Answer:
Time period of the osculation will be 0.0671 sec
Explanation:
It is given a vertical spring is stretched by 4 cm
So change in length of the spring x = 4 cm = 0.04 m
Mass which is hung from it m = 12 gram = 0.012 kg
Sprig force will be equal to weight of the mass
So 
k = 244.7 N/m
Now new mass is m = 28 gram = 0.028 kg
So time period with new mass will be
Answer:
55.96kJ
Explanation:
Energy = mass of diethyl ether × enthalpy of vaporization of diethyl ether
Volume (v) = 200mL, density (d) = 0.7138g/mL
Mass = d × v = 0.7138 × 200 = 142.76g
Enthalpy of vaporization of diethyl ether = 29kJ/mol
MW of diethyl ether (C2H5)2O = 74g/mol
Enthalpy in kJ/g = 29kJ/mol ÷ 74g/mol = 0.392kJ/g
Energy = 142.76g × 0.392kJ/g = 55.96kJ
Answer:
1. 610,000 lb ft
2. 490 J
Explanation:
1. First, convert mi/hr to ft/s:
100 mi/hr × (5280 ft / mi) × (1 hr / 3600 s) = 146.67 ft/s
Now find the kinetic energy:
KE = ½ mv²
KE = ½ (1825 lb / 32.2 ft/s²) (146.67 ft/s)²
KE = 610,000 lb ft
2. KE = ½ mv²
KE = ½ (5 kg) (14 m/s)²
KE = 490 J
Answer:
189 m/s
Explanation:
The pilot will experience weightlessness when the centrifugal force, F equals his weight, W.
So, F = W
mv²/r = mg
v² = gr
v = √gr where v = velocity, g = acceleration due to gravity = 9.8 m/s² and r = radius of loop = 3.63 × 10³ m
So, v = √gr
v = √(9.8 m/s² × 3.63 × 10³ m)
v = √(35.574 × 10³ m²/s²)
v = √(3.5574 × 10⁴ m²/s²)
v = 1.89 × 10² m/s
v = 189 m/s
When astronauts travel to the moon, their bodies experience a lower gravitational pull than on Earth, the type of force they are experiencing is <span>A. tension. Tension is the opposite of compression which is pulling of the astronaut from the ground or Earth</span>
