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seropon [69]
3 years ago
9

Astronauts often undergo special training in which they are subjected to extremely high centripetal accelerations. One device ha

s a radius of 15 m and can accelerate a person at 98 m/s2. What is the speed of the astronaut in this device?
Physics
1 answer:
sertanlavr [38]3 years ago
5 0

Answer:

 Speed of astronaut = 38.34 m/s

Explanation:

 Centripetal acceleration is given by the expression, a=\frac{v^2}{r}, where v is the velocity and r is the radius.

 Here centripetal acceleration is given, radius is given, we need to find velocity.

 Centripetal acceleration = 98 m/s²

 Radius = 15 m.

 We have,

     a=\frac{v^2}{r} \\ \\ 98=\frac{v^2}{15} \\ \\ v=38.34m/s

Speed of astronaut = 38.34 m/s

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A mass spectrometer is being used to separate common oxygen-16 from the much rarer oxygen-18, taken from a sample of old glacial
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Answer:

0.092 m

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A charged moving particle immersed in a region with magnetic field follows a circular trajectory at constant speed (uniform circular motion), since the magnetic forces acts perpendicular to the direction of motion of the particle.

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qvB=m\frac{v^2}{r}

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r=\frac{mv}{qB}

For the ion of oxygen-16, we have:

m_A=2.66\cdot 10^{-26}kg

q_A = 1.6\cdot 10^{-19}C (it is singly charged)

v_A=2.90\cdot 10^6 m/s

B_A=1.30 T

So the radius of its orbit is

r_A=\frac{m_A v_A}{q_A B_A}=\frac{(2.66\cdot 10^{-26})(2.90\cdot 10^6)}{(1.6\cdot 10^{-19})(1.30)}=0.371 m

For the ion of oxygen-18, we have:

m_B = \frac{18}{16}m_A = 2.99\cdot 10^{-26}kg

q_B = 1.6\cdot 10^{-19}C (it is singly charged)

v_B=2.90\cdot 10^6 m/s

B_B=1.30 T

So the radius of its orbit is

r_B=\frac{m_B v_B}{q_B B_B}=\frac{(2.99\cdot 10^{-26})(2.90\cdot 10^6)}{(1.6\cdot 10^{-19})(1.30)}=0.417 m

After each ion has travelled a semicircle, the separation between the two ions will be twice the difference in their radius, so:

d=2(r_B-r_A)=2(0.417-0.371)=0.092 m

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