Rod A and rod B are cylindrical rods made of the same metal. amd they differ only in size. Rod B has double the length and doubl

Answer:

The amount of heat that enters the gas throughout this two-step process totals 120 cal.

Explanation:

Given that,

Moles present = 3

Heat capacity at volume held constant = 4.9 cal/mol.K

Heat capacity at pressure held constant = 6.9 cal/mol.K

Starting temperature = 300 K

Ending temperature = 320 K

We are tasked with determining the heat absorbed by the gas at constant pressure

Employing the heat formula

Substituting the values into the equation

Next, we calculate the heat absorbed by the gas at constant volume

Using the corresponding heat formula

Insert the values into the formula

Now, it's necessary to evaluate the total heat flow into the gas during both steps

Using the total heat formula

Thus, the heat that transfers into the gas throughout this two-step process amounts to 120 cal.

Response:

v = 4.08 m/s²

Clarification:

Answer:   1m, 1m, 2m, 2m, 4m, 4m.

It’s important to remember that the masses attached do not influence the number of oscillations.

Explanation:

To determine the number of oscillations (complete cycles), we can apply the formula n = t / T ……equation 1

The variables that impact the period of a simple pendulum are solely its length and gravitational acceleration. The period remains unaffected by factors such as mass.

period (T)= 2 x π x √(L/g) ….equation 2

where π = 3.142, L= rope length, and g = 9.8 m/s (gravitational acceleration)

According to the question, the time (t) is 60 seconds.

By merging equations 1 and 2, we obtain  

number of oscillations = time / (2 x π x √(L/g))

Case 1: for L = 4m

number of oscillations = 60 / ( 2 x 3.142 x √(4/9.8))

= 14.9 = 14 complete cycles (the problem specifies complete cycles)

Case 2: for L = 2m

number of oscillations = 60 / ( 2 x 3.142 x √(2/9.8))

= 21.4 = 21 complete cycles

Case 3: where L = 4m, results in the same as case 1, yielding 14 complete cycles

Case 4: where L = 2m, mirrors the outcome in case 2, producing 21 complete cycles

Case 5: in the instance of L = 1m

number of oscillations = 60 / ( 2 x 3.142 x √(1/9.8))

= 30.1 = 30 complete cycles

Case 6: when L = 1m, which repeats case 5, also gives 30 complete cycles

From these findings, the order of the pendulums from the highest to lowest number of complete cycles is as follows: 1m, 2m, 2m, 4m, 4m.

Remember, the number of oscillations is independent of their respective masses.

Answer:

The rotational angular speed is measured at 1.34 rad/s.

Explanation:

Considering the following parameters,

Length = 3.40 m

Distance = 5.90 m

Angle = 45.0°

We are tasked with finding the angular speed of rotation

Using the balance equation

Horizontal component

Vertical component

Substituting the tension value

Substituting the value into the equation

Thus, the angular speed of rotation computes to 1.34 rad/s.

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.