Answer:
The resistance is 
Explanation:
Given that,
Diameter of tube = 8.5 mm
Length = 8 cm
Resistivity = 2.5 m
We need to calculate the resistance
The resistance is equal to the product of the resistivity and length divided by the area of cross section .
In mathematical form,

Where,
=resistivity
l = length
A = area of cross section
Put the value into the formula



Hence, The resistance is 
Yes!
I think there are two ways you could go with this answer:
1) Acceleration is the change in velocity over time, it can be negative or positive. If you have an object that is already moving forwards in a straight line and give it a constant negative acceleration, it will slow down and then start going in reverse.
2)Velocity is a vector, meaning it has both magnitude and direction. In the example above, the acceleration is due to a change in magnitude, or speed (from +ve to -ve) but not a change in direction. Something that has constant speed but is changing direction is also accelerating (like something that is orbiting). You could use the earth as an example, which is constantly accelerating due to moving in a circle around the sun. At any time in the year you can say that in half a year's time the earth's direction will be reversed.
The only answer that can justify being a hypothesis is C.
Explanation:
Average speed = distance / time
|v| = (7 km + 2 km) / (2 hr + 1 hr)
|v| = 3 km/hr
Average velocity = displacement / time
v = (7 km east + 2 km east) / (2 hr + 1 hr)
v = 3 km/hr east
This situation describes the Hooke's Law which states that "When an elastic object - such as a spring - is stretched, the increased length is called its extension. The extension of an elastic object is directly proportional to the force applied to it". The formula is <span>F = k × e , F for the force, k for spring constant expressed in N/m, e for extension in m. This equation works for as long the spring is not stretch too much because once it exceeded its limit, the spring will not return to its original length the moment the load is removed.</span>