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
Answer:
F = 5
Explanation:
F = m x v^2/r = 2 x 10^2/5 =200/5 = 40 (N)
Answer:
783 grams
Explanation:
Here mass is given in kg
Some of the prefixes of the SI units are
1 gram = 10⁻³ kilogram
1 milligram = 10⁻⁶ kilogram
1 microgram = 10⁻⁹ kilogram
1 nanogram = 10⁻¹² kilogram
The number is 0.783
Here, the only solution where the number of significant figures is three is gram. If any other prefix is chosen then the significant figures will increase


So, 0.783 kg = 783 grams
Answer:
W = 14523.6 J
Explanation:
Given,
Mass = 3.9 Kg
Vertical height , h = 380 m
Work done against gravitational force
W = m g h
g is acceleration due to gravity
W = 3.9 x 9.8 x 380
W = 14523.6 J
Hence, the work done by the gravitational force is equal to W = 14523.6 J