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lora16 [44]
3 years ago
6

Which of the following observations represent conclusive evidence of an interaction? (Select all that apply.)

Physics
1 answer:
9966 [12]3 years ago
6 0

Answer:

a,b,d and e are correct.

Explanation:

a) Change of temperature without change of velocity is a conclusive evidence of an interaction. As the temperature of a body will only change if heat energy is flows in or out of the body with surrounding. So, option a) is correct.

b) Change of the direction without change of speed is an evidence of an interaction. The direction changes when the object is under a perpendicular acceleration acceleration which clearly means the particle is interacting or external force applied. So, option b is correct.

d) Change of shape or configuration without change of velocity is conclusive evidence of an interaction. The shape can change only if external stress is applied on the particle. So, Option d is correct.

e) Change of identity without change of velocity is a conclusive evidence of an interaction. The identity of an object can only change if it interacts with surroundings. So, option e is correct

Therefore the only incorrect option is c .

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The magnetic field in a cyclotron is 1.25 T, and the maximum orbital radius of the circulating protons is 0.40 m. (a) What is th
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Answer:

1.92 x 10⁻¹²J

Explanation:

The magnetic force from the magnetic field gives the circulating protons gives the particle the necessary centripetal acceleration to keep it orbiting round the circular path. And from Newton's second law of motion, the force(F) is equal to the product of the mass(m) of the proton and the centripetal acceleration(a). i.e

F = ma

Where;

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=> F = m\frac{v^2}{r}           ------------(i)

We also know that the magnitude of this magnetic force experienced by the moving charge (proton) in a magnetic field is given by;

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Where;

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θ = the angle between the velocity and the magnetic field.

Combining equations (i) and (ii) gives

m\frac{v^2}{r} = q v B sin θ           [θ = 90° since the proton is orbiting at the maximum orbital radius]

=> m\frac{v^2}{r} = q v B sin 90°

=> m\frac{v^2}{r} = q v B

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=> m\frac{v}{r} = qB

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From the question;

B = 1.25T

m = mass of proton = 1.67 x 10⁻²⁷kg

r = 0.40m

q = charge of a proton = 1.6 x 10⁻¹⁹C

Substitute these values into equation(iii) as follows;

v = \frac{(1.6*10^{-19})(1.25)(0.4)}{(1.67*10^{-27})}

v = 4.79 x 10⁷m/s

Now, the kinetic energy, K, is given by;

K = \frac{1}{2}mv²

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K = \frac{1}{2}(1.67*10^{-27} * (4.79 * 10^7)^2 )

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