Answer:
average velocity include total displacement whereas average speed include total distance
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:
2081.65 m
Explanation:
We'll begin by calculating the time taken for the load to get to the target. This can be obtained as follow:
Height (h) = 3000 m
Acceleration due to gravity (g) = 10 m/s²
Time (t) =?
h = ½gt²
3000 = ½ × 10 × t²
3000 = 5 × t²
Divide both side by 5
t² = 3000 / 5
t² = 600
Take the square root of both side
t = √600
t = 24.49 s
Finally, we shall determine the distance from the target at which the load should be released. This can be obtained as follow:
Horizontal velocity (u) = 85 m/s
Time (t) = 24.49 s
Horizontal distance (s) =?
s = ut
s = 85 × 24.49
s = 2081.65 m
Thus, the load should be released from 2081.65 m.
Answer:
J for impulse
t for time
F for force
formula is J=F×t
Explanation:
putting values in eqs after rearranging
we need to find force so
F=J ÷t
F=400÷15
=26.67
=27(rounded off)
27N is the Force applied.