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rjkz [21]
3 years ago
11

A mass spectrometer is used to examine the isotopes of uranium. Ions in the beam emerge from the velocity selector at a speed of

2.81 ✕ 105 m/s and enter a uniform magnetic field of 0.605 T directed perpendicularly to the velocity of the ions. What is the distance between the impact points formed on the photographic plate by singly charged ions of 235U and 238U?
Physics
1 answer:
Vsevolod [243]3 years ago
6 0

Answer:

0.2 m

Explanation:

magnetic force = centripetal force

qvB = mv² / r

cancel v on both side and make r subject of the formula

r₁ = mv / qB where m = mass of ion, v = velocity of the ion

r₁  = ( 235 × 1.67 × 10⁻²⁶ × 2.81 × 10⁵) / ( 1.6 × 10⁻¹⁹ × 0.605 T) = 1139.24 × 10⁻² = 11.4 m

r₂ = ( 238 × 1.67 × 10⁻²⁶ × 2.81 × 10⁵) / ( 1.6 × 10⁻¹⁹ × 0.605 T)  = 1153.78 × 10⁻² = 11.5 m

D, distance between the impact points formed = d₂ - d₁ where d₂ and d₁ represent the diameters of the charged ions

D = (2 × 11.5 m) - ( 2 × 11.4) = 2 ( 11.5 - 11.4) = 0.2 m

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2 years ago
A satellite in outer space is moving at a constant velocity of 20.5 m/s in the +y direction when one of its on board thruster tu
Katyanochek1 [597]

Answer:

a)  V_f = 25.514 m/s

b)  Q =53.46 degrees CCW from + x-axis

Explanation:

Given:

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- Acceleration a = 0.31 i m/s^2

- Time duration for acceleration t = 49.0 s

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(a) What is the magnitude of the satellite's velocity when the thruster turns off?

(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis.

Solution:

- We can apply the kinematic equation of motion for our problem assuming a constant acceleration as given:

                                   V_f = V_i + a*t

                                   V_f = 20.5 j + 0.31 i *49

                                   V_f = 20.5 j + 15.19 i

- The magnitude of the velocity vector is given by:

                                   V_f = sqrt ( 20.5^2 + 15.19^2)

                                   V_f = sqrt(650.9861)

                                  V_f = 25.514 m/s

- The direction of the velocity vector can be computed by using x and y components of velocity found above:

                                 tan(Q) = (V_y / V_x)

                                 Q = arctan (20.5 / 15.19)

                                 Q =53.46 degrees

- The velocity vector is at angle @ 53.46 degrees CCW from the positive x-axis.

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