Ethylene glycol is termed as the primary ingredients in antifreeze.
The ethylene glycol molecular formula is C₂H₆O₂.
Molar mass of C₂H₆O₂ is = (2×12) +(6×1) + (216) = 62g/mol
Now that antifreeze by mass is 50%, then there is 1kg of ethylene glycol which is present in 1kg of water.
ΔTf = Kf×m
ΔTf = depression in the freezing point.
= freezing point of water freezing point of the solution
= O°c - Tf
= -Tf
Kf = depression in freezing constant of water = 1.86°C/m
M is the molarity of the solution.
=(mass/molar mass) mass of solvent in kg
=1000g/62 (g/mol) /1kg
=16.13m
If we plug the value we get
-Tf = 1.86 × 16.13 = 30
Tf = -30°c
Convert: 1km = 1000 m 200000 m * 0.01 s/m = 2000 s or 33 minutes and 20 seconds.
Answer:
50 N
Explanation:
Let the force in the horizontal rope be F₁ and the force in the diagonal rope be F₂:
The total force in the horizontal and vertical directions must be zero, since the object is at rest and is not accelerating.
The horizontal component of the forces:
F₁ + F₂ = -40N + F₂ = 0
F₂ = 40N
The vertical component of the forces:
F₁ + F₂ - mg = 0 + F₂ - mg = 0
F₂ = mg
If I assume the gravitational constant g = 10 m/s²:
F₂ = (3 kg) * (10 m/s²) = 30N
Adding the horizontal and vertical components of the force F₂:
F₂ = √((40N)² + (30N)²) = 50N
Answer:
(a) 0.177 m
(b) 16.491 s
(c) 25 cycles
Explanation:
(a)
Distance between the maximum and the minimum of the wave = 2A ............ Equation 1
Where A = amplitude of the wave.
Given: A = 0.0885 m,
Distance between the maximum and the minimum of the wave = (2×0.0885) m
Distance between the maximum and the minimum of the wave = 0.177 m.
(b)
T = 1/f ...................... Equation 2.
Where T = period, f = frequency.
Given: f = 4.31 Hz
T = 1/4.31
T = 0.23 s.
If 1 cycle pass through the stationary observer for 0.23 s.
Then, 71.7 cycles will pass through the stationary observer for (0.23×71.7) s.
= 16.491 s.
(c)
If 1.21 m contains 1 cycle,
Then, 30.7 m will contain (30.7×1)/1.21
= 25.37 cycles
Approximately 25 cycles.
