem 45AP

Owen and Dina are at rest in frame S′, which is moving with a speed of 0.600c with respect to frame S. They play a game of catch while Ed, at rest in frame S, watches the action (Fig. P26.45). Owen throws the ball to Dina with a speed of 0.800c (according to Owen) and their separation (measured in S′) is equal to 1.80 × 10¹² m. (a) According to Dina, how fast is the ball moving? (b) According to Dina, what time interval is required for the ball to reach her? According to Ed, (c) how far apart are Owen and Dina, and (d) how fast is the ball moving?

Chapter 26, Problem 45AP, Owen and Dina are at rest in frame S, which is moving with a speed of 0.600c with respect to frame

Figure. P26.45

(a)

Expert Solution

To determine

The speed of ball according to Dina.

Answer to Problem 45AP

The speed of the ball according to Dina is

$−0.800c$ .

Explanation of Solution

Given Info:

The speed of ball in $S'$ frame is $0.800c$ .

Dina and Owen are at rest in frame $S'$ . The speed of the ball relative to the frame will be the speed of the ball. The magnitude of speed of ball is $0.800c$ . The speed of the ball relative to Dina is $−0.800c$ .

Thus, the speed of the ball according to Dina is $−0.800c$ .

Conclusion:

The speed of the ball according to Dina is $−0.800c$ .

(b)

Expert Solution

To determine

The time interval took for the ball to reach Dina according to her.

Answer to Problem 45AP

The time interval took for the ball to reach Dina according to her is

$7.50×103 m$ .

Explanation of Solution

Given Info:

The distance between Dina and Owen measured at their rest frame is $1.80×1013 m$ .

The speed of ball in frame $S'$ is $0.800c$ .

The speed of light is $3.00×108 m/s$ .

Formula to calculate the time interval took for the ball to reach Dina according to her is,

$Δtp=Lpv'ball$

$Δtp$ is the time interval took for the ball to reach Dina according to her
$Lp$ is the distance between Dina and Owen measured at their rest frame
$v'ball$ is the speed of ball in frame $S'$
$c$ is the speed of light

Substitute $1.80×1013 m$ for $Lp$ and $0.800c$ for $v'ball$ to find $Δtp$ .

$Δtp=1.80×1013 m0.800c=1.80×1013 m0.800(3.00×108 m/s)=7.50×103 s$

Thus, the time interval took for the ball to reach Dina according to her is $7.50×103 m$ .

Conclusion:

The time interval took for the ball to reach Dina according to her is $7.50×103 m$ .

(c)

Expert Solution

To determine

The distance between Owen and Dina measured by Ed.

Answer to Problem 45AP

The distance between Owen and Dina measured by Ed is

$1.44×1012 m$ .

Explanation of Solution

Given Info:

The speed of frame $S'$ is $0.600c$ .

Formula to calculate the distance between Owen and Dina measured by Ed is,

$L=Lp1−(vc)2$

$L$ is the distance between Owen and Dina measured by Ed
$v$ is the speed of frame $S'$

Substitute $0.600c$ for $v$ and $1.80×1012 m$ for $Lp$ to find $L$ .

$L=(1.80×1012 m)1−(0.600cc)2=1.44×1012 m$

Thus, the distance between Owen and Dina measured by Ed is $1.44×1012 m$ .

Conclusion:

The distance between Owen and Dina measured by Ed is $1.44×1012 m$ .

(d)

Expert Solution

To determine

The speed of ball relative to Ed.

Answer to Problem 45AP

The speed of ball relative to Ed is

$0.385c$ .

Explanation of Solution

Given Info:

The speed of ball relative to Dina is $−0.800c$ .

The speed of Dina relative to Ed is $0.600c$ .

Formula to calculate distance covered by the cheetah is,

$vbE=vbD+vDE1+vbDvDEc2$

$vbE$ is the speed of ball relative to Ed
$vbD$ is the speed of ball relative to Dina
$vDE$ is the speed of Dina relative to Ed

Substitute $−0.800c$ for $vbD$ and $0.600c$ for $vDE$ to find $vbE$ .

$vbE=(−0.800c)+(0.600c)1+(−0.800c)(0.600c)c2=0.385c$

Thus, the speed of ball relative to Ed is $0.385c$ .

Conclusion:

The speed of ball relative to Ed is $0.385c$

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Answer 2

Textbook Question

Chapter 26, Problem 45AP

Owen and Dina are at rest in frame S′, which is moving with a speed of 0.600c with respect to frame S. They play a game of catch while Ed, at rest in frame S, watches the action (Fig. P26.45). Owen throws the ball to Dina with a speed of 0.800c (according to Owen) and their separation (measured in S′) is equal to 1.80 × 10¹² m. (a) According to Dina, how fast is the ball moving? (b) According to Dina, what time interval is required for the ball to reach her? According to Ed, (c) how far apart are Owen and Dina, and (d) how fast is the ball moving?

Chapter 26, Problem 45AP, Owen and Dina are at rest in frame S, which is moving with a speed of 0.600c with respect to frame

Figure. P26.45

(a)

Expert Solution

To determine

The speed of ball according to Dina.

Answer to Problem 45AP

The speed of the ball according to Dina is

$−0.800c$ .

Explanation of Solution

Given Info:

The speed of ball in $S'$ frame is $0.800c$ .

Dina and Owen are at rest in frame $S'$ . The speed of the ball relative to the frame will be the speed of the ball. The magnitude of speed of ball is $0.800c$ . The speed of the ball relative to Dina is $−0.800c$ .

Thus, the speed of the ball according to Dina is $−0.800c$ .

Conclusion:

The speed of the ball according to Dina is $−0.800c$ .

(b)

Expert Solution

To determine

The time interval took for the ball to reach Dina according to her.

Answer to Problem 45AP

The time interval took for the ball to reach Dina according to her is

$7.50×103 m$ .

Explanation of Solution

Given Info:

The distance between Dina and Owen measured at their rest frame is $1.80×1013 m$ .

The speed of ball in frame $S'$ is $0.800c$ .

The speed of light is $3.00×108 m/s$ .

Formula to calculate the time interval took for the ball to reach Dina according to her is,

$Δtp=Lpv'ball$

$Δtp$ is the time interval took for the ball to reach Dina according to her
$Lp$ is the distance between Dina and Owen measured at their rest frame
$v'ball$ is the speed of ball in frame $S'$
$c$ is the speed of light

Substitute $1.80×1013 m$ for $Lp$ and $0.800c$ for $v'ball$ to find $Δtp$ .

$Δtp=1.80×1013 m0.800c=1.80×1013 m0.800(3.00×108 m/s)=7.50×103 s$

Thus, the time interval took for the ball to reach Dina according to her is $7.50×103 m$ .

Conclusion:

The time interval took for the ball to reach Dina according to her is $7.50×103 m$ .

(c)

Expert Solution

To determine

The distance between Owen and Dina measured by Ed.

Answer to Problem 45AP

The distance between Owen and Dina measured by Ed is

$1.44×1012 m$ .

Explanation of Solution

Given Info:

The speed of frame $S'$ is $0.600c$ .

Formula to calculate the distance between Owen and Dina measured by Ed is,

$L=Lp1−(vc)2$

$L$ is the distance between Owen and Dina measured by Ed
$v$ is the speed of frame $S'$

Substitute $0.600c$ for $v$ and $1.80×1012 m$ for $Lp$ to find $L$ .

$L=(1.80×1012 m)1−(0.600cc)2=1.44×1012 m$

Thus, the distance between Owen and Dina measured by Ed is $1.44×1012 m$ .

Conclusion:

The distance between Owen and Dina measured by Ed is $1.44×1012 m$ .

(d)

Expert Solution

To determine

The speed of ball relative to Ed.

Answer to Problem 45AP

The speed of ball relative to Ed is

$0.385c$ .

Explanation of Solution

Given Info:

The speed of ball relative to Dina is $−0.800c$ .

The speed of Dina relative to Ed is $0.600c$ .

Formula to calculate distance covered by the cheetah is,

$vbE=vbD+vDE1+vbDvDEc2$

$vbE$ is the speed of ball relative to Ed
$vbD$ is the speed of ball relative to Dina
$vDE$ is the speed of Dina relative to Ed

Substitute $−0.800c$ for $vbD$ and $0.600c$ for $vDE$ to find $vbE$ .

$vbE=(−0.800c)+(0.600c)1+(−0.800c)(0.600c)c2=0.385c$

Thus, the speed of ball relative to Ed is $0.385c$ .

Conclusion:

The speed of ball relative to Ed is $0.385c$

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Answer 3

Textbook Question

Chapter 26, Problem 45AP

Owen and Dina are at rest in frame S′, which is moving with a speed of 0.600c with respect to frame S. They play a game of catch while Ed, at rest in frame S, watches the action (Fig. P26.45). Owen throws the ball to Dina with a speed of 0.800c (according to Owen) and their separation (measured in S′) is equal to 1.80 × 10¹² m. (a) According to Dina, how fast is the ball moving? (b) According to Dina, what time interval is required for the ball to reach her? According to Ed, (c) how far apart are Owen and Dina, and (d) how fast is the ball moving?

Chapter 26, Problem 45AP, Owen and Dina are at rest in frame S, which is moving with a speed of 0.600c with respect to frame

Figure. P26.45

(a)

Expert Solution

To determine

The speed of ball according to Dina.

Answer to Problem 45AP

The speed of the ball according to Dina is

$−0.800c$ .

Explanation of Solution

Given Info:

The speed of ball in $S'$ frame is $0.800c$ .

Dina and Owen are at rest in frame $S'$ . The speed of the ball relative to the frame will be the speed of the ball. The magnitude of speed of ball is $0.800c$ . The speed of the ball relative to Dina is $−0.800c$ .

Thus, the speed of the ball according to Dina is $−0.800c$ .

Conclusion:

The speed of the ball according to Dina is $−0.800c$ .

(b)

Expert Solution

To determine

The time interval took for the ball to reach Dina according to her.

Answer to Problem 45AP

The time interval took for the ball to reach Dina according to her is

$7.50×103 m$ .

Explanation of Solution

Given Info:

The distance between Dina and Owen measured at their rest frame is $1.80×1013 m$ .

The speed of ball in frame $S'$ is $0.800c$ .

The speed of light is $3.00×108 m/s$ .

Formula to calculate the time interval took for the ball to reach Dina according to her is,

$Δtp=Lpv'ball$

$Δtp$ is the time interval took for the ball to reach Dina according to her
$Lp$ is the distance between Dina and Owen measured at their rest frame
$v'ball$ is the speed of ball in frame $S'$
$c$ is the speed of light

Substitute $1.80×1013 m$ for $Lp$ and $0.800c$ for $v'ball$ to find $Δtp$ .

$Δtp=1.80×1013 m0.800c=1.80×1013 m0.800(3.00×108 m/s)=7.50×103 s$

Thus, the time interval took for the ball to reach Dina according to her is $7.50×103 m$ .

Conclusion:

The time interval took for the ball to reach Dina according to her is $7.50×103 m$ .

(c)

Expert Solution

To determine

The distance between Owen and Dina measured by Ed.

Answer to Problem 45AP

The distance between Owen and Dina measured by Ed is

$1.44×1012 m$ .

Explanation of Solution

Given Info:

The speed of frame $S'$ is $0.600c$ .

Formula to calculate the distance between Owen and Dina measured by Ed is,

$L=Lp1−(vc)2$

$L$ is the distance between Owen and Dina measured by Ed
$v$ is the speed of frame $S'$

Substitute $0.600c$ for $v$ and $1.80×1012 m$ for $Lp$ to find $L$ .

$L=(1.80×1012 m)1−(0.600cc)2=1.44×1012 m$

Thus, the distance between Owen and Dina measured by Ed is $1.44×1012 m$ .

Conclusion:

The distance between Owen and Dina measured by Ed is $1.44×1012 m$ .

(d)

Expert Solution

To determine

The speed of ball relative to Ed.

Answer to Problem 45AP

The speed of ball relative to Ed is

$0.385c$ .

Explanation of Solution

Given Info:

The speed of ball relative to Dina is $−0.800c$ .

The speed of Dina relative to Ed is $0.600c$ .

Formula to calculate distance covered by the cheetah is,

$vbE=vbD+vDE1+vbDvDEc2$

$vbE$ is the speed of ball relative to Ed
$vbD$ is the speed of ball relative to Dina
$vDE$ is the speed of Dina relative to Ed

Substitute $−0.800c$ for $vbD$ and $0.600c$ for $vDE$ to find $vbE$ .

$vbE=(−0.800c)+(0.600c)1+(−0.800c)(0.600c)c2=0.385c$

Thus, the speed of ball relative to Ed is $0.385c$ .

Conclusion:

The speed of ball relative to Ed is $0.385c$

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Answer 4

Textbook Question

Chapter 26, Problem 45AP

Owen and Dina are at rest in frame S′, which is moving with a speed of 0.600c with respect to frame S. They play a game of catch while Ed, at rest in frame S, watches the action (Fig. P26.45). Owen throws the ball to Dina with a speed of 0.800c (according to Owen) and their separation (measured in S′) is equal to 1.80 × 10¹² m. (a) According to Dina, how fast is the ball moving? (b) According to Dina, what time interval is required for the ball to reach her? According to Ed, (c) how far apart are Owen and Dina, and (d) how fast is the ball moving?

Chapter 26, Problem 45AP, Owen and Dina are at rest in frame S, which is moving with a speed of 0.600c with respect to frame

Figure. P26.45

(a)

Expert Solution

To determine

The speed of ball according to Dina.

Answer to Problem 45AP

The speed of the ball according to Dina is

$−0.800c$ .

Explanation of Solution

Given Info:

The speed of ball in $S'$ frame is $0.800c$ .

Dina and Owen are at rest in frame $S'$ . The speed of the ball relative to the frame will be the speed of the ball. The magnitude of speed of ball is $0.800c$ . The speed of the ball relative to Dina is $−0.800c$ .

Thus, the speed of the ball according to Dina is $−0.800c$ .

Conclusion:

The speed of the ball according to Dina is $−0.800c$ .

(b)

Expert Solution

To determine

The time interval took for the ball to reach Dina according to her.

Answer to Problem 45AP

The time interval took for the ball to reach Dina according to her is

$7.50×103 m$ .

Explanation of Solution

Given Info:

The distance between Dina and Owen measured at their rest frame is $1.80×1013 m$ .

The speed of ball in frame $S'$ is $0.800c$ .

The speed of light is $3.00×108 m/s$ .

Formula to calculate the time interval took for the ball to reach Dina according to her is,

$Δtp=Lpv'ball$

$Δtp$ is the time interval took for the ball to reach Dina according to her
$Lp$ is the distance between Dina and Owen measured at their rest frame
$v'ball$ is the speed of ball in frame $S'$
$c$ is the speed of light

Substitute $1.80×1013 m$ for $Lp$ and $0.800c$ for $v'ball$ to find $Δtp$ .

$Δtp=1.80×1013 m0.800c=1.80×1013 m0.800(3.00×108 m/s)=7.50×103 s$

Thus, the time interval took for the ball to reach Dina according to her is $7.50×103 m$ .

Conclusion:

The time interval took for the ball to reach Dina according to her is $7.50×103 m$ .

(c)

Expert Solution

To determine

The distance between Owen and Dina measured by Ed.

Answer to Problem 45AP

The distance between Owen and Dina measured by Ed is

$1.44×1012 m$ .

Explanation of Solution

Given Info:

The speed of frame $S'$ is $0.600c$ .

Formula to calculate the distance between Owen and Dina measured by Ed is,

$L=Lp1−(vc)2$

$L$ is the distance between Owen and Dina measured by Ed
$v$ is the speed of frame $S'$

Substitute $0.600c$ for $v$ and $1.80×1012 m$ for $Lp$ to find $L$ .

$L=(1.80×1012 m)1−(0.600cc)2=1.44×1012 m$

Thus, the distance between Owen and Dina measured by Ed is $1.44×1012 m$ .

Conclusion:

The distance between Owen and Dina measured by Ed is $1.44×1012 m$ .

(d)

Expert Solution

To determine

The speed of ball relative to Ed.

Answer to Problem 45AP

The speed of ball relative to Ed is

$0.385c$ .

Explanation of Solution

Given Info:

The speed of ball relative to Dina is $−0.800c$ .

The speed of Dina relative to Ed is $0.600c$ .

Formula to calculate distance covered by the cheetah is,

$vbE=vbD+vDE1+vbDvDEc2$

$vbE$ is the speed of ball relative to Ed
$vbD$ is the speed of ball relative to Dina
$vDE$ is the speed of Dina relative to Ed

Substitute $−0.800c$ for $vbD$ and $0.600c$ for $vDE$ to find $vbE$ .

$vbE=(−0.800c)+(0.600c)1+(−0.800c)(0.600c)c2=0.385c$

Thus, the speed of ball relative to Ed is $0.385c$ .

Conclusion:

The speed of ball relative to Ed is $0.385c$

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

The speed of ball according to Dina.

Answer to Problem 45AP

The speed of the ball according to Dina is

$−0.800c$ .

Explanation of Solution

Given Info:

The speed of ball in $S'$ frame is $0.800c$ .

Dina and Owen are at rest in frame $S'$ . The speed of the ball relative to the frame will be the speed of the ball. The magnitude of speed of ball is $0.800c$ . The speed of the ball relative to Dina is $−0.800c$ .

Thus, the speed of the ball according to Dina is $−0.800c$ .

Conclusion:

The speed of the ball according to Dina is $−0.800c$ .

(b)

Expert Solution

To determine

The time interval took for the ball to reach Dina according to her.

Answer to Problem 45AP

The time interval took for the ball to reach Dina according to her is

$7.50×103 m$ .

Explanation of Solution

Given Info:

The distance between Dina and Owen measured at their rest frame is $1.80×1013 m$ .

The speed of ball in frame $S'$ is $0.800c$ .

The speed of light is $3.00×108 m/s$ .

Formula to calculate the time interval took for the ball to reach Dina according to her is,

$Δtp=Lpv'ball$

$Δtp$ is the time interval took for the ball to reach Dina according to her
$Lp$ is the distance between Dina and Owen measured at their rest frame
$v'ball$ is the speed of ball in frame $S'$
$c$ is the speed of light

Substitute $1.80×1013 m$ for $Lp$ and $0.800c$ for $v'ball$ to find $Δtp$ .

$Δtp=1.80×1013 m0.800c=1.80×1013 m0.800(3.00×108 m/s)=7.50×103 s$

Thus, the time interval took for the ball to reach Dina according to her is $7.50×103 m$ .

Conclusion:

The time interval took for the ball to reach Dina according to her is $7.50×103 m$ .

(c)

Expert Solution

To determine

The distance between Owen and Dina measured by Ed.

Answer to Problem 45AP

The distance between Owen and Dina measured by Ed is

$1.44×1012 m$ .

Explanation of Solution

Given Info:

The speed of frame $S'$ is $0.600c$ .

Formula to calculate the distance between Owen and Dina measured by Ed is,

$L=Lp1−(vc)2$

$L$ is the distance between Owen and Dina measured by Ed
$v$ is the speed of frame $S'$

Substitute $0.600c$ for $v$ and $1.80×1012 m$ for $Lp$ to find $L$ .

$L=(1.80×1012 m)1−(0.600cc)2=1.44×1012 m$

Thus, the distance between Owen and Dina measured by Ed is $1.44×1012 m$ .

Conclusion:

The distance between Owen and Dina measured by Ed is $1.44×1012 m$ .

(d)

Expert Solution

To determine

The speed of ball relative to Ed.

Answer to Problem 45AP

The speed of ball relative to Ed is

$0.385c$ .

Explanation of Solution

Given Info:

The speed of ball relative to Dina is $−0.800c$ .

The speed of Dina relative to Ed is $0.600c$ .

Formula to calculate distance covered by the cheetah is,

$vbE=vbD+vDE1+vbDvDEc2$

$vbE$ is the speed of ball relative to Ed
$vbD$ is the speed of ball relative to Dina
$vDE$ is the speed of Dina relative to Ed

Substitute $−0.800c$ for $vbD$ and $0.600c$ for $vDE$ to find $vbE$ .

$vbE=(−0.800c)+(0.600c)1+(−0.800c)(0.600c)c2=0.385c$

Thus, the speed of ball relative to Ed is $0.385c$ .

Conclusion:

The speed of ball relative to Ed is $0.385c$

Answer 8

(a)

Expert Solution

To determine

The speed of ball according to Dina.

Answer to Problem 45AP

The speed of the ball according to Dina is

$−0.800c$ .

Explanation of Solution

Given Info:

The speed of ball in $S'$ frame is $0.800c$ .

Dina and Owen are at rest in frame $S'$ . The speed of the ball relative to the frame will be the speed of the ball. The magnitude of speed of ball is $0.800c$ . The speed of the ball relative to Dina is $−0.800c$ .

Thus, the speed of the ball according to Dina is $−0.800c$ .

Conclusion:

The speed of the ball according to Dina is $−0.800c$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The speed of ball according to Dina.

Answer 13

Answer to Problem 45AP

The speed of the ball according to Dina is

$−0.800c$ .

Answer 14

Explanation of Solution

Given Info:

The speed of ball in $S'$ frame is $0.800c$ .

Dina and Owen are at rest in frame $S'$ . The speed of the ball relative to the frame will be the speed of the ball. The magnitude of speed of ball is $0.800c$ . The speed of the ball relative to Dina is $−0.800c$ .

Thus, the speed of the ball according to Dina is $−0.800c$ .

Conclusion:

The speed of the ball according to Dina is $−0.800c$ .

Answer 15

(b)

Expert Solution

To determine

The time interval took for the ball to reach Dina according to her.

Answer to Problem 45AP

The time interval took for the ball to reach Dina according to her is

$7.50×103 m$ .

Explanation of Solution

Given Info:

The distance between Dina and Owen measured at their rest frame is $1.80×1013 m$ .

The speed of ball in frame $S'$ is $0.800c$ .

The speed of light is $3.00×108 m/s$ .

Formula to calculate the time interval took for the ball to reach Dina according to her is,

$Δtp=Lpv'ball$

$Δtp$ is the time interval took for the ball to reach Dina according to her
$Lp$ is the distance between Dina and Owen measured at their rest frame
$v'ball$ is the speed of ball in frame $S'$
$c$ is the speed of light

Substitute $1.80×1013 m$ for $Lp$ and $0.800c$ for $v'ball$ to find $Δtp$ .

$Δtp=1.80×1013 m0.800c=1.80×1013 m0.800(3.00×108 m/s)=7.50×103 s$

Thus, the time interval took for the ball to reach Dina according to her is $7.50×103 m$ .

Conclusion:

The time interval took for the ball to reach Dina according to her is $7.50×103 m$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The time interval took for the ball to reach Dina according to her.

Answer 20

Answer to Problem 45AP

The time interval took for the ball to reach Dina according to her is

$7.50×103 m$ .

Answer 21

Explanation of Solution

Given Info:

The distance between Dina and Owen measured at their rest frame is $1.80×1013 m$ .

The speed of ball in frame $S'$ is $0.800c$ .

The speed of light is $3.00×108 m/s$ .

Formula to calculate the time interval took for the ball to reach Dina according to her is,

$Δtp=Lpv'ball$

$Δtp$ is the time interval took for the ball to reach Dina according to her
$Lp$ is the distance between Dina and Owen measured at their rest frame
$v'ball$ is the speed of ball in frame $S'$
$c$ is the speed of light

Substitute $1.80×1013 m$ for $Lp$ and $0.800c$ for $v'ball$ to find $Δtp$ .

$Δtp=1.80×1013 m0.800c=1.80×1013 m0.800(3.00×108 m/s)=7.50×103 s$

Thus, the time interval took for the ball to reach Dina according to her is $7.50×103 m$ .

Conclusion:

The time interval took for the ball to reach Dina according to her is $7.50×103 m$ .

Answer 22

(c)

Expert Solution

To determine

The distance between Owen and Dina measured by Ed.

Answer to Problem 45AP

The distance between Owen and Dina measured by Ed is

$1.44×1012 m$ .

Explanation of Solution

Given Info:

The speed of frame $S'$ is $0.600c$ .

Formula to calculate the distance between Owen and Dina measured by Ed is,

$L=Lp1−(vc)2$

$L$ is the distance between Owen and Dina measured by Ed
$v$ is the speed of frame $S'$

Substitute $0.600c$ for $v$ and $1.80×1012 m$ for $Lp$ to find $L$ .

$L=(1.80×1012 m)1−(0.600cc)2=1.44×1012 m$

Thus, the distance between Owen and Dina measured by Ed is $1.44×1012 m$ .

Conclusion:

The distance between Owen and Dina measured by Ed is $1.44×1012 m$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

The distance between Owen and Dina measured by Ed.

Answer 27

Answer to Problem 45AP

The distance between Owen and Dina measured by Ed is

$1.44×1012 m$ .

Answer 28

Explanation of Solution

Given Info:

The speed of frame $S'$ is $0.600c$ .

Formula to calculate the distance between Owen and Dina measured by Ed is,

$L=Lp1−(vc)2$

$L$ is the distance between Owen and Dina measured by Ed
$v$ is the speed of frame $S'$

Substitute $0.600c$ for $v$ and $1.80×1012 m$ for $Lp$ to find $L$ .

$L=(1.80×1012 m)1−(0.600cc)2=1.44×1012 m$

Thus, the distance between Owen and Dina measured by Ed is $1.44×1012 m$ .

Conclusion:

The distance between Owen and Dina measured by Ed is $1.44×1012 m$ .

Answer 29

(d)

Expert Solution

To determine

The speed of ball relative to Ed.

Answer to Problem 45AP

The speed of ball relative to Ed is

$0.385c$ .

Explanation of Solution

Given Info:

The speed of ball relative to Dina is $−0.800c$ .

The speed of Dina relative to Ed is $0.600c$ .

Formula to calculate distance covered by the cheetah is,

$vbE=vbD+vDE1+vbDvDEc2$

$vbE$ is the speed of ball relative to Ed
$vbD$ is the speed of ball relative to Dina
$vDE$ is the speed of Dina relative to Ed

Substitute $−0.800c$ for $vbD$ and $0.600c$ for $vDE$ to find $vbE$ .

$vbE=(−0.800c)+(0.600c)1+(−0.800c)(0.600c)c2=0.385c$

Thus, the speed of ball relative to Ed is $0.385c$ .

Conclusion:

The speed of ball relative to Ed is $0.385c$

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

The speed of ball relative to Ed.

Answer 34

Answer to Problem 45AP

The speed of ball relative to Ed is

$0.385c$ .

Answer 35

Explanation of Solution

Given Info:

The speed of ball relative to Dina is $−0.800c$ .

The speed of Dina relative to Ed is $0.600c$ .

Formula to calculate distance covered by the cheetah is,

$vbE=vbD+vDE1+vbDvDEc2$

$vbE$ is the speed of ball relative to Ed
$vbD$ is the speed of ball relative to Dina
$vDE$ is the speed of Dina relative to Ed

Substitute $−0.800c$ for $vbD$ and $0.600c$ for $vDE$ to find $vbE$ .

$vbE=(−0.800c)+(0.600c)1+(−0.800c)(0.600c)c2=0.385c$

Thus, the speed of ball relative to Ed is $0.385c$ .

Conclusion:

The speed of ball relative to Ed is $0.385c$

Answer 36

Ch. 26.3 - Prob. 26.1QQ Ch. 26.4 - Suppose youre an astronaut being paid according to...Ch. 26.4 - True or False: People traveling near the speed of...Ch. 26.4 - You are packing for a trip to another star, and on...Ch. 26.4 - You observe a locket moving away from you. (i)...Ch. 26.7 - Prob. 26.6QQ Ch. 26.7 - Prob. 26.7QQ Ch. 26 - Choose the option from each pair that makes the...Ch. 26 - Choose the option that makes the following...Ch. 26 - Choose the option that makes the following...

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