Answer:
528 liter.
Explanation:
Volume of the tank(cuboid) = l*b*h
But volume of the water = l*b*h
Where
l= length of the tank
b = width of the tank
h = the length from the bottom of the tank,
3.55 in to m,
0.09017m
Length of the water in the tank = 0.570 - 0.09017
= 0.47983 m.
Volume = 0.47983*0.710*1.55
= 0.528 m3.
1 m3 = 1000 liter.
0.528 m3 = 0.528*1000
= 528 liter
Answer:
4 m/s
Explanation:
Momentum is defined as:
where
m is the mass of the object
v is its velocity
For the object in this problem, we know:
p = 200 kg m/s is the momentum
m = 50 kg is the mass
Solving for the velocity, we find:
Answer:
When the starting and ending points are the same, the total work is zero.
Explanation:
option ( D )correct
A force is said to be conservative when the work done by the force in moving a particle from a point A to a point B is independent of the path followed between A and B and is the same for all the paths. The work done depends only on the particles initial and final positions. And when the initial and final position in conservative field are same the work done is said to be zero.
Answer:
D
Explanation:
- The rate of the Diels-Alder is orders of magnitude faster if there is an electron-withdrawing group on the dienophile. For example, replacing a hydrogen on ethene with the electron-withdrawing group CN results in about a 10^5 increase in the reaction rate.
- Other common electron withdrawing functional groups that will accelerate the Diels Alder reaction of dienophiles include aldehydes, ketones, and esters.
- In short, any functional group conjugated with the pi bond which can act as a pi acceptor will accelerate a Diels-Alder reaction with a typical diene.
- See attachment for graphical explanation.
Answer: FR=2.330kN
Explanation:
Write down x and y components.
Fx= FSin30°
Fy= FCos30°
Choose the forces acting up and right as positive.
∑(FR) =∑(Fx )
(FR) x= 5-Fsin30°= 5-0.5F
(FR) y= Fcos30°-4= 0.8660-F
Use Pythagoras theorem
F2R= √F2-11.93F+41
Differentiate both sides
2FRdFR/dF= 2F- 11.93
Set dFR/dF to 0
2F= 11.93
F= 5.964kN
Substitute value back into FR
FR= √F2(F square) - 11.93F + 41
FR=√(5.964)(5.964)-11.93(5.964)+41
FR= 2.330kN
The minimum force is 2.330kN
