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.
Answer:
Explanation: This is done using the equation:

Because the Radius is a know value. We have the following.

Which is:
4188.7902 mm
A) 
Let's start by writing the equation of the forces along the directions parallel and perpendicular to the incline:
Parallel:
(1)
where
m is the mass
g = 9.8 m/s^2 the acceleration of gravity

is the coefficient of friction
R is the normal reaction
a is the acceleration
Perpendicular:
(2)
From (2) we find

And substituting into (1)

Solving for a,

B) 5.94 m/s
We can solve this part by using the suvat equation

where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the displacement
Here we have
v = ?
u = 0 (it starts from rest)

s = 8.70 m
Solving for v,

Answer:

Explanation:
Newton's 2nd law is given as
.
To find the acceleration in the horizontal direction, you need the horizontal component of the force being applied.
Using trigonometry to find the horizontal component of the force:

Use this horizontal component of the force to solve for for the acceleration of the object:
