Answer:
D
Explanation:
a weak base is the answer
(A) The total initial momentum of the system is
(1.30 kg) (27.0 m/s) + (23.0 kg) (0 m/s) = 35.1 kg•m/s
(B) Momentum is conserved, so that the total momentum of the system after the collision is
35.1 kg•m/s = (1.30 kg + 23.0 kg) <em>v</em>
where <em>v</em> is the speed of the combined blocks. Solving for <em>v</em> gives
<em>v</em> = (35.1 kg•m/s) / (24.3 kg) ≈ 1.44 m/s
(C) The kinetic energy of the system after the collision is
1/2 (1.30 kg + 23.0 kg) (1.44 m/s)² ≈ 25.4 J
and before the collision, it is
1/2 (1.30 kg) (27.0 m/s)² ≈ 474 J
so that the change in kinetic energy is
∆<em>K</em> = 25.4 J - 474 J ≈ -449 J
Answer:
Explanation:Abiotic is the answer.
Answer:
m1=914.9kg
m2=604.9kg
m3=864.75kg
Explanation
I think we are suppose to find the mass of the crate.
The effective force that moves the body in positive x direction is 3615N
ΣFx = Σma
Then Fx=3615N
Then the masses be m1, m2 and m3
Then,
ΣF = Σ(ma)
3615=(m1+m2+m3)a
Given that a=1.516
The masses are
m1+m2+m3=, 2384.56. Equation 1
Between mass 1 and mass 2 is, F12=1387.
The effective force that pull mass 1 is 1387.
F12=m1 ×a
Therefore,
m1=F12/a
m1=1387/1.516
m1=914.9kg.
The effective force that pulls crate 1 and crate 2 is F23
F23=(m1+m2)a
Therefore
2304=(m1+m2)a
Therefore, since a=1.516
m1+m2=2304/1.516
m1+m2=1519.8kg
Since m1=914.9kg
So, m2=1519.8-m1
m2=1519.8-914.9
m2=604.9kg
Also from equation 1
m1+m2+m3=2384.56
Since m1=914.9kg and m2=604.9kg
Then, m3=2384.56-604.9-914.9
m3=864.75kg
