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:
12m
Explanation:
To obtain the answer to the question given, we must observe the characteristics of image formed by a plane mirror.
The image formed by a plane mirror have the following characteristics:
1. Laterally inverted.
2. Same distance as the object from the mirror.
3. Same height as the object.
4. Virtual.
With the above information, we can calculate the distance between the boy and his image as follow:
Initially:
Object distance (u) = 4m
Image distance (v) = 4m
The boy moved 2m away, therefore:
Object distance (u) = 2 + 4 = 6m
Image distanc(v) = 2 + 4 = 6m
The distance between the boy and his image will be the sum of his distance (u) and image distance (v) i.e (u + v)
The distance between the boy and his image = 6 + 6 = 12m
Therefore, the distance between the boy and his image is 12m.
Answer:
Number of turns on the secondary coil of the adapter transformer is 10.
Explanation:
For a transformer,
where 
is the voltage induced in the secondary coil
is the voltage in the primary coil
is the number of turns of secondary coil
is the number of turns of primary coil
From the given question,
= 
⇒ = 
= 9.999733
∴ = 10 turns
Longshore drift has a very powerful influence on the shape and composition of the coastline. It changes the slopes of beaches and creates long, narrow shoals of land called spits, that extend out from shore. Longshore drift may also create or destroy entire “barrier islands” along a shoreline.
