Three moles of an ideal gas with a molar heat capacity at constant volume of 4.9 cal/(mol∙K) and a molar heat capacity at consta

The question lacks complete details or specifications. Here is the missing information: 1. impossible to establish 2. half of Isaac's 3. identical to Isaac's 4. double Isaac's The angular velocity of Feng will match that of Isaac. Thus, the correct choice is option 3.

Answer:

F = 0.535 N

Explanation:

We will apply energy concepts, considering both the peak and the bottom of the path.

Top

   Em₀ = U = mg y

Bottom

    = K = ½ m v²

    Emo =

    mg y = ½ m v²

    v = √ (2gy)

   y = L - L cos θ

  v = √ (2g L (1 - cos θ))

Next, we will employ Newton's second law at the lowest point where the acceleration is centripetal.

     F = ma

     a = v² / r

For the turning radius, the cable length is r = L.

    F = m 2g (1 - cos θ)

Now, let's find the result.

    F = 2  1.25  9.8 (1 - cos 12)

    F = 0.535 N

   

Answer:

Please include the diagrams and repost them.

Answer:

a) The jogger's acceleration is 1.5 m/s²

b) The car's acceleration is also 1.5 m/s²

c) Yes, the car covers a distance 76 m greater than the jogger.

Explanation:

a) Acceleration is the change in velocity over a given time interval:

a = (final velocity - initial velocity) / time

For the jogger:

a = (3.0 m/s - 0 m/s) / 2.0 s = 1.5 m/s²

b) For the car:

a = (41.0 m/s - 38.0 m/s) / 2.0 s = 1.5 m/s²

c) To find how far the car has traveled after 2 seconds, use the formula for position under acceleration along a straight path:

x = x₀ + v₀ t + ½ a t²

where

x = position at time t

x₀ = initial position

v₀ = initial velocity

t = elapsed time

a = acceleration

Assuming x₀ = 0 (origin at car's starting point):

x = 38.0 m/s × 2 s + ½ × 1.5 m/s² × (2.0 s)²

x = 79 m

Similarly, position of the jogger after 2 seconds is:

x = 0 m/s × 2 s + ½ × 1.5 m/s² × (2.0 s)² = 3 m

The difference traveled by the car compared to the jogger is 79 m - 3 m = 76 m

5.3 km Explanation: This scenario requires assessing continuous displacement in multiple directions. When complexity arises, vector analysis proves useful. Each individual displacement can be represented as a vector, with designated unit vectors established for the East and North directions. Simple calculations help derive these unit vectors for the North, South of West, and North of West directions. In conclusion, the total displacement vector equals the sum of individual displacement vectors, leading to the overall displacement.