Answer:
B = 6.18 10⁻⁶ T
the magnetic field is in the negative direction of the y axis
Explanation:
The magnetic force is given by
F = q v x B
as in the exercise indicate that the velocities perpendicular to the magnetic field,
F = q v B
Newton's second law is
F = m a
let's substitute
q v B = m a
B = m a / q v
let's calculate
B = 9.1 10⁻³¹ 2.50 10¹³ / (1.6 10⁻¹⁹ 2.30 10⁷)
B = 6.18 10⁻⁶ T
The direction of the field can be obtained with the right hand rule, where the thumb points in the direction of the velocity, the fingers extended in the direction of the magnetic field and the palm in the direction of the force for a positive charge.
In the exercise indicate that the velocity is the z axis
the acceleration and therefore the force in the x axis
therefore the magnetic field is in the negative direction of the y axis
Yup the correct answer is A jus finished the quiz :)
hope i helped^_^
Answer:
Part a)

Part b)

Explanation:
Part a)
If block is sliding up then net force must be zero and friction will be in opposite to the direction of motion of the block


so we have





Part b)
If block is sliding down then net force must be zero and friction will be in opposite to the direction of motion of the block


so we have





Answer:
(a) 0.613 m
(b) 0.385 m
(c) vₓ = 1.10 m/s, vᵧ = 3.50 m/s
v = 3.68 m/s², θ = 72.6° below the horizontal
Explanation:
(a) Take down to be positive.
Given in the y direction:
v₀ = 0 m/s
a = 10 m/s²
t = 0.350 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (0 m/s) (0.350 s) + ½ (10 m/s²) (0.350 s)²
Δy = 0.613 m
(b) Given in the x direction:
v₀ = 1.10 m/s
a = 0 m/s²
t = 0.350 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (1.10 m/s) (0.350 s) + ½ (0 m/s²) (0.350 s)²
Δx = 0.385 m
(c) Find: vₓ and vᵧ
vₓ = aₓt + v₀ₓ
vₓ = (0 m/s²) (0.350 s) + 1.10 m/s
vₓ = 1.10 m/s
vᵧ = aᵧt + v₀ᵧ
vᵧ = (10 m/s²) (0.350 s) + 0 m/s
vᵧ = 3.50 m/s
The magnitude is:
v² = vₓ² + vᵧ²
v = 3.68 m/s²
The direction is:
θ = atan(vᵧ / vₓ)
θ = 72.6° below the horizontal
If the motorcycle can go from 0 to 50mph in 3 seconds the person is describing the motorcycle's acceleration.