Digital bits on a 12.0-cm diameter audio CD are encoded along an outward spiraling path that starts at radius R1=2.5cm and finis

Answer:

σ₁ = C/m²

σ₂ =  C/m²

Explanation:

Provided Information:

i) Smaller sphere's radius ( r ) = 5 cm.

ii) Larger sphere's radius ( R ) = 12 cm.

iii) Electric field at the larger sphere's surface  ( E₁ ) = 358 kV/m, which is equivalent to 358 * 1000 v/m

Charge (Q₁) = 572.8 C

Since the electric field inside a conductor is zero, the electric potential ( V ) remains constant.

V = constant

∴

  = C

Surface charge density ( σ₁ ) for the larger sphere.

Calculated Area ( A₁ )  = 4 * π * R²  = 4 * 3.14 * 0.12 = 0.180864 m².

σ₁  = = =  C/m².

Surface charge density ( σ₂ ) for the smaller sphere.

Calculated Area ( A₂ )  = 4 * π * r²  = 4 * 3.14 * 0.05²  = 0.0314 m².

σ₂ = = = C/m²

Let's consider a few possibilities. 1. The lowest velocity of the paratrooper would be just before hitting the ground. 2. Given that the jump originated from a relatively short height, the paratrooper utilized a static line, allowing the parachute to deploy almost instantly after leaping. Hence, we will convert 100 mi/h to ft/s: 100 mi/h * 5280 ft/mi / 3600 s/h = 146.67 ft/sec. Based on the first assumption, the maximum distance fallen by the paratrooper would equate to 8 seconds at 146.67 ft/s, translating to 8 s * 146.67 ft/s = 1173.36 ft. This calculated distance is nearly on par with the jump height, validating both assumptions 1 and 2. Thus, this scenario seems plausible. Moreover, considering the terminal velocity for a parachutist in a freefall position with limbs spread out typically reaches 120 mi/h, which is slightly above the 100 mi/h mentioned in the article. This as well aligns with the notion of the parachute acting like a flag, adding some air resistance.

Answer:

b) TA = TB = TC

Explanation:

• When the blocks are brought into contact and isolated from the environment, they will exchange heat until they achieve thermal equilibrium.
• During this exchange, the hotter body will lose heat, which will be gained by the cooler body.
• The equilibrium state will be established once this equation is satisfied:

       

• Substituting the initial temperatures T₀A = 300º C and T₀B = 400ºC, while simplifying for equal block masses mA = mB, enables us to solve for the final temperature, Tfin:

       

• At equilibrium, when both blocks combine, they will yield a uniform final temperature of 350ºC.
• When block C, also at this temperature, makes contact, all three blocks will simultaneously reflect this final temperature of 350 ºC.
• Therefore, option b) is correct.

Answer:

The distance covered by the minutes hand is 39.60 cm.

Explanation:

Note: A clock has a circular shape, where the minutes hand acts as the radius, and its motion creates an arc.

Length of an arc is calculated as ∅/360(2πr)

L = ∅/360(2πr).................... Equation 1π

Here, L represents the arc’s length, ∅ is the angle made by the arc, and r is the arc’s radius.

Given: ∅ = 252°, r = 9 cm, π = 3.143.

Upon substituting these values into equation 1,

L = 252/360(2×3.143×9)

L = 0.7×2×3.143×9

L = 39.60 cm.

Thus, the distance traversed by the minutes hand is 39.60 cm.

Answer:

Insufficient details provided; please clarify further.