<u>Answer:</u>
<em>1. A NaCl solution with a concentration of 50g/100mL of water at 40°C:</em> The NaCl solution with a given concentration is saturated at this temperature .As the temperature increases the solution will more dissolves.
<em>2. A sugar solution with a concentration of 200g/100mL of water at 40°C: </em>The sugar solution with a given concentration is saturated at this temperature. As the temperature increases the solution will more dissolves.
<em>3. A sugar solution with a concentration of 240g/100mL of water at 40°C:</em> The sugar solution with a given concentration is saturated at given temperature.
Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.
By Newton's second law,
- the net horizontal force acting on the beam is

where
are the magnitudes of the tensions in ropes 1 and 2, respectively;
- the net vertical force acting on the beam is

where
and
.
Eliminating
, we have





Solve for
.



Answer:
A) 21.2 kg.m/s at 39.5 degrees from the x-axis
Explanation:
Mass of the smaller piece = 200g = 200/1000 = 0.2 kg
Mass of the bigger piece = 300g = 300/1000 = 0.3 kg
Velocity of the small piece = 82 m/s
Velocity of the bigger piece = 45 m/s
Final momentum of smaller piece = 0.2 × 82 = 16.4 kg.m/s
Final momentum of bigger piece = 0.3 × 45 = 13.5 kg.m/s
since they acted at 90oc to each other (x and y axis) and also momentum is vector quantity; then we can use Pythagoras theorems
Resultant momentum² = 16.4² + 13.5² = 451.21
Resultant momentum = √451.21 = 21.2 kg.m/s at angle 39.5 degrees to the x-axis ( tan^-1 (13.5 / 16.4)
Answer:
25 mm = 0 deg C
200 mm = 100 deg C
200 - 25 = 175 = change in thread per 100 deg C
95 - 25 = 70 mm - change in thread from 0 deg C
70 / 175 * 100 = 40 deg C final temperature at 95 mm