Answer:
The answer to your question is the letter A) F = 9.23 x 10⁻⁷ N
Explanation:
Data
q₁ = -6.25 x 10⁻⁹ C
q₂ = -6.25 x 10⁻⁹ C
d = 0.617 m
k = 9 x 10⁹ Nm²/C²
F = ?
Formula
F = k q₁q₂ /r²
-Substitution
F = (9 x 10⁹)(-6.25 x 10⁻⁹)(-6.25 x 10⁻⁹) / (0.617)²
-Simplification
F = 3.512 x 10⁻⁷ / 0.381
-Result
F = 9.227 x 10⁻⁷ N ≈ 9.23 x 10⁻⁷ N
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:
1.23 m/s²
Explanation:
Given:
v₀ = 0 m/s
v = 11.1 m/s
t = 9 s
Find: a
Equation:
v = at + v₀
Plug in:
11.1 m/s = a (9 s) + 0 m/s
a = 1.23 m/s²
The runner's acceleration is 1.23 m/s².
It would be C as the law says "<span>Formally stated, </span>Newton's third law<span> is: For every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object."</span>