Answer:
a) a = 4.9 m / s², N = 16.97 N and b) F = 9.8 N
Explanation:
a) For this exercise we will use Newton's second law, we write a reference system with the x axis parallel to the plane, see attached, in this system the only force we have to break down is weight, let's use trigonometry
sin 30 = Wx / W
cos 30 = Wy / W
Wx = W sin30
Wy = W cos 30
Let's write the equations on each axis
X axis
Wx = ma
Y Axis
N- Wy = 0
N = Wy = mg cos 30
N = 2.0 9.8 cos 30
N = 16.97 N
We calculate the acceleration
a = Wx / m
a = mg sin 30 / m
a = g sin 30
a =9.8 sin 30
a = 4.9 m / s²
b) For the block to move with constant speed, the acceleration must be zero, so the force applied must be equal to the weight component
F -Wx = 0
F = Wx
F = m g sin 30
F = 2.0 9.8 sin 30
F = 9.8 N
Answer:
L = 1.15 m
Explanation:
The diffraction phenomenon is described by the equation
a sin θ = m λ
Where a is the width of the slit, λ the wavelength and m is an integer, the order of diffraction is left.
The diffraction measurements are made on a screen that is far from the slit, and the angles in the experiment are very small, let's use trigonometry
tan θ = y / L
tan θ = sint θ / cos θ≈ sin θ
We substitute in the first equation
a (y / L) = m λ
The first maximum occurs for m = 1
The distance is measured from the center point of maximum, which coincides with the center of the slit, in this case the distance is the total width of the central maximum, so the distance (y) measured from the center is
y = 1.15 / 2 = 0.575 cm
y = 0.575 10⁻² m
Let's clear the distance to the screen (L)
L = a y / λ
Let's calculate
L = 115 10⁻⁶ 0.575 10⁻² / 575 10⁻⁹
L = 1.15 m
Answer:
i. The radius 'r' of the electron's path is 4.23 × 
m.
ii. The frequency 'f' of the motion is 455.44 KHz.
Explanation:
The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.
r = 
Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.
From the question, B = 1.63 × T, v = 121 m/s, Θ = 
(since it enters perpendicularly to the field), q = e = 1.6 × 
C and m = 9.11 × 
Kg.
Thus,
r = ÷ sinΘ
But, sinΘ = sin = 1.
So that;
r =
= (9.11 × × 121) ÷ (1.6 × × 1.63 × )
= 1.10231 × 
÷ 2.608 × 
= 4.2266 ×
= 4.23 × m
The radius 'r' of the electron's path is 4.23 × m.
B. The frequency 'f' of the motion is called cyclotron frequency;
f = 
= (1.6 × × 1.63 × ) ÷ (2 ×
× 9.11 × )
= 2.608 × ÷ 5.7263 × 
= 455442.4323
f = 455.44 KHz
The frequency 'f' of the motion is 455.44 KHz.
electromagnet
When a electric current is passed through an insulated wire that is coiled around an iron core, like a nail, an electromagnet is created.
