Answer:34 cm
Explanation:
Given
mass of meter stick m=80 gm
stick is balanced when support is placed at 51 cm mark
Let us take 5 gm tack is placed at x cm on meter stick so that balancing occurs at x=50 cm mark
balancing torque
When an airplane is flying straight and level at a constant speed, the lift it produces balances its weight, and the thrust it produces balances its drag. However, this balance of forces changes as the airplane rises and descends, as it speeds up and slows down, and as it turns.
To solve the problem, we enumerate all the given first. Then the required and lastly the solution.
Given:
V1= 1.56x10^3 L = 1560 L P2 = 44.1 kPa
P1 = 98.9 kPa
Required: V2
Solution:
Assuming the gas is ideal. Ideal gas follows Boyle's Law which states that at a given temperature the product of pressure and volume of a gas is constant. In equation,
PV = k
Applying to the problem, we have
P1*V1 = P2*V2
(98.9 kPa)*(1560 L) = (44.1 kPa)*V2
V2 = 3498.5 L
<em>ANSWER: V2 = 3498.5 L</em>
The velocity would switch on the cars
