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KonstantinChe [14]
3 years ago
5

Singly charged uranium-238 ions are accelerated through a potential difference of 2.90 kV and enter a uniform magnetic field of

magnitude 1.29 T directed perpendicular to their velocities. (a) Determine the radius of their circular path. (Give your answer to three significant figures.)
Physics
1 answer:
Lera25 [3.4K]3 years ago
6 0

Answer: 0.091 m

Explanation:

r = 1/B * √(2mV/e), where

r = radius of their circular path

B = magnitude of magnetic field = 1.29 T

m = mass of Uranium -238 ion = 238 * amu = 238 * 1.6*10^-27 kg

V = potential difference = 2.9 kV

e = charge of the Uranium -238 ion = 1.6*10^-19 C

r = 1/1.29 * √[(2 * 238 * 1.6*10^-27 * 2900) / 1.6*10^-19]

r = 1/1.29 * √(2.21*10^-21 / 1.6*10^-19)

r = 1/1.29 * √0.0138

r = 1/1.29 * 0.117

r = 0.091 m

Therefore, the radius of their circular path is 0.091 m

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Answer:

c

Explanation:

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3 years ago
Question: Point B is 25 km due east of point A. Starting from point A, a camel walks 20 km in
Korolek [52]

<em>Resultant angle; θ = 25.59°  </em>

This question is dealing with bearings and distance.

We are told that from point A, the camel walks 20 km at 15° in the south of east direction.

Thus, d_s,e = 20 km

Resolving along the horizontal east direction gives; d_e = 20 cos 15

d_e = 19.32 km

Also, resolving along the vertical south direction gives; d_s = 20 sin 15

d_s = 5.18 km

Net vertical distance; d_vert = 8km - 5.18km = 2.72 km

Net horizontal distance; d_hor = 25km - 19.32 km = 5.68 km

Now, the resultant angle is given by;

tan θ = d_vert/d_hor

tan θ = 2.72/5.68

tan θ = 0.4789

θ = tan^(-1) 0.4789

θ = 25.59°

Read more at; brainly.com/question/22518031

8 0
3 years ago
A(n) 1400-kg car going at 6.32 m/s in the positive x direction collides with a 2900-kg truck at rest. The collision is totally i
tresset_1 [31]

Answer:

a) Acceleration of the car is given as

a_{car} = -21 m/s^2

b) Acceleration of the truck is given as

a_{truck} = 10.15 m/s^2

Explanation:

As we know that there is no external force in the direction of motion of truck and car

So here we can say that the momentum of the system before and after collision must be conserved

So here we will have

m_1v_1 + m_2v_2 = (m_1 + m_2)v

now we have

1400 (6.32) + 2900(0) = (1400 + 2900) v

v = 2.06 m/s

a) For acceleration of car we know that it is rate of change in velocity of car

so we have

a_{car} = \frac{v_f - v_i}{t}

a_{car} = \frac{2.06 - 6.32}{0.203}

a_{car} = -21 m/s^2

b) For acceleration of truck we will find the rate of change in velocity of the truck

so we have

a_{truck} = \frac{v_f - v_i}{t}

a_{truck} = \frac{2.06 - 0}{0.203}

a_{truck} = 10.15 m/s^2

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