Answer:
The answer is 1.87nm/s.
Explanation:
The 
water loss must be replaced by of sap. 110g of sap corresponds to a volume of
thus rate of sap replacement is
The volume of sap in the vessel of length 
is
,
where 
is the cross sectional area of the vessel.
For 2000 such vessels, the volume is
taking the derivative of both sides we get:
on the left-hand-side 
is the velocity 
of the sap, and on right-hand-side 
; therefore,
and since the cross-sectional area is
;
therefore,
solving for we get:
which is the upward speed of the sap in each vessel.
The force acting in the front direction is the 130N.
The frictional force is acting backwards 30N.
1) The net force is 130N - 30N = 100N
2) s = ut + (1/2)at^2 u = 0, Start from rest, s = 25m t =5.
25 = 0*5 + (1/2)* a * 5^2.
25 = 0 + 25/2 * a.
25 = (25/2)a. Divide 25 from both sides.
1 = (1/2)* a. Cross multiply.
2 = a.
a = 2 m/s^2.
3) Mass of the box
Net Force, F = ma
100 = m*2. Divide both sides by 2.
100/2 = m
50 = m.
m = 50 kg.
4) Final velocity , v = u + at.
v = 0 + 2*5 = 10 m/s.
Kinetic Energy, K = (1/2) * mv^2.
= 1/2 * 50 * 10 * 10.
= 2500 J.
Answer:
1.) 1620 km/h^2
2.) 2.7 km
Explanation:
1.) Given that the car start from rest. The initial velocity U will be equal to zero. That is,
U = 0.
Final velocity V = 54 km/h
Time t = 2 minute = 2/60 = 1/30 hour
Acceleration a will be change in velocity per time taken. That is,
a = ( V - U )/ t
Substitute V, U and t into the formula
a = 54 ÷ 1/30
a = 54 × 30 = 1620 km/h ^2
2.) Distance travelled S by the car during the time can be calculated by using the 2nd equation of motion.
S = Ut + 1/2at^2
Substitute all the parameters into the formula
S = 54 × 1/30 + 1/2 × 1620 × (1/30)^2
S = 54/30 + 810 × 1/900
S = 54/30 + 810/900
S = (1620+810)/900
S = 2430/900
S = 2.7 km.
Therefore, distance travelled by the car during this time is 2.7 km
