Answer:
the speed of the bullet before striking the block is 302.3 m/s.
Explanation:
Given;
mass of the bullet, m₁ = 28.3 g = 0.0283 kg
mass of the wooden block, m₂ = 5004 g = 5.004 kg
initial velocity of the block, u₂ = 0
final velocity of the bullet-wood system, v = 1.7 m/s
let the initial velocity of the bullet before striking the block = u₁
Apply the principle of conservation of linear momentum to determine the initial velocity of the bullet.
m₁u₁ + m₂u₂ = v(m₁ + m₂)
0.0283u₁ + 5.004 x 0 = 1.7(0.0283 + 5.004)
0.0283u₁ = 8.5549
u₁ = 8.5549 / 0.0283
u₁ = 302.3 m/s
Therefore, the speed of the bullet before striking the block is 302.3 m/s.
Answer: A.E.
Explanation: To understand the distinction between mass and weight and to be able to calculate the weight of an object from its mass and Newton's law of gravitation. The concepts of mass and weight are often confused. In fact, in everyday conversations, the word "weight" often replaces "mass," as in "My weight is seventy-five kilograms"
Answer: I am getting a result around the 54,800 area, so I am selecting the 55,000 HP answer.
I don’t know but I am pretty sure b
Answer:
x(t) = ⅟₁₀₈t⁴ + 10t + 24
v(t) = ⅟₂₇t³ + 10
Explanation:
a(t) = C₁t²
velocity is the integral of acceleration
v(t) = ⅓C₁t³ + C₂
position is the integral of velocity
x(t) = (⅟₁₂C₁)t⁴ + C₂t + C₃
x(0) = 24 = (⅟₁₂C₁)0⁴ + C₂0 + C₃
C₃ = 24
x(6) = 96 = (⅟₁₂C₁)6⁴ + C₂6 + 24
72 = 108C₁ + 6C₂
C₂ = 12 - 18C₁
v(6) = 18 = ⅓C₁6³ + C₂
18 = 72C₁ + C₂
18 = 72C₁ + (12 - 18C₁)
6 = 54C₁
C₁ = 1/9
C₂ = 12 - 18(1/9)
C₂ = 10
