Answer:
C. a rolling bowling ball
I just answered this question on my quiz.
Answer:

Explanation:
The equation for work is:

We can substitute the given values into the equation:

Answer:
Shield volcanoes, the third type of volcano, are built almost entirely of fluid lava flows. Flow after flow pours out in all directions from a central summit vent, or group of vents, building a broad, gently sloping cone of flat, domical shape, with a profile much like that of a warrior's shield.
Explanation:
Answer:
52 rad
Explanation:
Using
Ф = ω't +1/2αt²................... Equation 1
Where Ф = angular displacement of the object, t = time, ω' = initial angular velocity, α = angular acceleration.
Since the object states from rest, ω' = 0 rad/s.
Therefore,
Ф = 1/2αt²................ Equation 2
make α the subject of the equation
α = 2Ф/t².................. Equation 3
Given: Ф = 13 rad, t = 2.5 s
Substitute into equation 3
α = 2(13)/2.5²
α = 26/2.5
α = 4.16 rad/s².
using equation 2,
Ф = 1/2αt²
Given: t = 5 s, α = 4.16 rad/s²
Substitute into equation 2
Ф = 1/2(4.16)(5²)
Ф = 52 rad.
If both bars are made of a good conductor, then their specific heat capacities must be different. If both are metals, specific heat capacities of different metals can vary by quite a bit, eg, both are in kJ/kgK, Potassium is 0.13, and Lithium is very high at 3.57 - both of these are quite good conductors.
If one of the bars is a good conductor and the other is a good insulator, then, after the surface application of heat, the temperatures at the surfaces are almost bound to be different. This is because the heat will be rapidly conducted into the body of the conducting bar, soon achieving a constant temperature throughout the bar. Whereas, with the insulator, the heat will tend to stay where it's put, heating the bar considerably over that area. As the heat slowly conducts into the bar, it will also start to cool from its surface, because it's so hot, and even if it has the same heat capacity as the other bar, which might be possible, it will eventually reach a lower, steady temperature throughout.