Give me some answer choices and i will be happy to help
Answer
given,
focal length of lens A = 5.77 cm
focal length of lens B= 27.9 cm
flies distance from mirror = 11.3 m
now,
Using lens formula
q =11.79 cm
image of lens A is object of lens B
distance of lens = 59.9 - 11.79 = 48.11
now, Again applying lens formula
q' =66.41 cm
hence, the image distance from the second lens is equal to q' =66.41 cm
I believe it would be an unbalanced force. Because the forces are unbalanced, one side is stronger and, therefore, the object will move.
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.
