Answer:
The work done, W = 19.6 J
Explanation:
It’s provided that
The mass of the block, m = 5 kg
The velocity of the block, v = 10 m/s
The coefficient of kinetic friction between the block and rough surface is 0.2
Distance traveled by the block, d = 2 m
As the block traverses the rough section, it loses energy equal to the work done by the kinetic energy.
W = 19.6 J
Thus, the change in kinetic energy of the block moving through the rough section is 19.6 J. Consequently, this is the required answer.
Answer:
(1) Utilize the information provided in Table R2 and the error propagation principle to calculate the travel time ratio (with errors) of the other objects compared to the hollow cylinder? ℎ?. Complete Table R5 below. [6] Table R5 Solid cylinder Billiard ball Racquetball?? ℎ? ± ± ± (2) Examine how the solid cylinder's ratio to the hollow cylinder supports or contradicts the theoretical ratio in Eq. (8) stated in the manual. Compute the percentage error and discuss. [4] Answer: (3) Based on the travel time ratio, determine (i) if the billiard ball is solid or hollow, and (ii) if the racquetball is solid or hollow. Provide your reasoning. (Answers may vary if your measurements lack sufficient clarity.) [4]
Subscribe to unlock
Image of page 4
PHYS2125 Physics Laboratory I ©2018 Kuei Sun The University of Texas at Dallas 5 Answer: (4) Identify the object in Table R2 with the highest SEOM. Provide reasoning for the relatively high SEOM and suggest improvements. [3] (5) Discuss TWO potential systematic errors in measurement. [3] Answer: **Please attach your calculation details. Use as many pages as needed; calculations that reflect your understanding may earn partial credit. **Ensure your workspace and equipment are identical to how you left them.
Explanation:
J(r) = Br. We know that the area of a small segment, dA, is represented as 2 π dr. Thus, I = J A and dI = J dA. Plugging in the values gives us dI = B r. 2 π dr which simplifies to dI= 2π Br² dr. Now, integrating the above equation: Given that B= 2.35 x 10⁵ A/m³, with r₁ = 2 mm and r₂ equal to 2 + 0.0115 mm, or 2.0115 mm.
The string does not experience any force of tension, as it balances two forces acting in the same direction. Hence, the tension is zero.
Explanation:
If tension existed in the string, it would mean that two equal but opposite forces are exerting pull in contrary directions.
When a force of f newtons is applied from the right and another force of f newtons from the left, the resulting action occurs through one force. Because there is action on the same string in opposing directions, the tension in the string can only be equal to the magnitude of the string itself.
Therefore, the string indeed has no tension since it is dealing with two forces acting in the same direction. Thus, the tension is zero.
