From tables,
SVP at 30°C = 4.24 kPa
From ideal gas expressions;
n = PV/RT = (4.24*1000*450)/(8.314*303) = 757.4 moles
Now, 75% of 757.4 moles will evaporate leaving 20%. Then, 25% of 757.5 moles...
25% of 757.4 moles = 25/100*757.4 = 189.35 moles
Mass of 189.35 moles = 189.35 moles*18 g/mol = 3408.3 g ≈ 3.4 kg
Answer:
The minimum speed of the box bottom of the incline so that it will reach the skier is 8.19 m/s.
Explanation:
It is given that,
Mass of the box, m = 2.2 kg
The box is inclined at an angle of 30 degrees
Vertical distance, d = 3.1 m
The coefficient of friction, 
Using the work energy theorem, the loss of kinetic energy is equal to the sum of gain in potential energy and the work done against friction.


W is the work done by the friction.







v = 8.19 m/s
So, the speed of the box is 8.19 m/s. Hence, this is the required solution.
Let
be the direction the swimmer must swim relative to east. Then her velocity relative to the water is

The current has velocity vector (relative to the Earth)

The swimmer's resultant velocity (her velocity relative to the Earth) is then


We want the resultant vector to be pointing straight north, which means its horizontal component must be 0:

which is approximately 41º west of north.
Answer:
51.2 mi/h
Explanation:
Total distance, d = 100 miles
First 60 miles with speed 55 mi/h
Next 40 miles with speed 75 mi/h
Time taken for first 60 miles, t1 = 60 / 55 = 1.09 h
Time taken for 40 miles, t2 = 40 / 75 = 0.533 h
Time spent to get stuck, t3 = 20 min = 0.33 h
Total time, t = t1 + t2 + t3 = 1.09 + 0.533 + 0.33 = 1.953 h
The average speed is defined as the ratio of total distance traveled to the total time taken.
Average speed = 
Thus, the average speed of the journey is 51.2 mi/h.
We actually don't need to know how far he/she is standing from the net, as we know that the ball reaches its maximum height (vertex) at the net. At the vertex, it's vertical velocity is 0, since it has stopped moving up and is about to come back down, and its displacement is 0.33m. So we use v² = u² + 2as (neat trick I discovered just then for typing the squared sign: hold down alt and type 0178 on ur numpad wtih numlock on!!!) ANYWAY....... We apply v² = u² + 2as in the y direction only. Ignore x direction.
IN Y DIRECTION: v² = u² + 2as 0 = u² - 2gh u = √(2gh) (Sub in values at the very end)
So that will be the velocity in the y direction only. But we're given the angle at which the ball is hit (3° to the horizontal). So to find the velocity (sum of the velocity in x and y direction on impact) we can use: sin 3° = opposite/hypotenuse = (velocity in y direction only) / (velocity) So rearranging, velocity = (velocity in y direction only) / sin 3° = √(2gh)/sin 3° = (√(2 x 9.8 x 0.33)) / sin 3° = 49 m/s at 3° to the horizontal (2 sig figs)