Answer:
t = 4 s
Explanation:
As we know that the particle A starts from Rest with constant acceleration
So the distance moved by the particle in given time "t"



Now we know that B moves with constant speed so in the same time B will move to another distance

now we know that B is already 349 cm down the track
so if A and B will meet after time "t"
then in that case


on solving above kinematics equation we have

Answer:
No, it is not proper to use an infinitely long cylinder model when finding the temperatures near the bottom or top surfaces of a cylinder.
Explanation:
A cylinder is said to be infinitely long when is of a sufficient length. Also, when the diameter of the cylinder is relatively small compared to the length, it is called infinitely long cylinder.
Cylindrical rods can also be treated as infinitely long when dealing with heat transfers at locations far from the top or bottom surfaces. However, it not proper to treat the cylinder as being infinitely long when:
* When the diameter and length are comparable (i.e have the same measurement)
When finding the temperatures near the bottom or top of a cylinder, it is NOT PROPER TO USE AN INFINITELY LONG CYLINDER because heat transfer at those locations can be two-dimensional.
Therefore, the answer to the question is NO, since it is not proper to use an infinitely long cylinder when finding temperatures near the bottom or top of a cylinder.
Answer
Given,
Time period of star,T = 3.37 x 10⁷ s
Radius of circular orbit,R = 1.04 x 10¹¹ m
a) Angular speed of the planet

b) tangential speed

v = 1.94 x 10⁴ m/s
c) centripetal acceleration magnitude

a = 3.62 x 10⁻³ m/s²
Resistance ∞ (proportional) length
resistance ∞ 1/ area
therefore,
(the constant that we take is known as the resistivity)
resistance = (resistivity*length )/ area
resistivity = (resistance * area ) / length
= (3 * 45) / 3 = 135/3 = 45 Ωm
in short your answer is 45 Ωm