As we move from Radiowaves to UV waves in electromagnetic spectrum we know that the wavelength is maximum for radio waves while its frequency is minimum
PART (i)
So as we move from radiowaves to UV region
the Frequency will increase
Part (ii)
As we move from Radiowaves to UV region all the parts must have same speed in same medium
All parts will move with speed of light
so Speed will remain the same
Explanation:
(a) Hooke's law:
F = kx
7.50 N = k (0.0300 m)
k = 250 N/m
(b) Angular frequency:
ω = √(k/m)
ω = √((250 N/m) / (0.500 kg))
ω = 22.4 rad/s
Frequency:
f = ω / (2π)
f = 3.56 cycles/s
Period:
T = 1/f
T = 0.281 s
(c) EE = ½ kx²
EE = ½ (250 N/m) (0.0500 m)²
EE = 0.313 J
(d) A = 0.0500 m
(e) vmax = Aω
vmax = (0.0500 m) (22.4 rad/s)
vmax = 1.12 m/s
amax = Aω²
amax = (0.0500 m) (22.4 rad/s)²
amax = 25.0 m/s²
(f) x = A cos(ωt)
x = (0.0500 m) cos(22.4 rad/s × 0.500 s)
x = 0.00919 m
(g) v = dx/dt = -Aω sin(ωt)
v = -(0.0500 m) (22.4 rad/s) sin(22.4 rad/s × 0.500 s)
v = -1.10 m/s
a = dv/dt = -Aω² cos(ωt)
a = -(0.0500 m) (22.4 rad/s)² cos(22.4 rad/s × 0.500 s)
a = -4.59 m/s²
Answer:
1.1648×10⁻¹¹ N
Explanation:
Using
F = qvBsinФ..................... Equation 1
Where F = Force on the proton, q = charge, v = velocity, B = magnetic Field, Ф = angle between the magnetic Field and the velocity.
Note: The angle between v and B = 90°
Given: v = 5.2×10⁷ m/s, B = 1.4 T, q = 1.6×10⁻¹⁹ C, Ф = 90°
Substitute into equation 1
F = 1.6×10⁻¹⁹(5.2×10⁷)(1.4)sin90°
F = 11.648×10⁻¹²
F = 1.1648×10⁻¹¹ N.
Answer:
Washign machine and swingset
Explanation:
