The distance between the astronaut and asteroid.

Question

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Answer 1

Question

Chapter 4, Problem 173P

(a)

To determine

The distance between the astronaut and asteroid.

(a)

Expert Solution

Answer to Problem 173P

The distance between the astronaut and asteroid is $19 m$ .

Explanation of Solution

The mass of the astronaut is $60.0 kg$ , the initial speed of both astronaut and asteroid is $0 m/s$ , the mass of the asteroid is $40.0 kg$ , the force on the asteroid and astronaut is $250 N$ , the time taken during the application of force is $0.35 s$ and time taken after the force is $5.00 s$ .

Write the expression to calculate the acceleration of the astronaut.

$a=FM$

Here, a is the acceleration of the astronaut, F is the force on the astronaut and M is the mass of the astronaut.

Substitute $250 N$ for F and $60.0 kg$ for M in the above equation to calculate a.

$a=250 N60.0 kg=4.17 m/s2$

Write the expression to calculate the acceleration of the asteroid.

$a′=F′m$

Here, $a′$ is the acceleration of the asteroid, $F′$ is the force on the asteroid and m is the mass of the asteroid.

Substitute $250 N$ for $F′$ and $40.0 kg$ for a in the above equation to calculate $a′$ .

$a′=250 N40.0 kg=6.25 m/s2$

Write the expression to calculate the distance travelled by the astronaut.

$x=ut+12at2$

Here, x is the distance moved by the astronaut, u is the initial speed of the astronaut and t is the time taken during the application of force.

Substitute $0 m/s$ for u, $4.17 m/s2$ for a and $0.35 s$ for t in the above equation to calculate x.

$x=(0 m/s)(0.35 s)+12(4.17 m/s2)(0.35 s)2=0 m+0.255 m=0.255 m$

Write the expression to calculate the distance travelled by the asteroid.

$x′=u′t+12a′t2$

Here, $x′$ is the distance moved by the asteroid, $u′$ is the initial speed of the asteroid and t is the time taken during the application of force.

Substitute $0 m/s$ for $u′$ , $6.25 m/s2$ for $a′$ and $0.35 s$ for t in the above equation to calculate $x′$ .

$x′=(0 m/s)(0.35 s)+12(6.25 m/s2)(0.35 s)2=0 m+0.383 m=0.383 m$

Write the expression to calculate the speed of the astronaut.

$v=u+at$

Here, v is the speed of the astronaut.

Substitute $0 m/s$ for u, $4.17 m/s2$ for a and $0.35 s$ for t in the above equation to calculate v.

$v=0 m/s+(4.17 m/s2)0.35 s=1.46 m/s$

Write the expression to calculate the speed of the asteroid.

$v′=u′+a′t$

Here, $v′$ is the speed of the astronaut.

Substitute $0 m/s$ for $u′$ , $6.25 m/s2$ for $a′$ and $0.35 s$ for t in the above equation to calculate $v′$ .

$v′=0 m/s+(6.25 m/s2)0.35 s=2.19 m/s$

Write the expression to calculate the distance moved by the astronaut for $5.00 s$ .

$s=vT$

Here, s is the distance moved by the astronaut and T is the time taken after the force.

Substitute $5.00 s$ for T and $1.46 m/s$ for v in the above equation to calculate s.

$s=1.46 m/s(5.00 s)=7.30 m$

Write the expression to calculate the distance moved by the asteroid for $5.00 s$ .

$s′=v′T$

Here, $s′$ is the distance moved by the asteroid.

Substitute $5.00 s$ for T and $2.19 m/s$ for $v′$ in the above equation to calculate $s′$ .

$s′=2.19 m/s(5.00 s)=10.95 m$

Write the expression to calculate the distance between the astronaut and asteroid.

$X=x+x′+s+s′$

Here, X is the distance between astronaut and asteroid.

Substitute $0.255 m$ for x, $0.383 m$ for $x′$ , $7.30 m$ for s and $10.95 m$ for $s′$ in the above equation to calculate X.

$X=0.255 m+0.383 m+7.30 m+10.95 m=18.888 m∼19 m$

Conclusion:

Therefore, the distance between the astronaut and asteroid is $19 m$ .

(b)

To determine

The relative speed between astronaut and asteroid.

(b)

Expert Solution

Answer to Problem 173P

The relative speed between astronaut and asteroid is $3.65 m/s$ .

Explanation of Solution

Both are travelling in opposite direction. Thus the speed with respect one another is the sum of the speed of individual one.

Write the expression to calculate the relative speed between astronaut and asteroid.

$V=v+v′$

Here, V is the relative speed between astronaut and asteroid.

Substitute $1.46 m/s$ for v and $2.19 m/s$ for $v′$ in the above equation to calculate V.

$V=1.46 m/s+2.19 m/s=3.65 m/s$

Conclusion:

Therefore, the relative speed between astronaut and asteroid is $3.65 m/s$ .

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Answer 2

Question

Chapter 4, Problem 173P

(a)

To determine

The distance between the astronaut and asteroid.

(a)

Expert Solution

Answer to Problem 173P

The distance between the astronaut and asteroid is $19 m$ .

Explanation of Solution

The mass of the astronaut is $60.0 kg$ , the initial speed of both astronaut and asteroid is $0 m/s$ , the mass of the asteroid is $40.0 kg$ , the force on the asteroid and astronaut is $250 N$ , the time taken during the application of force is $0.35 s$ and time taken after the force is $5.00 s$ .

Write the expression to calculate the acceleration of the astronaut.

$a=FM$

Here, a is the acceleration of the astronaut, F is the force on the astronaut and M is the mass of the astronaut.

Substitute $250 N$ for F and $60.0 kg$ for M in the above equation to calculate a.

$a=250 N60.0 kg=4.17 m/s2$

Write the expression to calculate the acceleration of the asteroid.

$a′=F′m$

Here, $a′$ is the acceleration of the asteroid, $F′$ is the force on the asteroid and m is the mass of the asteroid.

Substitute $250 N$ for $F′$ and $40.0 kg$ for a in the above equation to calculate $a′$ .

$a′=250 N40.0 kg=6.25 m/s2$

Write the expression to calculate the distance travelled by the astronaut.

$x=ut+12at2$

Here, x is the distance moved by the astronaut, u is the initial speed of the astronaut and t is the time taken during the application of force.

Substitute $0 m/s$ for u, $4.17 m/s2$ for a and $0.35 s$ for t in the above equation to calculate x.

$x=(0 m/s)(0.35 s)+12(4.17 m/s2)(0.35 s)2=0 m+0.255 m=0.255 m$

Write the expression to calculate the distance travelled by the asteroid.

$x′=u′t+12a′t2$

Here, $x′$ is the distance moved by the asteroid, $u′$ is the initial speed of the asteroid and t is the time taken during the application of force.

Substitute $0 m/s$ for $u′$ , $6.25 m/s2$ for $a′$ and $0.35 s$ for t in the above equation to calculate $x′$ .

$x′=(0 m/s)(0.35 s)+12(6.25 m/s2)(0.35 s)2=0 m+0.383 m=0.383 m$

Write the expression to calculate the speed of the astronaut.

$v=u+at$

Here, v is the speed of the astronaut.

Substitute $0 m/s$ for u, $4.17 m/s2$ for a and $0.35 s$ for t in the above equation to calculate v.

$v=0 m/s+(4.17 m/s2)0.35 s=1.46 m/s$

Write the expression to calculate the speed of the asteroid.

$v′=u′+a′t$

Here, $v′$ is the speed of the astronaut.

Substitute $0 m/s$ for $u′$ , $6.25 m/s2$ for $a′$ and $0.35 s$ for t in the above equation to calculate $v′$ .

$v′=0 m/s+(6.25 m/s2)0.35 s=2.19 m/s$

Write the expression to calculate the distance moved by the astronaut for $5.00 s$ .

$s=vT$

Here, s is the distance moved by the astronaut and T is the time taken after the force.

Substitute $5.00 s$ for T and $1.46 m/s$ for v in the above equation to calculate s.

$s=1.46 m/s(5.00 s)=7.30 m$

Write the expression to calculate the distance moved by the asteroid for $5.00 s$ .

$s′=v′T$

Here, $s′$ is the distance moved by the asteroid.

Substitute $5.00 s$ for T and $2.19 m/s$ for $v′$ in the above equation to calculate $s′$ .

$s′=2.19 m/s(5.00 s)=10.95 m$

Write the expression to calculate the distance between the astronaut and asteroid.

$X=x+x′+s+s′$

Here, X is the distance between astronaut and asteroid.

Substitute $0.255 m$ for x, $0.383 m$ for $x′$ , $7.30 m$ for s and $10.95 m$ for $s′$ in the above equation to calculate X.

$X=0.255 m+0.383 m+7.30 m+10.95 m=18.888 m∼19 m$

Conclusion:

Therefore, the distance between the astronaut and asteroid is $19 m$ .

(b)

To determine

The relative speed between astronaut and asteroid.

(b)

Expert Solution

Answer to Problem 173P

The relative speed between astronaut and asteroid is $3.65 m/s$ .

Explanation of Solution

Both are travelling in opposite direction. Thus the speed with respect one another is the sum of the speed of individual one.

Write the expression to calculate the relative speed between astronaut and asteroid.

$V=v+v′$

Here, V is the relative speed between astronaut and asteroid.

Substitute $1.46 m/s$ for v and $2.19 m/s$ for $v′$ in the above equation to calculate V.

$V=1.46 m/s+2.19 m/s=3.65 m/s$

Conclusion:

Therefore, the relative speed between astronaut and asteroid is $3.65 m/s$ .

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Answer 3

Question

Chapter 4, Problem 173P

(a)

To determine

The distance between the astronaut and asteroid.

(a)

Expert Solution

Answer to Problem 173P

The distance between the astronaut and asteroid is $19 m$ .

Explanation of Solution

The mass of the astronaut is $60.0 kg$ , the initial speed of both astronaut and asteroid is $0 m/s$ , the mass of the asteroid is $40.0 kg$ , the force on the asteroid and astronaut is $250 N$ , the time taken during the application of force is $0.35 s$ and time taken after the force is $5.00 s$ .

Write the expression to calculate the acceleration of the astronaut.

$a=FM$

Here, a is the acceleration of the astronaut, F is the force on the astronaut and M is the mass of the astronaut.

Substitute $250 N$ for F and $60.0 kg$ for M in the above equation to calculate a.

$a=250 N60.0 kg=4.17 m/s2$

Write the expression to calculate the acceleration of the asteroid.

$a′=F′m$

Here, $a′$ is the acceleration of the asteroid, $F′$ is the force on the asteroid and m is the mass of the asteroid.

Substitute $250 N$ for $F′$ and $40.0 kg$ for a in the above equation to calculate $a′$ .

$a′=250 N40.0 kg=6.25 m/s2$

Write the expression to calculate the distance travelled by the astronaut.

$x=ut+12at2$

Here, x is the distance moved by the astronaut, u is the initial speed of the astronaut and t is the time taken during the application of force.

Substitute $0 m/s$ for u, $4.17 m/s2$ for a and $0.35 s$ for t in the above equation to calculate x.

$x=(0 m/s)(0.35 s)+12(4.17 m/s2)(0.35 s)2=0 m+0.255 m=0.255 m$

Write the expression to calculate the distance travelled by the asteroid.

$x′=u′t+12a′t2$

Here, $x′$ is the distance moved by the asteroid, $u′$ is the initial speed of the asteroid and t is the time taken during the application of force.

Substitute $0 m/s$ for $u′$ , $6.25 m/s2$ for $a′$ and $0.35 s$ for t in the above equation to calculate $x′$ .

$x′=(0 m/s)(0.35 s)+12(6.25 m/s2)(0.35 s)2=0 m+0.383 m=0.383 m$

Write the expression to calculate the speed of the astronaut.

$v=u+at$

Here, v is the speed of the astronaut.

Substitute $0 m/s$ for u, $4.17 m/s2$ for a and $0.35 s$ for t in the above equation to calculate v.

$v=0 m/s+(4.17 m/s2)0.35 s=1.46 m/s$

Write the expression to calculate the speed of the asteroid.

$v′=u′+a′t$

Here, $v′$ is the speed of the astronaut.

Substitute $0 m/s$ for $u′$ , $6.25 m/s2$ for $a′$ and $0.35 s$ for t in the above equation to calculate $v′$ .

$v′=0 m/s+(6.25 m/s2)0.35 s=2.19 m/s$

Write the expression to calculate the distance moved by the astronaut for $5.00 s$ .

$s=vT$

Here, s is the distance moved by the astronaut and T is the time taken after the force.

Substitute $5.00 s$ for T and $1.46 m/s$ for v in the above equation to calculate s.

$s=1.46 m/s(5.00 s)=7.30 m$

Write the expression to calculate the distance moved by the asteroid for $5.00 s$ .

$s′=v′T$

Here, $s′$ is the distance moved by the asteroid.

Substitute $5.00 s$ for T and $2.19 m/s$ for $v′$ in the above equation to calculate $s′$ .

$s′=2.19 m/s(5.00 s)=10.95 m$

Write the expression to calculate the distance between the astronaut and asteroid.

$X=x+x′+s+s′$

Here, X is the distance between astronaut and asteroid.

Substitute $0.255 m$ for x, $0.383 m$ for $x′$ , $7.30 m$ for s and $10.95 m$ for $s′$ in the above equation to calculate X.

$X=0.255 m+0.383 m+7.30 m+10.95 m=18.888 m∼19 m$

Conclusion:

Therefore, the distance between the astronaut and asteroid is $19 m$ .

(b)

To determine

The relative speed between astronaut and asteroid.

(b)

Expert Solution

Answer to Problem 173P

The relative speed between astronaut and asteroid is $3.65 m/s$ .

Explanation of Solution

Both are travelling in opposite direction. Thus the speed with respect one another is the sum of the speed of individual one.

Write the expression to calculate the relative speed between astronaut and asteroid.

$V=v+v′$

Here, V is the relative speed between astronaut and asteroid.

Substitute $1.46 m/s$ for v and $2.19 m/s$ for $v′$ in the above equation to calculate V.

$V=1.46 m/s+2.19 m/s=3.65 m/s$

Conclusion:

Therefore, the relative speed between astronaut and asteroid is $3.65 m/s$ .

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Answer 4

Question

Chapter 4, Problem 173P

(a)

To determine

The distance between the astronaut and asteroid.

(a)

Expert Solution

Answer to Problem 173P

The distance between the astronaut and asteroid is $19 m$ .

Explanation of Solution

The mass of the astronaut is $60.0 kg$ , the initial speed of both astronaut and asteroid is $0 m/s$ , the mass of the asteroid is $40.0 kg$ , the force on the asteroid and astronaut is $250 N$ , the time taken during the application of force is $0.35 s$ and time taken after the force is $5.00 s$ .

Write the expression to calculate the acceleration of the astronaut.

$a=FM$

Here, a is the acceleration of the astronaut, F is the force on the astronaut and M is the mass of the astronaut.

Substitute $250 N$ for F and $60.0 kg$ for M in the above equation to calculate a.

$a=250 N60.0 kg=4.17 m/s2$

Write the expression to calculate the acceleration of the asteroid.

$a′=F′m$

Here, $a′$ is the acceleration of the asteroid, $F′$ is the force on the asteroid and m is the mass of the asteroid.

Substitute $250 N$ for $F′$ and $40.0 kg$ for a in the above equation to calculate $a′$ .

$a′=250 N40.0 kg=6.25 m/s2$

Write the expression to calculate the distance travelled by the astronaut.

$x=ut+12at2$

Here, x is the distance moved by the astronaut, u is the initial speed of the astronaut and t is the time taken during the application of force.

Substitute $0 m/s$ for u, $4.17 m/s2$ for a and $0.35 s$ for t in the above equation to calculate x.

$x=(0 m/s)(0.35 s)+12(4.17 m/s2)(0.35 s)2=0 m+0.255 m=0.255 m$

Write the expression to calculate the distance travelled by the asteroid.

$x′=u′t+12a′t2$

Here, $x′$ is the distance moved by the asteroid, $u′$ is the initial speed of the asteroid and t is the time taken during the application of force.

Substitute $0 m/s$ for $u′$ , $6.25 m/s2$ for $a′$ and $0.35 s$ for t in the above equation to calculate $x′$ .

$x′=(0 m/s)(0.35 s)+12(6.25 m/s2)(0.35 s)2=0 m+0.383 m=0.383 m$

Write the expression to calculate the speed of the astronaut.

$v=u+at$

Here, v is the speed of the astronaut.

Substitute $0 m/s$ for u, $4.17 m/s2$ for a and $0.35 s$ for t in the above equation to calculate v.

$v=0 m/s+(4.17 m/s2)0.35 s=1.46 m/s$

Write the expression to calculate the speed of the asteroid.

$v′=u′+a′t$

Here, $v′$ is the speed of the astronaut.

Substitute $0 m/s$ for $u′$ , $6.25 m/s2$ for $a′$ and $0.35 s$ for t in the above equation to calculate $v′$ .

$v′=0 m/s+(6.25 m/s2)0.35 s=2.19 m/s$

Write the expression to calculate the distance moved by the astronaut for $5.00 s$ .

$s=vT$

Here, s is the distance moved by the astronaut and T is the time taken after the force.

Substitute $5.00 s$ for T and $1.46 m/s$ for v in the above equation to calculate s.

$s=1.46 m/s(5.00 s)=7.30 m$

Write the expression to calculate the distance moved by the asteroid for $5.00 s$ .

$s′=v′T$

Here, $s′$ is the distance moved by the asteroid.

Substitute $5.00 s$ for T and $2.19 m/s$ for $v′$ in the above equation to calculate $s′$ .

$s′=2.19 m/s(5.00 s)=10.95 m$

Write the expression to calculate the distance between the astronaut and asteroid.

$X=x+x′+s+s′$

Here, X is the distance between astronaut and asteroid.

Substitute $0.255 m$ for x, $0.383 m$ for $x′$ , $7.30 m$ for s and $10.95 m$ for $s′$ in the above equation to calculate X.

$X=0.255 m+0.383 m+7.30 m+10.95 m=18.888 m∼19 m$

Conclusion:

Therefore, the distance between the astronaut and asteroid is $19 m$ .

(b)

To determine

The relative speed between astronaut and asteroid.

(b)

Expert Solution

Answer to Problem 173P

The relative speed between astronaut and asteroid is $3.65 m/s$ .

Explanation of Solution

Both are travelling in opposite direction. Thus the speed with respect one another is the sum of the speed of individual one.

Write the expression to calculate the relative speed between astronaut and asteroid.

$V=v+v′$

Here, V is the relative speed between astronaut and asteroid.

Substitute $1.46 m/s$ for v and $2.19 m/s$ for $v′$ in the above equation to calculate V.

$V=1.46 m/s+2.19 m/s=3.65 m/s$

Conclusion:

Therefore, the relative speed between astronaut and asteroid is $3.65 m/s$ .

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Answer 5

Chapter 4, Problem 173P

(a)

To determine

The distance between the astronaut and asteroid.

(a)

Expert Solution

Answer to Problem 173P

The distance between the astronaut and asteroid is $19 m$ .

Explanation of Solution

The mass of the astronaut is $60.0 kg$ , the initial speed of both astronaut and asteroid is $0 m/s$ , the mass of the asteroid is $40.0 kg$ , the force on the asteroid and astronaut is $250 N$ , the time taken during the application of force is $0.35 s$ and time taken after the force is $5.00 s$ .

Write the expression to calculate the acceleration of the astronaut.

$a=FM$

Here, a is the acceleration of the astronaut, F is the force on the astronaut and M is the mass of the astronaut.

Substitute $250 N$ for F and $60.0 kg$ for M in the above equation to calculate a.

$a=250 N60.0 kg=4.17 m/s2$

Write the expression to calculate the acceleration of the asteroid.

$a′=F′m$

Here, $a′$ is the acceleration of the asteroid, $F′$ is the force on the asteroid and m is the mass of the asteroid.

Substitute $250 N$ for $F′$ and $40.0 kg$ for a in the above equation to calculate $a′$ .

$a′=250 N40.0 kg=6.25 m/s2$

Write the expression to calculate the distance travelled by the astronaut.

$x=ut+12at2$

Here, x is the distance moved by the astronaut, u is the initial speed of the astronaut and t is the time taken during the application of force.

Substitute $0 m/s$ for u, $4.17 m/s2$ for a and $0.35 s$ for t in the above equation to calculate x.

$x=(0 m/s)(0.35 s)+12(4.17 m/s2)(0.35 s)2=0 m+0.255 m=0.255 m$

Write the expression to calculate the distance travelled by the asteroid.

$x′=u′t+12a′t2$

Here, $x′$ is the distance moved by the asteroid, $u′$ is the initial speed of the asteroid and t is the time taken during the application of force.

Substitute $0 m/s$ for $u′$ , $6.25 m/s2$ for $a′$ and $0.35 s$ for t in the above equation to calculate $x′$ .

$x′=(0 m/s)(0.35 s)+12(6.25 m/s2)(0.35 s)2=0 m+0.383 m=0.383 m$

Write the expression to calculate the speed of the astronaut.

$v=u+at$

Here, v is the speed of the astronaut.

Substitute $0 m/s$ for u, $4.17 m/s2$ for a and $0.35 s$ for t in the above equation to calculate v.

$v=0 m/s+(4.17 m/s2)0.35 s=1.46 m/s$

Write the expression to calculate the speed of the asteroid.

$v′=u′+a′t$

Here, $v′$ is the speed of the astronaut.

Substitute $0 m/s$ for $u′$ , $6.25 m/s2$ for $a′$ and $0.35 s$ for t in the above equation to calculate $v′$ .

$v′=0 m/s+(6.25 m/s2)0.35 s=2.19 m/s$

Write the expression to calculate the distance moved by the astronaut for $5.00 s$ .

$s=vT$

Here, s is the distance moved by the astronaut and T is the time taken after the force.

Substitute $5.00 s$ for T and $1.46 m/s$ for v in the above equation to calculate s.

$s=1.46 m/s(5.00 s)=7.30 m$

Write the expression to calculate the distance moved by the asteroid for $5.00 s$ .

$s′=v′T$

Here, $s′$ is the distance moved by the asteroid.

Substitute $5.00 s$ for T and $2.19 m/s$ for $v′$ in the above equation to calculate $s′$ .

$s′=2.19 m/s(5.00 s)=10.95 m$

Write the expression to calculate the distance between the astronaut and asteroid.

$X=x+x′+s+s′$

Here, X is the distance between astronaut and asteroid.

Substitute $0.255 m$ for x, $0.383 m$ for $x′$ , $7.30 m$ for s and $10.95 m$ for $s′$ in the above equation to calculate X.

$X=0.255 m+0.383 m+7.30 m+10.95 m=18.888 m∼19 m$

Conclusion:

Therefore, the distance between the astronaut and asteroid is $19 m$ .

(b)

To determine

The relative speed between astronaut and asteroid.

(b)

Expert Solution

Answer to Problem 173P

The relative speed between astronaut and asteroid is $3.65 m/s$ .

Explanation of Solution

Both are travelling in opposite direction. Thus the speed with respect one another is the sum of the speed of individual one.

Write the expression to calculate the relative speed between astronaut and asteroid.

$V=v+v′$

Here, V is the relative speed between astronaut and asteroid.

Substitute $1.46 m/s$ for v and $2.19 m/s$ for $v′$ in the above equation to calculate V.

$V=1.46 m/s+2.19 m/s=3.65 m/s$

Conclusion:

Therefore, the relative speed between astronaut and asteroid is $3.65 m/s$ .

Answer 6

Chapter 4, Problem 173P

(a)

To determine

The distance between the astronaut and asteroid.

(a)

Expert Solution

Answer to Problem 173P

The distance between the astronaut and asteroid is $19 m$ .

Explanation of Solution

The mass of the astronaut is $60.0 kg$ , the initial speed of both astronaut and asteroid is $0 m/s$ , the mass of the asteroid is $40.0 kg$ , the force on the asteroid and astronaut is $250 N$ , the time taken during the application of force is $0.35 s$ and time taken after the force is $5.00 s$ .

Write the expression to calculate the acceleration of the astronaut.

$a=FM$

Here, a is the acceleration of the astronaut, F is the force on the astronaut and M is the mass of the astronaut.

Substitute $250 N$ for F and $60.0 kg$ for M in the above equation to calculate a.

$a=250 N60.0 kg=4.17 m/s2$

Write the expression to calculate the acceleration of the asteroid.

$a′=F′m$

Here, $a′$ is the acceleration of the asteroid, $F′$ is the force on the asteroid and m is the mass of the asteroid.

Substitute $250 N$ for $F′$ and $40.0 kg$ for a in the above equation to calculate $a′$ .

$a′=250 N40.0 kg=6.25 m/s2$

Write the expression to calculate the distance travelled by the astronaut.

$x=ut+12at2$

Here, x is the distance moved by the astronaut, u is the initial speed of the astronaut and t is the time taken during the application of force.

Substitute $0 m/s$ for u, $4.17 m/s2$ for a and $0.35 s$ for t in the above equation to calculate x.

$x=(0 m/s)(0.35 s)+12(4.17 m/s2)(0.35 s)2=0 m+0.255 m=0.255 m$

Write the expression to calculate the distance travelled by the asteroid.

$x′=u′t+12a′t2$

Here, $x′$ is the distance moved by the asteroid, $u′$ is the initial speed of the asteroid and t is the time taken during the application of force.

Substitute $0 m/s$ for $u′$ , $6.25 m/s2$ for $a′$ and $0.35 s$ for t in the above equation to calculate $x′$ .

$x′=(0 m/s)(0.35 s)+12(6.25 m/s2)(0.35 s)2=0 m+0.383 m=0.383 m$

Write the expression to calculate the speed of the astronaut.

$v=u+at$

Here, v is the speed of the astronaut.

Substitute $0 m/s$ for u, $4.17 m/s2$ for a and $0.35 s$ for t in the above equation to calculate v.

$v=0 m/s+(4.17 m/s2)0.35 s=1.46 m/s$

Write the expression to calculate the speed of the asteroid.

$v′=u′+a′t$

Here, $v′$ is the speed of the astronaut.

Substitute $0 m/s$ for $u′$ , $6.25 m/s2$ for $a′$ and $0.35 s$ for t in the above equation to calculate $v′$ .

$v′=0 m/s+(6.25 m/s2)0.35 s=2.19 m/s$

Write the expression to calculate the distance moved by the astronaut for $5.00 s$ .

$s=vT$

Here, s is the distance moved by the astronaut and T is the time taken after the force.

Substitute $5.00 s$ for T and $1.46 m/s$ for v in the above equation to calculate s.

$s=1.46 m/s(5.00 s)=7.30 m$

Write the expression to calculate the distance moved by the asteroid for $5.00 s$ .

$s′=v′T$

Here, $s′$ is the distance moved by the asteroid.

Substitute $5.00 s$ for T and $2.19 m/s$ for $v′$ in the above equation to calculate $s′$ .

$s′=2.19 m/s(5.00 s)=10.95 m$

Write the expression to calculate the distance between the astronaut and asteroid.

$X=x+x′+s+s′$

Here, X is the distance between astronaut and asteroid.

Substitute $0.255 m$ for x, $0.383 m$ for $x′$ , $7.30 m$ for s and $10.95 m$ for $s′$ in the above equation to calculate X.

$X=0.255 m+0.383 m+7.30 m+10.95 m=18.888 m∼19 m$

Conclusion:

Therefore, the distance between the astronaut and asteroid is $19 m$ .

Answer 7

Chapter 4, Problem 173P

(a)

To determine

The distance between the astronaut and asteroid.

Answer 8

(a)

Expert Solution

Answer to Problem 173P

The distance between the astronaut and asteroid is $19 m$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The mass of the astronaut is $60.0 kg$ , the initial speed of both astronaut and asteroid is $0 m/s$ , the mass of the asteroid is $40.0 kg$ , the force on the asteroid and astronaut is $250 N$ , the time taken during the application of force is $0.35 s$ and time taken after the force is $5.00 s$ .

Write the expression to calculate the acceleration of the astronaut.

$a=FM$

Here, a is the acceleration of the astronaut, F is the force on the astronaut and M is the mass of the astronaut.

Substitute $250 N$ for F and $60.0 kg$ for M in the above equation to calculate a.

$a=250 N60.0 kg=4.17 m/s2$

Write the expression to calculate the acceleration of the asteroid.

$a′=F′m$

Here, $a′$ is the acceleration of the asteroid, $F′$ is the force on the asteroid and m is the mass of the asteroid.

Substitute $250 N$ for $F′$ and $40.0 kg$ for a in the above equation to calculate $a′$ .

$a′=250 N40.0 kg=6.25 m/s2$

Write the expression to calculate the distance travelled by the astronaut.

$x=ut+12at2$

Here, x is the distance moved by the astronaut, u is the initial speed of the astronaut and t is the time taken during the application of force.

Substitute $0 m/s$ for u, $4.17 m/s2$ for a and $0.35 s$ for t in the above equation to calculate x.

$x=(0 m/s)(0.35 s)+12(4.17 m/s2)(0.35 s)2=0 m+0.255 m=0.255 m$

Write the expression to calculate the distance travelled by the asteroid.

$x′=u′t+12a′t2$

Here, $x′$ is the distance moved by the asteroid, $u′$ is the initial speed of the asteroid and t is the time taken during the application of force.

Substitute $0 m/s$ for $u′$ , $6.25 m/s2$ for $a′$ and $0.35 s$ for t in the above equation to calculate $x′$ .

$x′=(0 m/s)(0.35 s)+12(6.25 m/s2)(0.35 s)2=0 m+0.383 m=0.383 m$

Write the expression to calculate the speed of the astronaut.

$v=u+at$

Here, v is the speed of the astronaut.

Substitute $0 m/s$ for u, $4.17 m/s2$ for a and $0.35 s$ for t in the above equation to calculate v.

$v=0 m/s+(4.17 m/s2)0.35 s=1.46 m/s$

Write the expression to calculate the speed of the asteroid.

$v′=u′+a′t$

Here, $v′$ is the speed of the astronaut.

Substitute $0 m/s$ for $u′$ , $6.25 m/s2$ for $a′$ and $0.35 s$ for t in the above equation to calculate $v′$ .

$v′=0 m/s+(6.25 m/s2)0.35 s=2.19 m/s$

Write the expression to calculate the distance moved by the astronaut for $5.00 s$ .

$s=vT$

Here, s is the distance moved by the astronaut and T is the time taken after the force.

Substitute $5.00 s$ for T and $1.46 m/s$ for v in the above equation to calculate s.

$s=1.46 m/s(5.00 s)=7.30 m$

Write the expression to calculate the distance moved by the asteroid for $5.00 s$ .

$s′=v′T$

Here, $s′$ is the distance moved by the asteroid.

Substitute $5.00 s$ for T and $2.19 m/s$ for $v′$ in the above equation to calculate $s′$ .

$s′=2.19 m/s(5.00 s)=10.95 m$

Write the expression to calculate the distance between the astronaut and asteroid.

$X=x+x′+s+s′$

Here, X is the distance between astronaut and asteroid.

Substitute $0.255 m$ for x, $0.383 m$ for $x′$ , $7.30 m$ for s and $10.95 m$ for $s′$ in the above equation to calculate X.

$X=0.255 m+0.383 m+7.30 m+10.95 m=18.888 m∼19 m$

Conclusion:

Therefore, the distance between the astronaut and asteroid is $19 m$ .

Answer 13

(b)

To determine

The relative speed between astronaut and asteroid.

(b)

Expert Solution

Answer to Problem 173P

The relative speed between astronaut and asteroid is $3.65 m/s$ .

Explanation of Solution

Both are travelling in opposite direction. Thus the speed with respect one another is the sum of the speed of individual one.

Write the expression to calculate the relative speed between astronaut and asteroid.

$V=v+v′$

Here, V is the relative speed between astronaut and asteroid.

Substitute $1.46 m/s$ for v and $2.19 m/s$ for $v′$ in the above equation to calculate V.

$V=1.46 m/s+2.19 m/s=3.65 m/s$

Conclusion:

Therefore, the relative speed between astronaut and asteroid is $3.65 m/s$ .

Answer 14

(b)

To determine

The relative speed between astronaut and asteroid.

Answer 15

(b)

Expert Solution

Answer to Problem 173P

The relative speed between astronaut and asteroid is $3.65 m/s$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Both are travelling in opposite direction. Thus the speed with respect one another is the sum of the speed of individual one.

Write the expression to calculate the relative speed between astronaut and asteroid.

$V=v+v′$

Here, V is the relative speed between astronaut and asteroid.

Substitute $1.46 m/s$ for v and $2.19 m/s$ for $v′$ in the above equation to calculate V.

$V=1.46 m/s+2.19 m/s=3.65 m/s$

Conclusion:

Therefore, the relative speed between astronaut and asteroid is $3.65 m/s$ .

Answer 20

Ch. 4.1 - CHECKPOINT 4.1A Identify the forces acting on the...Ch. 4.1 - Prob. 4.1PP Ch. 4.1 - Prob. 4.1BCP Ch. 4.1 - Prob. 4.2PP Ch. 4.2 - Prob. 4.3PP Ch. 4.2 - Prob. 4.2CP Ch. 4.2 - Prob. 4.4PP Ch. 4.4 - Prob. 4.5PP Ch. 4.4 - Prob. 4.4CP Ch. 4.5 - Practice Problem 4.6 A Creative Defense After an...

The distance between the astronaut and asteroid.

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The distance between the astronaut and asteroid.

Answer to Problem 173P

Explanation of Solution

The relative speed between astronaut and asteroid.

Answer to Problem 173P

Explanation of Solution

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