Work,
in thermodynamics, is the amount of energy that is transferred from one system
to another system without transfer of entropy. It is equal to the external
pressure multiplied by the change in volume of the system. It is expressed as
follows:<span>
W = PdV
Integrating and assuming that P is not affected
by changes in V or it is constant, then we will have:
W = P (V2 - V1)
Substituting the given values:
P = 1.0 atm = 101325 Pa
(V2 - V1) = 0.50 L =
W = 101325 N/m^3 ( 0.50) (1/1000) m^3
W = 50.66 N-m or 50.66 J
<span>
So, in the expansion process about 50.66 J of work is being done.
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False ...................
Answer:
a) v = 0.4799 m / s, b) K₀ = 1600.92 J, K_f = 5.46 J
Explanation:
a) How the two players collide this is a momentum conservation exercise. Let's define a system formed by the two players, so that the forces during the collision are internal and also the system is isolated, so the moment is conserved.
Initial instant. Before the crash
p₀ = m v₁ + M v₂
where m = 95 kg and his velocity is v₁ = -3.75 m / s, the other player's data is M = 111 kg with velocity v₂ = 4.10 m / s, we have selected the direction of this player as positive
Final moment. After the crash
p_f = (m + M) v
as the system is isolated, the moment is preserved
p₀ = p_f
m v₁ + M v₂ = (m + M) v
v =
let's calculate
v =
v = 0.4799 m / s
b) let's find the initial kinetic energy of the system
K₀ = ½ m v1 ^ 2 + ½ M v2 ^ 2
K₀ = ½ 95 3.75 ^ 2 + ½ 111 4.10 ^ 2
K₀ = 1600.92 J
the final kinetic energy
K_f = ½ (m + M) v ^ 2
k_f = ½ (95 + 111) 0.4799 ^ 2
K_f = 5.46 J
Ep is gravitational potential energy
m is mass (kg)
g is gravitational field strength (N/kg)
h is height (m)
Ep= mgh
= 80kg*9.8N/kg*100m
= 78 400 J
Answer:
Diameter
Explanation:
I know that circumference is dependent, so diameter has to be independent