Trees. Every time the wind blows there is a wave of motion which is movement
Answer:
The transverse wave will travel with a speed of 25.5 m/s along the cable.
Explanation:
let T = 2.96×10^4 N be the tension in in the steel cable, ρ = 7860 kg/m^3 is the density of the steel and A = 4.49×10^-3 m^2 be the cross-sectional area of the cable.
then, if V is the volume of the cable:
ρ = m/V
m = ρ×V
but V = A×L , where L is the length of the cable.
m = ρ×(A×L)
m/L = ρ×A
then the speed of the wave in the cable is given by:
v = √(T×L/m)
= √(T/A×ρ)
= √[2.96×10^4/(4.49×10^-3×7860)]
= 25.5 m/s
Therefore, the transverse wave will travel with a speed of 25.5 m/s along the cable.
Answer:
F = - 3.56*10⁵ N
Explanation:
To attempt this question, we use the formula for the relationship between momentum and the amount of movement.
I = F t = Δp
Next, we try to find the time that the average speed in the contact is constant (v = 600m / s), so we say
v = d / t
t = d / v
Given that
m = 26 g = 26 10⁻³ kg
d = 50 mm = 50 10⁻³ m
t = d/v
t = 50 10⁻³ / 600
t = 8.33 10⁻⁵ s
F t = m v - m v₀
This is so, because the bullet bounces the speed sign after the crash is negative
F = m (v-vo) / t
F = 26*10⁻³ (-500 - 640) / 8.33*10⁻⁵
F = - 3.56*10⁵ N
The negative sign is as a result of the force exerted against the bullet
