See the picture below
Heat is added to an open pan of water at 100.0°c, vaporizing the water. the expanding steam that results does 43.0 kj of work, a
2 answers:
Answer:
Heat transferred to the gas is given as
Q = 647 kJ
Explanation:
As per first law of thermodynamics we know that
now here we know that
change in internal energy of the gas is
Also the steam expands so we will have
since volume increases to here work is done by the gas
now from above equation the heat given to the system is
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(a) 0.714 cm
First of all, we need to find the spring constant of the spring. This can be done by using Hooke's law:
where
F is the force applied on the spring
k is the spring constant
x is the stretching of the spring
At the beginning, the force applied is the weight of the block of m = 4.20 kg hanging on the spring, therefore:
The stretching of the spring due to this force is
x = 2.00 cm = 0.02 m
Therefore, the spring constant is
Now, a new object of 1.50 kg is hanging on the spring instead of the previous one. So, the weight of this object is
And so, the stretching of th spring in this case is
(b) 1.65 J
The work done on a spring is given by:
where
k is the spring constant
x is the stretching of the spring
In this situation,
k = 2060 N/m
x = 4.00 cm = 0.04 m is the stretching due to the external agent
So, the work done is
Answer:
t₂=6.35min
Explanation:
t₁ = first observed time (=5.1 min)
t₂ = unknown; this is the quantity we want to find
V₁ = observer's initial speed (=0.84c)
V₂ = observer's final speed (=0.90c)
Lorentz factors for V₁ and V₂:
γ₁ = 1/√(1−(V₁/c)²)
γ₂ = 1/√(1−(V₂/c)²)
The "proper time" (the time measured by the person filling her car) is:
t′ = t₁/γ₁
The proper time is stated to be the same for both observations, so we also have:
t′ = t₂/γ₂
Combine those two equations and solve for t₂
t₂ = t₁(γ₂/γ₁)
t₂= t₁√((1−(V₁/c)²)/(1−(V₂/c)²))