Answer:
7,166 hrs =430 minutes
Explanation:
Since both train are on the same track, going one towards the other, the relative speed is the addition of both, then the time they need to meet, and consistently crash, is the time that (65mph + 55 mph)=120mph need to travel the total distance of 860 miles, of course in this case one part is traveled by the first train and the rest by the other. Then to find the time we use a three rule
1 h --->120mi
X ---->860mi, then X=(860 mi* 1h)/120 mi = 43/6 hrs= 7,16666 hrs, turning this into minutes need that we notice 1h=60min, then 43/6 hrs *60 min/hrs = 430 minutes.
Answer:
t = 0.319 s
Explanation:
With the sudden movement of the athlete a pulse is formed that takes time to move along the rope, the speed of the rope is given by
v = √T/λ
Linear density is
λ = m / L
λ = 4/20
λ = 0.2 kg / m
The tension in the rope is equal to the athlete's weight, suppose it has a mass of m = 80 kg
T = W = mg
T = 80 9.8
T = 784 N
The pulse rate is
v = √(784 / 0.2)
v = 62.6 m / s
The time it takes to reach the hook can be searched with kinematics
v = x / t
t = x / v
t = 20 / 62.6
t = 0.319 s
Since U=0,
h=1/2gt^2 (h= ut+1/2gt^2, U=0)
h=1/2*10*4*4
h=80m
Atoms of the same element having equal numbers of protons, but different numbers of neutrons is called isotope.
Under the assumption that the tires do not change in volume, apply Gay-Lussac's law:
P/T = const.
P = pressure, T = temperature, the quotient of P/T must stay constant.
Initial P and T values:
P = 210kPa + 101.325kPa
P = 311.325kPa (add 101.325 to change gauge pressure to absolute pressure)
T = 25°C = 298.15K
Final P and T values:
P = ?, T = 0°C = 273.15K
Set the initial and final P/T values equal to each other and solve for the final P:
311.325/298.15 = P/273.15
P = 285.220kPa
Subtract 101.325kPa to find the final gauge pressure:
285.220kPa - 101.325kPa = 183.895271kPa
The final gauge pressure is 184kPa or 26.7psi.
