Two forces with magnitudes of 6 pounds and 18 pounds are applied to an object. The magnitude of the resultant is 13 pounds. Find
1 answer:
Answer: the angle between the resultant vector and the vector of the 18 pound is 28°
Explanation:
given that data in the question; as its interpreted in the diagram below;
from the cosine rule, we know that;
a² = b² + c² - 2bc
so
(13)² = (6)² + (18)² - (2 × 6 × 18 ) cos∅
169 = 36 + 324 - 216cos∅
169 = 360 - 216cos∅
216cos∅ = 360 - 169
216cos∅ = 191
cos∅ = 0.8842
∅ = cos⁻¹ ( 0.8842 )
∅ = 27.8° ≈ 28° {nearest whole number}
Therefore the angle between the resultant vector and the vector of the 18 pound is 28°
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Answer:
block velocity v = 0.09186 = 9.18 10⁻² m/s and speed bollet v₀ = 11.5 m / s
Explanation:
We will solve this problem using the concepts of the moment, let's try a system formed by the two bodies, the bullet and the block; In this system all scaffolds during the crash are internal, consequently, the moment is preserved.
Let's write the moment in two moments before the crash and after the crash, let's call the mass of the bullet (m) and the mass of the Block (M)
Before the crash
p₀ = m v₀ + 0
After the crash
= (m + M) v
p₀ =
m v₀ = (m + M) v (1)
Now let's lock after the two bodies are joined, in this case the mechanical energy is conserved, write it in two moments after the crash and when you have the maximum compression of the spring
Initial
Em₀ = K = ½ m v2
Final
E = Ke = ½ k x2
Emo = E
½ m v² = ½ k x²
v² = k/m x²
Let's look for the spring constant (k), with Hook's law
F = -k x
k = -F / x
k = - 0.75 / -0.25
k = 3 N / m
Let's calculate the speed
v = √(k/m) x
v = √ (3/8.00) 0.15
v = 0.09186 = 9.18 10⁻² m/s
This is the spped of the block plus bullet rsystem right after the crash
We substitute calculate in equation (1)
m v₀ = (m + M) v
v₀ = v (m + M) / m
v₀ = 0.09186 (0.008 + 0.992) /0.008
v₀ = 11.5 m / s