The sketch of the system is: two strings, 1 and 2, are attached to the ceiling and to a third string, 3.The third string holds the bag of cement.
The free body diagram of the weight with the string 3, drives to the tension T3 = weihgt => T3 = 325 N
The other free body diagram is around the joint of the three strings.
In this case, you can do the horizontal forces equilibrium equation as:
T1* cos(60) - T2*cos(40) = 0
And the vertical forces equilibrium equation:
Ti sin(60) + T2 sin(40) = T3 = 325 N
Then you have two equations with two unknown variables, T1 and T2
0.5 T1 - 0.766 T2 = 0
0.866 T1 + 0.643T2 = 325
When you solve it you get, T1 = 252.8 N and T2 = 165 N
Answer: T1 = 252.8 N, T2 = 165N, and T3 = 325N
The answer is A. Trust me!
Answer: The first answer for the first problem, and the 2nd answer for the second problem
Explanation: For the first one, if it is absolute zero, the molecules would not move at all.
For the second one, the temperature of the sample will increase due to the movement.
Answer:
25.71 kgm/s
Explanation:
Let K₁ and K₂ be the initial and final kinetic energies of object A and v₁ and v₂ its initial and final speeds.
Given that K₂ = 0.7K₁
1/2mv₂² = 0.7(1/2mv₁²)
v₂ = √0.7v₁ = √0.7 × 20 m/s = ±16.73 m/s
Since A rebounds, its velocity = -16.73 m/s and its momentum change, p₂ = mΔv = m(v₂ - v₁) = 0.7 kg (-16.73 - 20) m/s = 0.7( -36.73) = -25.71 kgm/s.
Th magnitude of object A's momentum change is thus 25.71 kgm/s
Answer:
b) -10 m/s
Explanation:
In perfectly elastic head on collisions of identical masses, the velocities are exchanged with one another.
