Answer:
- The difference in length for steel is 2.46 x 10⁻⁴ m
- The difference in length for invar is 1.845 x 10⁻⁵ m
Explanation:
Given;
original length of steel, L₁ = 1.00 m
original length of invar, L₁ = 1.00 m
coefficients of volume expansion for steel, 
= 3.6 × 10⁻⁵ /°C
coefficients of volume expansion for invar, 
= 2.7 × 10⁻⁶ /°C
temperature rise in both meter stick, θ = 20.5°C
Difference in length, can be calculated as:
L₂ = L₁ (1 + αθ)
L₂ = L₁ + L₁αθ
L₂ - L₁ = L₁αθ
ΔL = L₁αθ
Where;
ΔL is difference in length
α is linear expansivity = 
Difference in length, for steel at 20.5°C:
ΔL = L₁αθ
Given;
L₁ = 1.00 m
θ = 20.5°C
ΔL = 1 x 1.2 x 10⁻⁵ x 20.5 = 2.46 x 10⁻⁴ m
Difference in length, for invar at 20.5°C:
ΔL = L₁αθ
Given;
L₁ = 1.00 m
θ = 20.5°C
ΔL = 1 x 0.9 x 10⁻⁶ x 20.5 = 1.845 x 10⁻⁵ m
Complete Question:
Gauss's law:
Group of answer choices
A. can always be used to calculate the electric field.
B. relates the electric field throughout space to the charges distributed through that space.
C. only applies to point charges.
D. relates the electric field at points on a closed surface to the net charge enclosed by that surface.
E. relates the surface charge density to the electric field.
Answer:
D. relates the electric field at points on a closed surface to the net charge enclosed by that surface.
Explanation:
Gauss's law states that the total (net) flux of an electric field at points on a closed surface is directly proportional to the electric charge enclosed by that surface.
This ultimately implies that, Gauss's law relates the electric field at points on a closed surface to the net charge enclosed by that surface.
This electromagnetism law was formulated in 1835 by famous scientists known as Carl Friedrich Gauss.
Mathematically, Gauss's law is given by this formula;
ϕ = (Q/ϵ0)
Where;
ϕ is the electric flux.
Q represents the total charge in an enclosed surface.
ε0 is the electric constant.
Answer:
a) A = 4.0 m
, b) w = 3.0 rad / s
, c) f = 0.477 Hz
, d) T = 20.94 s
Explanation:
The equation that describes the oscillatory motion is
x = A cos (wt + fi)
In the exercise we are told that the expression is
x = 4.0 cos (3.0 t + 0.10)
let's answer the different questions
a) the amplitude is
A = 4.0 m
b) the frequency or angular velocity
w = 3.0 rad / s
c) angular velocity and frequency are related
w = 2π f
f = w / 2π
f = 3 / 2π
f = 0.477 Hz
d) the period
frequency and period are related
T = 1 / f
T = 1 / 0.477
T = 20.94 s
e) the phase constant
Ф = 0.10 rad
f) velocity is defined by
v = dx / dt
v = - A w sin (wt + Ф)
speed is maximum when sine is + -1
v = A w
v = 4 3
v = 12 m / s
g) the angular velocity is
w² = k / m
k = m w²
k = 1.2 3²
k = 10.8 N / m
h) the total energy of the oscillator is
Em = ½ k A²
Em = ½ 10.8 4²
Em = 43.2 J
i) the potential energy is
Ke = ½ k x²
for t = 0 x = 4 cos (0 + 0.1)
x = 3.98 m
j) kinetic energy
K = ½ m v²
for t = 00.1
²
v = A w sin 0.10
v = 4 3 sin 0.10
v = 1.98 m / s
Answer:
25.08m/s
Explanation:
mgh1 + 0.5mv1² = mgh2 + 0.5mv2²
h1 = 0m
v1 = u
h2 = 5m
v2 = 23m/s
putting the values into the formula above;
m(10)(0) + 0.5m(u²) = m(10)(5) + 0.5m(23²)
0 + 0.5mu² = 50m + 264.5m
0.5mu² = 314.5m
dividing through by m
0.5u² = 314.5
u² = 629
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