The event in the life of a star that begins its expansion into a giant is its core that was hot enough for fusion reaction.
<h3>What is fusion reaction?</h3>
Nuclear fusion is a type of reaction in which two or more atomic nuclei are fuse to form one or more different atomic nuclei with the release or the absorption of energy.
So we can conclude that the event in the life of a star that begins its expansion into a giant is its core that was hot enough for fusion reaction.
Learn more about reaction here: brainly.com/question/26018275
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Answer:
Technician B is correct
Explanation:
Freezing is a method of conversion of substance in its liquid state to solid state. It is the process by which a liquid substance changes to a solid at a particular temperature.
Increasing the pressure and decreasing the temperature of a liquid increases its freezing point. For example, in other to freeze water i.e to change water to ice, it has to be kept in a fridge at a temperature lower than the temperature of the water. The essence of covering the fridge after placing the water in the fridge is to increase the pressure of the liquid hence increasing its freezing rate.
Based on the above explanation, it can be concluded that technician B is correct.
<span>K.E = 0.5 * m * v^2 ( m = mass(Kg), V = Velocity(m/s)
= 0.5 * 8 * 5^2
= 4 * 25
= 100 J </span>
Answer:
27.1 m/s
Explanation:
Given that at a race car driving event, a staff member notices that the skid marks left by the race car are 9.06 m long. The very experienced staff member knows that the deceleration of a car when skidding is -40.52 m/s2.
Using third equation of motion,
V^2 = U^2 + 2aS
Since the car is decelerating, the final velocity V = 0
Substitute all the parameter into the equation above,
0 = U^2 - 2 * 40.52 * 9.06
U^2 = 734.22
U = 
U = 27.096
U = 27.1 m/s approximately
Therefore, the staff member can estimate for the original speed of the race car to be 27.1 m/s if it came to a stop during the skid
we know that center of mass is given as
r = (m₁ 
+ m₂ 
)/(m₁ + m₂)
taking derivative both side relative to "t"
dr/dt = (m₁ d/dt + m₂ d/dt)/(m₁ + m₂)
v = (m₁ 
+ m₂ 
)/(m₁ + m₂)
taking derivative again relative to "t" both side
dv/dt = (m₁ d/dt + m₂ d/dt)/(m₁ + m₂)
a= (m₁ 
+ m₂ 
)/(m₁ + m₂)
