Answer:
Explanation:
Let fuel is released at the rate of dm / dt where m is mass of the fuel
thrust created on rocket
= d ( mv ) / dt
= v dm / dt
this is equal to force created on the rocket
= 220 dv / dt
so applying newton's law
v dm / dt = 220 dv / dt
v dm = 220 dv
dv / v = dm / 220
integrating on both sides
∫ dv / v = ∫ dm / 220
lnv = ( m₂ - m₁ ) / 220
ln4000 - ln 2500 = ( m₂ - m₁ ) / 220
( m₂ - m₁ ) = 220 x ( ln4000 - ln 2500 )
( m₂ - m₁ ) = 220 x ( 8.29 - 7.82 )
= 103.4 kg .
Answer:
Both Thomson and Rutherford used charged particles in their experiments.
Explanation:
Ok so this is simple projectile motion problem.
if we have an object falling in free fall it is subject to gravity of -9.80m/s^2
so it says it takes 6 sec to fall and we know initial velocity was zero so we know that h=vt+1/2gt^2 so we get h=0+1/2*9.80*6^2 = 176.4m
so solving for final speed we get KE=PE = 1/2mv^2=mgh = 1/2v^2=gh so
v=sqrt(2*g*h) = sqrt(2*9.8*176.4m) = 58.8m/s final speed when it hits the ground
hope this helps you! Thanks!!
c adding research resources during an investigation
