A bonds with elements from b
Answer:
323 m/s²
Explanation:
Given:
x₀ = 0 m
y₀ = 0 m
x = 29500 cos 65°
y = 29500 sin 65°
v₀x = 1810 cos 20°
v₀y = 1810 sin 20°
t = 9.20
Find:
ax, ay, θ
First, in the x direction:
x = x₀ + v₀ t + ½ at²
29500 cos 32° = 0 + (1810 cos 20°) (9.20) + ½ ax (9.20)²
25017 = 15648 + 42.32 ax
ax ≈ 221.4
And in the y direction:
y = y₀ + v₀ t + ½ at²
29500 sin 32° = 0 + (1810 sin 20°) (9.20) + ½ ay (9.20)²
15633 = 5695 + 42.32 ay
ay ≈ 234.8
Therefore, the magnitude of the acceleration is:
a² = ax² + ay²
a² = (221.4)² + (234.8)²
a ≈ 322.7
Rounded to 3 significant figures, the magnitude of the acceleration is approximately 323 m/s².
remember
change distance 150cm to 1.5m
putting values
f =
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Answer:
m = 0.255kg
Explanation:
from the formular of a mass - spring system
T = 2π√m/k
making m as the subject of formular
m = T² k/4π²
T =5.91s
k = 0.288 N/m
m = 10.059/39.489
m = 0.255kg
Answer: heat to mechanical to electrical
Explanation:
