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3 years ago

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An electron is moving at a speed of 4.8 × 10 6 m/s at an angle of 30.0° with respect to a uniform magnetic field of 9.2 × 10 –4

T. What is the radius of the resulting helical path? ( me = 9.11 × 10 –31 kg, qe = 1.60 × 10 –19 C)

1 answer:

omeli [17]3 years ago

6 0

Answer:

r = 5.94 10⁻² m

Explanation:

The magnetic force is given by the ratio

F =q v x B

The bold indicate vectors and the vector product determines that the force is perpendicular to the other two vectors, so the modulus of the velocity does not change, but its direction, therefore the acceleration in Newton's second law is centripetal

F = m a

The magnitude of the force is

F = q v B sin θ

The centripetal acceleration is

a = v² / r

Let's replace

q v B sin θ = m v² / r

r = m v / (q B sin θ)

Let's calculate

r = 9.11 10⁻³¹ 4.8 10⁶ / (1.60 10⁻¹⁹ 9.2 10⁻⁴ sin 30)

r = 5.94 10⁻² m

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4 years ago

1) A car accelerates uniformly from rest and reaches a speed of 19.8 m/s in 7 s. The diameter of a tire is 38.5 cm. Find the num

andre [41]

Answer:

1) 57 revolutions.

2) 4.52 m

3) 1531.2 m

Explanation:

Question 1:

Given:

Initial velocity, $u = 0$

Final velocity, $v = 19.8\ m/s$

Time, $t = 7\ s$

Acceleration, $a = \frac{\textrm{Final velocity-Initial velocity}}{Time}=\frac{19.8}{7}=2.83\ m/s^2$

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$S=ut+\frac{a}{2}t^2\\S=0+\frac{2.83}{2}7^2=69.335\ m$

Displacement of tire in 1 revolution is equal to its circumference.

Therefore, displacement in 1 revolution = $\pi\times(Diameter)=\pi \times 38.5\times 10^{-2}=1.2095\ m$

Now, number of revolutions is given as:

$N=\frac{Total\ displacement}{Displacement\ per\ revolution}\\N=\frac{69.335}{1.2095}=57$

Therefore, the number of revolutions are 57.

Question 2:

Given:

Radius of the wheel is, $R=4.8\ m$

Angle of rotation is, $\theta=54$ °

Converting degree to radians, we get:

$\theta=54\times \frac{\pi}{180}=0.3\pi$

Now, path length is given as:

$L=R\theta=(4.8)(0.3\pi)=1531.2\ m$

Therefore, the path length of a point on the wheel is 4.52 m

Question 3:

Radius of the wheel is, $R=4.8\ m$

Angle of rotation is, $\theta=319$ radians

Now, path length is given as:

$L=R\theta=(4.8)(319)=4.52\ m$

Therefore, the path length of a point on the wheel is 1531.2 m.

7 0

3 years ago