Explanation :
The heating curve shows how water changes from one state of matter to another based on temperature and the addition or removal of heat over time.
Initially, ice is heated until its temperature reaches 
and changes to liquid state.
From the attached graph it is clear that until 
the temperature will rise steadily. Here, the liquid begins to vaporize. Vaporization is the state of matter at which liquid state changes to the gaseous state.
So, E is the point which shows the gaseous state.
It often requires STANDARD reaction time
Your answer is 311.29271 lbs
Answer:
a) v = 2.4125 m / s , b) Em_{f} / Em₀ = 0.89
Explanation:
a) This is an inelastic crash problem, the system is made up of the four carriages, so the forces during the crash are internal and the moment is conserved
Initial
p₀ = m v₁ + 3 m v₂
Final
= (4 m) v
p₀ =p_{f}
m (v₁ + 3 v₂) = 4 m v
v = (v₁ +3 v₂) / 4
Let's calculate
v = (3.86 + 3 1.93) / 4
v = 2.4125 m / s
b) the initial mechanical energy is
Em₀ = K₁ + 3 K₂
Em₀ = ½ m v₁² + ½ 3m v₂²
The final mechanical energy
= K
Em_{f} = ½ 4 m v²
The fraction of energy lost is
Em_{f} / Em₀ = ½ 4m v² / ½ m (v₁² +3 v₂²)
Em_{f} / Em₀ = 4 v₂ / (v₁² + 3 v₂²)
Em_{f} / Em₀ = 4 2.4125² / (3.86² + 3 1.93²)
Em_{f} / em₀ = 23.28 / 26.07
Em_{f} / Em₀ = 0.89
The current in the ideal diode with forward biased voltage drop of 65V is 132.6 mA.
To find the answer, we have to know more about the ideal diode.
<h3>
What is an ideal diode?</h3>
- A type of electronic component known as an ideal diode has two terminals, only permits the flow of current in one direction, and has less zero resistance in one direction and infinite resistance in another.
- A semiconductor diode is the kind of diode that is used the most commonly.
- It is a PN junction-containing crystalline semiconductor component that is wired to two electrical terminals.
<h3>How to find the current in ideal diode?</h3>
- Here we have given with the values,
- We have the expression for current in mA of the ideal diode with forward biased voltage drop as,
Thus, we can conclude that, the current in mA of the ideal diode with forward biased voltage drop of 65 V is 132.6.
Learn more about the ideal diode here:
brainly.com/question/14988926
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