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
Answer:
H = start height (v = 0)
h = present height
v = present speed
assuming no friction
total energy = PE + KE
mgH = mgh + .5mv^2
if PE = KE then
mgH = mgh + mgh
h = H/2
potential energy = kinetic energy when object is at half its start height.
Explanation:
I believe it would be 2m/s.
Answer:
The fraction of the protons would have no electrons 
Explanation:
We are given that
Amoeba has total number of protons=
Net charge, Q=0.300pC
Electrons are fewer than protons=
We have to find the fraction of protons would have no electrons.
The fraction of the protons would have no electrons
=
The fraction of the protons would have no electrons
=
Hence, the fraction of the protons would have no electrons
A homozygous dominant
B: homozygous recessive
C: heterozygous
I hope this helps you :)
