Answer:
0.358Kg
Explanation:
The potential energy in the spring at full compression = the initial kinetic energy of the bullet/block system
0.5Ke^2 = 0.5Mv^2
0.5(205)(0.35)^2 = 12.56 J = 0.5(M + 0.0115)v^2
Using conservation of momentum between the bullet and the block
0.0115(265) = (M + 0.0115)v
3.0475 = (M + 0.0115)v
v = 3.0475/(M + 0.0115)
plugging into Energy equation
12.56 = 0.5(M + 0.0115)(3.0475)^2/(M + 0.0115)^2
12.56 = 0.5 × 3.0475^2 / ( M + 0.0115 )
12.56 = 0.5 × 9.2872/ M + 0.0115
12.56 = 4.6436/ M + 0.0115
12.56 ( M + 0.0115 ) = 4.6436
12.56M + 0.1444 = 4.6436
12.56M = 4.6436 - 0.1444
12.56 M = 4.4992
M = 4.4992÷12.56
M = 0.358 Kg
Explanation:
its either a or d however i say the best choice is d
Answer: The minimum acceleration for the air plane is 2.269m/s2.
Explanation: To solve such problem the equation of motion are applicable.
The initial velocity is 0 since the airplane was initially standing. We are going to use this equation
V^2=U^2+2as
33^2=0+2a (240)
a= 2.269m/s2
Answer:
one im so sry i have no idea. I have been researshing for about 30min and i cant find anything im so sry:/
Explanation:
Answer:
4.41 m/s^2
Explanation:
(v_f)^2 - (v_i)^2 = 2a * change in distance
(21)^2 - (0)^2 = 2a * 50
a = (21^2)/(2*50)
a = 4.41 m/s^2
