Explanation:
(a) Given:
Δx = 150 m
v₀ = 27 m/s
v = 54 m/s
Find: a
v² = v₀² + 2aΔx
(54 m/s)² = (27 m/s)² + 2a (150 m)
a = 7.29 m/s²
(b) Given:
Δx = 150 m
v₀ = 0 m/s
a = 7.29 m/s²
Find: t
Δx = v₀ t + ½ at²
150 m = (0 m/s) t + ½ (7.29 m/s²) t²
t = 6.42 s
(c) Given:
v₀ = 0 m/s
v = 27 m/s
a = 7.29 m/s²
Find: t
v = at + v₀
27 m/s = (7.29 m/s²) t + 0 m/s
t = 3.70 s
(d) Given:
v₀ = 0 m/s
v = 27 m/s
a = 7.29 m/s²
Find: Δx
v² = v₀² + 2aΔx
(27 m/s)² = (0 m/s)² + 2 (7.29 m/s²) Δx
Δx = 50 m
Answer:
Explanation:
Given,
mass of the bar = 1.1 Kg
length of rod, l = 0.40 m
diameter of the rod, d = 2 cm
frequency, f = 1.5 MHz
time, t = 0.12 ms
wavelength of the shock wave = ?
Speed of the wave =
v = 3333.33 m/s
wavelength of the wave
Answer:
along the lead, at right angles to the lead at the middle, and at right angles to the lead at one end.
Is the first option (hope so)
Plugs exposed to liquid such as water; too many things plugged in at once
