Answer:
v= 13 m/s
Explanation:
Velocity is defined as the derivative of displacement with respect to time
v= ds/dt
Known data
s(t) = 5t + 2t² : distance that the ball has rolled after t seconds
vi= 5 m/s : initial velocity
t= 2 s
Problem develoment
s(t) = 5t + 2t²
v= ds/dt= 5 + 4t : velocity of the ball in function of the time
We replace t =2 s in the equation of velocity
v= 5 + 4(2)
v= 13 m/s : velocity after 2 seconds
In stars more massive than the sun, the core temperature is hotter, which allows for fusion of more complex elements.
Most of the fusion occurs in the core.
In stars more massive than the sun, fusion continues through Deuterium, Carbon, and finally reaching iron/nickel.
Up to this point, the fusion reaction was endothermic, which means that the energy expended to produce the fusion reaction was exceeded by the energy produced in the reaction.
Fusion past iron is exothermic, and therefore the star will be able to survive by fusing elements heavier than iron.
After the core is almost entirely iron, the star is no longer in the Main Sequence.
So, fusion in stars more massive than the sun continue fusing until the core is almost entirely <em>iron</em>.
Answer: The velocity at different marked time points are given as
t1 = -
t2 = +
t3 = +
t4 = -
t5 = 0
Explanation:
The slope of the tangent of the curve indicates the instantaneous velocity. So if the slope of the tangent is positive, that Is, the tangent makes a positive angle (above the horizontal axis) with the horizontal
axis, then the velocity at this point is positive, and if the slope of the tangent is negative, that is the tangent makes a negative angle with the horizontal axis (below the horizontal axis), then the velocity at this point is negative.
When the tangent of the line is parallel to the horizontal axis, the velocity is 0.
From the position-time graph attached, the sign on the instantaneous velocity for each time marked on the graph is given below
t1 = -
t2 = +
t3 = +
t4 = -
t5 = 0
QED!
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