Answer: The velocity of what?
(5 L/m)x(1,440 m/da)x(365 da/yr)x(20yr)
That's 52,560,000 Liters, or 52,560 m^3.
Basically its just an air like fluid with chemicals.... just like air it just goes as it pleases and fills space
Answer:
Explanation:
Total frictional force on boxes
= total weight x coefficient of friction
= ( 11 + 7 ) x 9.8 x .02 = 3.53 N
Net force on boxes
= 8.9 - 3.53
= 5.37 N
acceleration = 5.37 /( 11 + 7 )
= 0.3 m / s ²
b ) Let us consider movement of block A ( 11 kg )
acceleration a = .3 m/s²
friction force on block A
11 X 9.8 X .02
= 2.156 N
Net force on block A
8.9 - friction force - reaction force by block B
= 8.9 - 2.156 - F_ c
force = mass x acceleration
8.9 - 2.156 - F_ c = 11 x .3
F_ c = 3.444 N
c )
If force is applied from the side of box B
We consider all forces on box B
frictional force on it
= 7 x 9.8 x .02
= 1.372 N
Net force on it
8.9 - 1.372 - F_c
force = mass x acceleration
= 8.9 - 1.372 - F_c = 7 x .3 ( Acceleration of box B will be the same that is 0.3 m/s² )
F_c = 5.428 N
Answer:
s=0.36m part c lands
Explanation:
from conservation of linear momentum
we know that
the momentum before Collision(explosion) is equal to the momentum after collision( explosion)
0=maua+mbub+mcuc
ma=.5kg
mb=.5kg
mc=9.8-(0.5+0.5)
mc=8.8kg
since the fragments all fall to the ground at the same time, we assume same time for all
t=0.27s
the velocity of a
v=u +at
since it falling stariaght down
v=0+9.81*.27
v=2.64m/s is the vertical side of its velocity
, since its falling straight down , the horizontal velocity is zero
the velocity of b
s=ut horizontal component of the distance
6.5=.27*ub
ub=24.07m/s
from
0=maua+mbub+mcuc
0=0.5*0+0.5*24.07+8.8*cu
uc=1.36
the distance from a will be
s=ut
s=1.36*.27
s=0.36m