Answer:
Amplitude—distance between the resting position and the maximum displacement of the wave
Frequency—number of waves passing by a specific point per second
Period—time it takes for one wave cycle to complete
wavelength λ - the distance between adjacent identical parts of a wave, parallel to the direction of propagation.
Tension - described as the pulling force transmitted axially by the means of a string, a cable, chain, or similar one-dimensional continuous object, or by each end of a rod, truss member, or similar three-dimensional object
This problem involves Newton's universal law of gravitation and the equation to follow would be.
F = GM₁M₂/r²
Given: M₁ = 0.890 Kg; M₂ = 0.890 Kg; F = 8.06 x 10⁻¹¹ N; G = 6.673 X 10⁻¹¹ N m²/Kg²
Solving for distance r = ?
r = √GM₁M₂/F
r = √(6.673 x 10⁻¹¹ N m₂/Kg²)(0.890 Kg)(0.890 Kg)/ 8.06 x 10⁻¹¹ N
r = 0.81 m
Answer:
a) 4.485 kg b) 3.94 kg
Explanation:
since the maximum tension the line can stand is 44 N and for question a the speed is constant (acceleration must be zero since the velocity or speed is not changing), F(tension) = mass * acceleration due to gravity (g) .
44 = m * 9.81m/s^2
m = 44/9.81 = 4.485kg
b) F(tension) = ma + mg ( where a is the acceleration of the body and g is the acceleration of the gravity)
44 = m (a +g)
44 = m (1.37 + 9.81)
44/11.18 = m
m = 3.94 kg
Answer:
I = 21.13 mA ≈ 21 mA
Explanation:
If
I₁ = 5 mA
L₁ = L₂ = L
V₁ = V₂ = V
ρ₁ = 1.68*10⁻⁸ Ohm-m
ρ₂ = 1.59*10⁻⁸ Ohm-m
D₁ = D
D₂ = 2D
S₁ = 0.25*π*D²
S₂ = 0.25*π*(2*D)² = π*D²
If we apply the equation
R = ρ*L / S
where (using Ohm's Law):
R = V / I
we have
V / I = ρ*L / S
If V and L are the same
V / L = ρ*I / S
then
(V / L)₁ = (V / L)₂ ⇒ ρ₁*I₁ / S₁ = ρ₂*I₂ / S₂
If
S₁ = 0.25*π*D² and
S₂ = 0.25*π*(2*D)² = π*D²
we have
ρ₁*I₁ / (0.25*π*D²) = ρ₂*I₂ / (π*D²)
⇒ I₂ = 4*ρ₁*I₁ / ρ₂
⇒ I₂ = 4*1.68*10⁻⁸ Ohm-m*5 mA / 1.59*10⁻⁸ Ohm-m
⇒ I₂ = 21.13 mA
