The force = 0.3 * 9.8 = 2.94
x = 0.13
k = 2.94/0.13
k = 22.6 N/m
Answer:
<h3>A. 1.62×10⁵m/s²</h3>
Explanation:
Given
velocity of the bullet = 521m/s
distance = 0.840m
required
acceleration of the bullet a
Using the equation of motion:
v² = u²+2aS where:
v is the final velocity
u is the initial velocity
a is the acceleration
S is the distance
Substitute the given parameters into the equation to get a:
521² = 0²+2a(0.840)
271,441 = 0+1.68a
271,441 = 1.68a
a = 271,441/1.68
a = 161,572.02
a = 1.62×10⁵m/s²
<em>Hence the acceleration of the bullet is 1.62×10⁵m/s²</em>
