C. Thick wire and cold temperature.
Explanation:
The resistance of a wire is given by: R = (ρL)/A
where ρ is the resistivity of the material, L is the length of the wire, A is the cross-sectional area of the wire.
From the formula, we see that the thicker the wire, the larger A, therefore the smaller the resistivity. so, a thick wire will have lower resistivity.
Moreover, the resistance of a wire increases with the temperature. In fact, high temperatures mean more motion of the atoms/electrons inside the wire, so more resistance to the flow of current through it. Therefore, colder temperature means lower resistance.
So, the correct option is thick wire and cold temperature.
Observe that the object below moves in the negative direction with a changing velocity. An object which moves in the negative direction has a negative velocity. If the object is slowing down then its acceleration vector is directed in the opposite direction as its motion (in this case, a positive acceleration). The dot diagram shows that each consecutive dot is not the same distance apart (i.e., a changing velocity). The position-time graph shows that the slope is changing (meaning a changing velocity) and negative (meaning a negative velocity). The velocity-time graph shows a line with a positive (upward) slope (meaning that there is a positive acceleration); the line is located in the negative region of the graph (corresponding to a negative velocity). The acceleration-time graph shows a horizontal line in the positive region of the graph (meaning a positive acceleration).
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Answer:
Explanation:
If friction is neglected, the wheel cannot roll and can only slide frictionlessly and will have the same velocity at the bottom of the ramp as if it had been in free fall as it has converted the same amount of potential energy.
mgh = ½mv²
v = √(2gh) = √(2(9.81)(2.00)) = 6.26418... = 6.26 m/s
However if we do not ignore all friction and the wheel rolls without slipping down the slope, the potential energy becomes linear and rotational kinetic energy
mgh = ½mv² + ½Iω²
mgh = ½mv² + ½(½mR²)(v/R)²
2gh = v² + ½v²
2gh = 3v²/2
v = √(4gh/3) =√(4(9.81)(2.00)/3) = 5.11468... = 5.11 m/s
Answer:
Cement and sand will do
Explanation:
because without it you can't build
Answer:
Plug in the given values and solve for the final velocity. Remember, when the ball is on the ground it has a height of zero.
Explanation:
