On a hypothetical scale X The ice point is 40° and steam point is 120°.

a) 0.13*τ; b) 2.08*τ. To describe the discharging process of a capacitor through a resistor, consider the following: Q(t) = Qo * exp(-t/τ) to signify a loss of 1/8 of its charge. In this scenario, Q(t) = 7/8 * Qo = 7/8 * exp(-t/τ). By rearranging, we have ln(7/8)*τ = -t, thus t = -ln(7/8)*τ = 0.13. For a loss of 7/8 of its charge, we use Q(t) = 1/7 * Qo * exp(-t/τ), leading to t = -ln(1/8)*τ = 2.08.

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]

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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:

1) The projectile's motion follows
,
In order to determine the velocity, we must compute the derivative of h(t): Next, we will compute the speed at t=2 s and t=4 s: The negative value of the second speed suggests that the projectile has already attained its highest point and is now descending. 2) The maximum height of the projectile occurs when its speed equals zero: Thus, we have And solving yields


3) To determine the maximum height, we substitute the time at which the projectile reaches this peak into h(t), specifically t=2.30 s: 4) The time at which the projectile lands is when the height reaches zero; h(t)=0, which leads to This results in a second-degree equation, producing two answers: the negative root can be disregarded as it lacks physical significance; the second root is

, which indicates the landing time of the projectile. 5) The moment the projectile impacts the ground corresponds to the velocity at time t=4.68 s:

, carrying a negative sign to denote a downward direction.

The sheet's thickness is calculated to be t= 0.0003 mm. The mass provided is m = 230 mg, equivalent to 0.23 g. The area of the sheet is A= 23 x 17 cm², totaling A= 391 cm². Given the density of gold is ρ = 19.32 g/cm³, we assume the sheet's thickness is t cm. From the equation Mass = Density x Volume, we know m = ρ A t. Substituting the values results in: 0.23 = 19.32 × 391 × t, leading to t = 0.000030 cm, or equivalently, 0.0003 mm as the final thickness.

Applying the formula of force equals mass times acceleration, and considering acceleration approximated as g at 10 m/s², the brick possessing the greater mass will hit the ground first due to its superior force that counteracts resistive forces like air resistance.