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vovangra [49]
3 years ago
6

Now assume that Alice and Bob are twins, and Alice left Earth and Bob stayed behind fixing his spaceship. If Alice spent some ti

me moving near the speed of light before returning to Earth, which statement is correct when Alice returns to Earth
Physics
1 answer:
alexgriva [62]3 years ago
8 0
我們的確認為我可以幫你們解決問題,我們要不要買不起房屋貸款土地
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What is the speed of a wave that has a frequency of 125 Hz and a wavelength of 1.25 meters?
kirill115 [55]
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Your "v" or </span>velocity = 156.25 meters/second
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3 years ago
Can anyone pls help me out in dis i am struggling in dis!
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3 0
3 years ago
Having aced your Physics 2111 class, you get a sweet summer-job working in the International Space Station. Your room-mate, Cosm
Sphinxa [80]

Answer:

a

The speed is   s =  5.857 m/s

b

The distance is  D = 22.4  \  m

Explanation:

From the question we are told that

     The speed of the banana is  v =  16 \ m/s

   The distance from my  location is  d =  8.2  \ m  

     The time taken is  t = 1.4 \ s

The speed of the ice cream is

          s =  \frac{d}{t}

substituting values

        s =  \frac{8.4}{1.4}

        s =  5.857 m/s

The distance of separation between i and Valdimir is the same as the distance covered by the banana

   So  

          D =  v * t

substituting values

        D = 16 *  1.4

        D = 22.4  \  m

     

3 0
3 years ago
When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius
Marina CMI [18]

To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

\tau = I\alpha

where

I = Moment of inertia \rightarrow \frac{1}{2}mr^2 For a disk

\alpha = Angular acceleration

The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

2 \alpha \theta = \omega_f^2-\omega_i^2

Where

\omega_{f,i} = Final and Initial Angular velocity

\alpha = Angular acceleration

\theta = Angular displacement

Our values are given as

\omega_i = 0 rad/s

\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})

\omega_f = 47.12rad/s

\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad

r = 7cm = 7*10^{-2}m

m = 17g = 17*10^{-3}kg

Using the expression of angular acceleration we can find the to then find the torque, that is,

2\alpha\theta=\omega_f^2-\omega_i^2

\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}

\alpha = \frac{47.12^2-0^2}{2*6\pi}

\alpha = 58.89rad/s^2

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

\tau = I\alpha

\tau = (\frac{1}{2}mr^2)\alpha

\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)

\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m

Therefore the torque exerted on it is 2.45*10^{-3}N\cdot m

3 0
3 years ago
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