(a) The object moves with uniform velocity from A to B.
(b) The object moves with constant velocity from B to C.
(c) The object moves with increasing velocity from C to D.
<h3>
Velocity of the object from point A to B</h3>
V(A to B) = (6 - 0)/(4 - 0) = 1.5 m/s
<h3>
Velocity of the object from point B to C</h3>
V(B to C) = (6 - 6)/(11 - 4) = 0 m/s
<h3>
Velocity of the object from point C to D</h3>
V(C to D) = (7 - 6)/(12 - 11) = 1 m/s
final velocity = 1 + 1.5 m/s = 2.5 m/s
Thus, we can conclude the following;
The object moves with uniform velocity from A to B.
The object moves with constant velocity from B to C.
The object moves with increasing velocity from C to D.
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Answer:
the work is done by the gas on the environment -is W= - 3534.94 J (since the initial pressure is lower than the atmospheric pressure , it needs external work to expand)
Explanation:
assuming ideal gas behaviour of the gas , the equation for ideal gas is
P*V=n*R*T
where
P = absolute pressure
V= volume
T= absolute temperature
n= number of moles of gas
R= ideal gas constant = 8.314 J/mol K
P=n*R*T/V
the work that is done by the gas is calculated through
W=∫pdV= ∫ (n*R*T/V) dV
for an isothermal process T=constant and since the piston is closed vessel also n=constant during the process then denoting 1 and 2 for initial and final state respectively:
W=∫pdV= ∫ (n*R*T/V) dV = n*R*T ∫(1/V) dV = n*R*T * ln (V₂/V₁)
since
P₁=n*R*T/V₁
P₂=n*R*T/V₂
dividing both equations
V₂/V₁ = P₁/P₂
W= n*R*T * ln (V₂/V₁) = n*R*T * ln (P₁/P₂ )
replacing values
P₁=n*R*T/V₁ = 2 moles* 8.314 J/mol K* 300K / 0.1 m3= 49884 Pa
since P₂ = 1 atm = 101325 Pa
W= n*R*T * ln (P₁/P₂ ) = 2 mol * 8.314 J/mol K * 300K * (49884 Pa/101325 Pa) = -3534.94 J
Answer:
The battery can supply 130 W for 11.75 h
Explanation:
In order to discover the time in wich the battery can supply this energy we need to find how much current is being drawn from it, we do that by using the equation for real power that is P = V*I, since we have V and P we can solve for I as seen bellow:
I = P/V = 130/12 = 10.834 A
We can use this value to find how many hours the power can supply said current. We do that by dividing the current capacity of the battery by the current drawn:
t = 141/12 = 11.75 h
