Answer:
0.182 m/s
Explanation:
m1 = 30,000 kg, m2 = 110,000 kg, u1 = 0.85 m/s
let the velocity of loaded freight car is v
Use the conservation of momentum
m1 x u1 + m2 x 0 = (m1 + m2) x v
30,000 x 0.85 = (30,000 + 110,000) x v
v = 0.182 m/s
Answer:
Explanation:
The velocity of a wave in a string is equal to:
v = √(T / (m/L))
where T is the tension and m/L is the mass per length.
To find the mass per length, we need to find the cross-sectional area of the thread.
A = πr² = π/4 d²
A = π (3.0×10⁻⁶ m)²
A = 2.83×10⁻¹¹ m²
So the mass per length is:
m/L = ρA
m/L = (1300 kg/m³) (2.83×10⁻¹¹ m²)
m/L = 3.68×10⁻⁸ kg/m
So the wave velocity is:
v = √(T / (m/L))
v = √(7.0×10⁻³ N / (3.68×10⁻⁸ kg/m))
v ≈ 440 m/s
The speed of sound in air at sea level is around 340 m/s. So the spider will feel the vibration in the thread before it hears the sound.
Answer:
V = I * R
R = 2 / 3.5 = .571 ohms maximum resistance of wire
R = ρ L / A where R is proportional to L and inversely proportional to A
A = ρ L / R minimum area of wire
ρ = 1 / μ = 1.67E-8 ohm-m resistivity inverse of conductivity
A = 1.67E-8 ohm-m * 225 m / .571 ohm = 6.68E-6 m^2
A = 6.68 mm^2 since 1 mm^2 = 10-6 m^2 or 1 mm = 10-3 m
A = Π r^2 = 6.68 mm^2
r = (6.68 / 3.14)^1/2 mm = 2.13 mm radius of wire
d = 2 * r = 4.26 mm
D. How high do the sound waves go
