Engineers have designed experimental cars that use liquid hydrogen for fuel instead of gasoline. this is an example of Substitution.
Substitution is the process where one thing replaces other for a particular purpose. here liquid hydrogen is replacing the gasoline for fuel in car. hence the correct word to fill is Substitution.
Answer:
a) k = 120 N / m
, b) f = 0.851 Hz
, c) v = 1,069 m / s
, d) x = 0
, e) a = 5.71 m / s²
, f) x = 0.200 m
, g) Em = 2.4 J
, h) v = -1.01 m / s
Explanation:
a) Hooke's law is
F = k x
k = F / x
k = 24.0 / 0.200
k = 120 N / m
b) the angular velocity of the simple harmonic movement is
w = √ k / m
w = √ (120 / 4.2)
w = 5,345 rad / s
Angular velocity and frequency are related.
w = 2π f
f = w / 2π
f = 5.345 / 2π
f = 0.851 Hz
c) the equation that describes the movement is
x = A cos (wt + Ф)
As the body is released without initial velocity, Ф = 0
x = 0.2 cos wt
Speed is
v = dx / dt
v = -A w sin wt
The speed is maximum for sin wt = ±1
v = A w
v = 0.200 5.345
v = 1,069 m / s
d) when the function sin wt = -1 the function cos wt = 0, whereby the position for maximum speed is
x = A cos wt = 0
x = 0
e) the acceleration is
a = d²x / dt² = dv / dt
a = - Aw² cos wt
The acceleration is maximum when cos wt = ± 1
a = A w²
a = 0.2 5.345
a = 5.71 m / s²
f) the position for this acceleration is
x = A cos wt
x = A
x = 0.200 m
g) Mechanical energy is
Em = ½ k A²
Em = ½ 120 0.2²
Em = 2.4 J
h) the position is
x = 1/3 A
Let's calculate the time to reach this point
x = A cos wt
1/3 A = A cos 5.345t
t = 1 / w cos⁻¹(1/3)
The angles are in radians
t = 1.23 / 5,345
t = 0.2301 s
Speed is
v = -A w sin wt
v = -0.2 5.345 sin (5.345 0.2301)
v = -1.01 m / s
i) acceleration
a = -A w² sin wt
a = - 0.2 5.345² cos (5.345 0.2301)
a = -1.91 m / s²
There are two units of sound: intensity and in decibels. Decibels are not additive, you must convert it first to units of intensity (W/m²) using this formula:
dB = 10 log(I/10⁻¹²)
A. 100 dB = 10 log(I/10⁻¹²)
Solving for I,
I = 0.01 W/m²
90 dB = 10 log(I/10⁻¹²)
Solving for I,
I = 0.001
Ratio = 0.01/0.001 = 10
<em>Thus,the choir is 10 times more intense than the soloist.</em>
B. Since there are 90 singers, there would be 9 groups of 10-person choir that produces 100 dB or 0.01 W/m². The total intensity would be
Total intensity = 0.01 W/m² (original choir) + 0.001 W/m² (soloist) + 10(0.01 W/m²) (additional 90 singers) = 0.111 W/m²
dB = 10 log(0.111/10⁻¹²) = <em>110.45 dB</em>
C. Rock concert:
120 dB = 10 log(I/10⁻¹²)
Solving for I,
I = 1 W/m²
Ratio = 1/0.111 = 9
<em>Therefore, the rock concert is 9 times more intense than the choir concert.</em>
Answer:
0.2631 N/C
Explanation:
Given that:
The radius of the wire r = 0.22 mm = 0.22 × 10⁻³ m
The radius of the thick wire r' = 0.55 mm = 0.55 × 10⁻³ m
The numbers of electrons passing through B, N = 6.0 × 10¹⁸ electrons
Electron mobility μ = 6.0 x 10-4 (m/s)/(N/C)
= 0.0006
The number of electron flow per second is calculated as follows:
The magnitude of the electric field is:
E = 
E = 
E = 
E = 0.2631 N/C
