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Rainbow [258]
3 years ago
6

One of the primary goals of the Kepler space telescope is to search for Earth-like planets. Data gathered by the telescope indic

ates the existence of one such planet named Boralis orbiting a star 135 ly from our solar system. Consider an interstellar spaceship leaving Earth to travel to Boralis. The ship can reach 0.80c almost instantly and can also decelerate almost instantly. The ship is 143 m long in its reference frame.
1. What is the length of the moving ship (in m) as measured by observers on Earth?


2. How much time (in years) will it take for the spaceship to travel from Earth to Boralis, as measured by an observer on Earth?


3. How much time (in years) will it take for the spaceship to travel from Earth to Boralis, as measured by astronauts in the spaceship?
Physics
1 answer:
amid [387]3 years ago
8 0

Answer:

85.62 m

168.75 years

101.04 years

Explanation:

L_0 = Length of ship = 143 m

v = Velocity of ship = 0.8c

c = Speed of light

s = Distance to Boralis orbit = 135 ly

Gamma value

\gamma=\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}\\\Rightarrow \gamma=\dfrac{1}{\sqrt{1-\dfrac{0.8^2c^2}{c^2}}}\\\Rightarrow \gamma=1.67

Length contraction is given by

L=\dfrac{L_0}{\gamma}\\\Rightarrow L=\dfrac{143}{1.67}\\\Rightarrow L=85.62\ m

The length is 85.62 m

Time taken

t=\dfrac{s}{v}\\\Rightarrow t=\dfrac{135}{0.8}\\\Rightarrow t=168.75\ years

Time taken from the perspective one Earth is 168.75 years

Time dilation is given by

t'=\dfrac{t}{\gamma}\\\Rightarrow t'=\dfrac{168.75}{1.67}\\\Rightarrow t'=101.04\ years

The time taken from the perspective of the ship is 101.04 years

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A torque of 36.5 N · m is applied to an initially motionless wheel which rotates around a fixed axis. This torque is the result
vivado [14]

Answer:

21.6\ \text{kg m}^2

3.672\ \text{Nm}

54.66\ \text{revolutions}

Explanation:

\tau = Torque = 36.5 Nm

\omega_i = Initial angular velocity = 0

\omega_f = Final angular velocity = 10.3 rad/s

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I = Moment of inertia

From the kinematic equations of linear motion we have

\omega_f=\omega_i+\alpha_1 t\\\Rightarrow \alpha_1=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha_1=\dfrac{10.3-0}{6.1}\\\Rightarrow \alpha_1=1.69\ \text{rad/s}^2

Torque is given by

\tau=I\alpha_1\\\Rightarrow I=\dfrac{\tau}{\alpha_1}\\\Rightarrow I=\dfrac{36.5}{1.69}\\\Rightarrow I=21.6\ \text{kg m}^2

The wheel's moment of inertia is 21.6\ \text{kg m}^2

t = 60.6 s

\omega_i = 10.3 rad/s

\omega_f = 0

\alpha_2=\dfrac{0-10.3}{60.6}\\\Rightarrow \alpha_1=-0.17\ \text{rad/s}^2

Frictional torque is given by

\tau_f=I\alpha_2\\\Rightarrow \tau_f=21.6\times -0.17\\\Rightarrow \tau=-3.672\ \text{Nm}

The magnitude of the torque caused by friction is 3.672\ \text{Nm}

Speeding up

\theta_1=0\times t+\dfrac{1}{2}\times 1.69\times 6.1^2\\\Rightarrow \theta_1=31.44\ \text{rad}

Slowing down

\theta_2=10.3\times 60.6+\dfrac{1}{2}\times (-0.17)\times 60.6^2\\\Rightarrow \theta_2=312.03\ \text{rad}

Total number of revolutions

\theta=\theta_1+\theta_2\\\Rightarrow \theta=31.44+312.03=343.47\ \text{rad}

\dfrac{343.47}{2\pi}=54.66\ \text{revolutions}

The total number of revolutions the wheel goes through is 54.66\ \text{revolutions}.

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