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zysi [14]
3 years ago
14

An astronaut finds herself in a predicament in which she has become untethered from her shuttle. She figures that she could get

back to her shuttle by throwing one of three objects she possesses in the opposite direction of the shuttle. The masses of the objects are 5.3 kg, 7.9 kg, and 10.5 kg, respectively. She is able to throw the first object with a speed of 15.00 m/s, the second with a speed of 11.2 m/s, and the third with a speed of 7.0 m/s. If the mass of the astronaut and her remaining gear is 75.0 kg, determine the final speed of the astronaut with respect to the shuttle if she were to throw each object successively, starting with the least massive and ending with the most massive. Assume that the speeds described are those measured in the rest frame of the astronaut.
Physics
1 answer:
Blizzard [7]3 years ago
7 0

In order to solve the problem, it is necessary to apply the concepts related to the conservation of momentum, especially when there is an impact or the throwing of an object.

The equation that defines the linear moment is given by

mV_i = (m-m_O)V_f - m_OV_O

where,

m=Total mass

m_O = Mass of Object

V_i = Velocity before throwing

V_f = Final Velocity

V_O = Velocity of Object

Our values are:

m_1=5.3kgm_2=7.9kg\\m_3=10.5kg\\m_A=75kg\\m_{Total}=m=98.7Kg

Solving to find the final speed, after throwing the object we have

V_f=\frac{mV_0+m_TV_O}{m-m_O}

We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.

That way during each section the equations should be modified depending on the previous one, let's start:

A) 5.3Kg\rightarrow 15m/s

V_{f1}=\frac{mV_0+m_TV_O}{m-m_O}

V_{f1}=\frac{(98.7)*0+5.3*15}{98.7-5.3}

V_{f1}=0.8511m/s

B) 7.9Kg\rightarrow 11.2m/s

V_{f2}=\frac{mV_{f1}+m_TV_O}{m-m_O}

V_{f2}=\frac{(98.7)(0.8511)+(7.9)(11.2)}{98.7-5.3-7.9}

V_{f2} = 2.0173m/s

C) 10.5Kg\rightarrow 7m/s

V_{f3}=\frac{mV_{f2}+m_TV_O}{m-m_O}

V_{f3}=\frac{(98.7)(2.0173)+(10.5)(7)}{98.7-5.3-7.9-10.5}

V_{f3} = 3.63478m/s

Therefore the final velocity of astronaut is 3.63m/s

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Spaceship A is 10 meters long and approaching you from the south at a speed of.7c while spaceship B, which is also 10 meters lon
Valentin [98]

Answer:

a) 0.94 C

b) 7.14 m

c) 3.11 m

d) 1.40 s

e) 2.93 s

Explanation:

First we need to set up a coordinate system. This will have the positive X axis pointing north. So spaceship A has positive speed, and spaceship B has negative speed.

The Lorentz transformation for speed is:

u' = \frac{u - v}{1 - \frac{u*v}{c^2}}

u: speed of spaceship A as observed by you

v: speed of spaceship B as observed by you

In the case of the speed of spaceship A as observed by spaceship B:

u' = \frac{0.7c - (-0.7c)}{1 - \frac{0.7c*(-0.7c)}{c^2}} = 0.94c

The transform for lengths is:

L = L0 * \sqrt{1 - \frac{v^2}{c^2}}

For the case of spaceship A as observed by you:

L = 10 m * \sqrt{1 - \frac{(0.7c)^2}{c^2}} = 7.14 m

For the case of spaceship A as observed by spaceship B:

L = 10 m * \sqrt{1 - \frac{(0.94c)^2}{c^2}} = 3.12 m

The time dilation equation is:

T = \frac{T0}{\sqrt{1-\frac{v^2}{c^2}}}

For the case of the event as observed by you:

\frac{1 s}{\sqrt{1-\frac{(0.7c)^2}{c^2}}} = 1.40 s

For the case of the event as observed by spaceship B:

\frac{1 s}{\sqrt{1-\frac{(0.94c)^2}{c^2}}} = 2.93 s

3 0
3 years ago
A small earthquake starts a lamppost vibrating back and forth. The amplitude of the vibration of the top of the lamppost is 7.0
krok68 [10]

Answer:

(a) T=6.07s

(b) x_{max|4.0s}=3.62cm

Explanation:

For Part (a)

The initial amplitude is given as A=7.0 cm

Apply the equation x_{max}(t)=Ae^{-t/T} with  x_{max}(7.6s)=2.0cm we have:

x_{max}(t)=Ae^{-t/T}\\2.0cm=(7.0cm)e^{-7.6s/T}\\T=-\frac{7.6s}{ln(\frac{2.0cm}{7.0cm} )} \\T=6.07s

For Part (b)

Apply x_{max}(t)=Ae^{-t/T}  with t=4.0s and T=6.07s we have

x_{max}(t)=Ae^{-t/T}\\x_{max}(4.0s)=(7.0cm)e^{-4.0/6.07}\\x_{max|4.0s}=3.62cm

8 0
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