According to the net force, the acceleration of the book is 16.47 m/s².
We need to know about force to solve this problem. According to second Newton's Law, the force applied to an object will be proportional to mass and acceleration. Hence, it can be written as
∑F = m . a
where F is force, m is mass and a is acceleration
From the question above, we know that
m = 3 kg
g = 9.8 m/s²
F1 = 20 N
Find the net force
∑F = F1 + W
∑F = 20 + m . g
∑F = 20 + 3 . 9.8
∑F = 20 + 29.4
∑F = 49.4 N
Find the acceleration
∑F = m . a
49.4 = 3 . a
a = 16.47 m/s²
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Answer:
longitudinal and transverse.
Explanation:
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Directly proportional to pressure
Answer:
50 W
Explanation:
Case 1
Power = V * I
100 = 220 * I
I =
A
Case 2
P = V * I
P = 110 * 
P = 50 W
I think the answer is 50 W
Hope it helps
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