To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.
If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,
So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb
Answer:
-5 m/s
Explanation:
The linear velocity of B is equal and opposite the linear velocity of E.
vB = -vE
vB = -ωE rE
10 m/s = -ωE (12 m)
ωE = -0.833 rad/s
The angular velocity of E is the same as the angular velocity of D.
ωE = ωD
ωD = -0.833 rad/s
The linear velocity of Q is the same as the linear velocity of D.
vQ = vD
vQ = ωD rD
vQ = (-0.833 rad/s) (6 m)
vQ = -5 m/s
Its simple use formuila ,
PV=nRT
n,R is constant as the both have same moles.
so,
(p1v1)/T1 = (p2v2)/T2
so, 128.53338kpa
Answer: The work done in J is 324
Explanation:
To calculate the amount of work done for an isothermal process is given by the equation:
W = amount of work done = ?
P = pressure = 732 torr = 0.96 atm (760torr =1atm)
= initial volume = 5.68 L
= final volume = 2.35 L
Putting values in above equation, we get:
To convert this into joules, we use the conversion factor:
So,
The positive sign indicates the work is done on the system
Hence, the work done for the given process is 324 J
Answer:
a bowling ball because it has the most mass.