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3 years ago

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A bicycle rider traveling east at 10 km/hr sees a blue car pass her appearing to travel west at 50 km/hr. What is the blue car's

apparent speed for an observer riding in a red car traveling 15 km/hr west relative to the Earth? g

1 answer:

BigorU [14]3 years ago

8 0

Answer:

$v_{bR} = 25 km/h$ towards west

Explanation:

As we know that the speed of the blue car as appear to the bicycle rider is given as

$v_{bc} = 50 km/h$ towards west

also it is given that bicycle is moving at speed of 10 km/h towards East

so here we have

$v_{bc} = v_b - v_c$

so we have

$v_b = v_{bc} + v_c$

$v_b = -50 + 10 = 40 km/h$ towards west

now speed of the red car is given as 15 km/h towards west

so here the relative speed of blue car with respect to red car is given as

$v_{bR} = v_b - v_R$

$v_{bR} = 40 - 15 = 25 km/h$ towards west

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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

Nataly_w [17]

Answer:

0.092 m

Explanation:

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.

Since the magnetic force acts as centripetal force, we can write:

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

where

q is the charge of the particle

v is its velocity

B is the strength of the magnetic field

m is the mass of the particle

r is the radius of the orbit

Solving the equation for r,

$r=\frac{mv}{qB}$

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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The answer is C.

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Answer:

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