The shot putter should get out of the way before the ball returns to the launch position.
Assume that the launch height is the reference height of zero.
u = 11.0 m/s, upward launch velocity.
g = 9.8 m/s², acceleration due to gravity.
The time when the ball is at the reference position (of zero) is given by
ut - (1/2)gt² = 0
11t - 0.5*9.8t² = 0
t(11 - 4.9t) = 0
t = 0 or t = 4.9/11 = 0.45 s
t = 0 corresponds to when the ball is launched.
t = 0.45 corresponds to when the ball returns to the launch position.
Answer: 0.45 s
The concept of this problem is the Law of Conservation of Momentum. Momentum is the product of mass and velocity. To obey the law, the momentum before and after collision should be equal:
m₁ v₁ + m₂v₂ = m₁v₁' + m₂v₂', where
m₁ and m₂ are the masses of the proton and the carbon nucleus, respectively,
v₁ and v₂ are the velocities of the proton and the carbon nucleus before collision, respectively,
v₁' and v₂' are the velocities of the proton and the carbon nucleus after collision, respectively,
m(164) + 12m(0) = mv₁' + 12mv₂'
164 = v₁' + 12v₂' --> equation 1
The second equation is the coefficient of restitution, e, which is equal to 1 for perfect collision. The equation is
(v₂' - v₁')/(v₁ - v₂) = 1
(v₂' - v₁')/(164 - 0) = 1
v₂' - v₁'=164 ---> equation 2
Solving equations 1 and 2 simultaneously, v₁' = -138.77 m/s and v₂' = +25.23 m/s. This means that after the collision, the proton bounced to the left at 138.77 m/s, while the stationary carbon nucleus move to the right at 25.23 m/s.
In the circuit outside of the battery the electrons have to expend all of their energy on the internal resistance of the battery which causes heating
Answer:
Technician A only
Explanation:
Both high-side pressures and low-side pressures are low with the engine running and the selector set to the air-conditioning position. Technician A says that the system is undercharged. Technician B says the cooling fan could be inoperative. Which technician is correct?
usually . An overcharged system will result in lower than normal low side pressures
An undercharged system will not enable the compressor to create pressure. As a result of the low amount of refrigerant, the cooling ability is reduced. When we say undercharged, we mean the refrigerant in the system is low, so the both the high side pressures and low side pressures will be low.
Answer:
74.86°C
Explanation:
P₂ = Vapour pressure of water at sea level = 760 mmHg
P₁ = Pressure at base camp = 296 mmHg
T₂ = Temperature of water = 373 K
ΔH°vap for H2O = 40.7 kJ/mol = 40700 J/mol
R = Gas constant = 8.314 J/mol K
From Claussius Clapeyron equation
T₁ = 347.996 K = 74.86°C
∴Water will boil at 74.86°C
