A spring is stretched 6 in by a mass that weighs 8 lb. The mass is attached to a dashpot mechanism that has a damping constant o

Response:

Once it has crossed, the locomotive requires 17.6 seconds to achieve a speed of 32 m/s.

Details:

  The locomotive's acceleration is 1.6

  The duration taken to pass the crossing is 2.4 seconds.

  We can apply the motion equation, v = u + at, where v represents final velocity, u indicates initial velocity, a denotes acceleration, and t signifies time.

  When the speed reaches 32 m/s, we have v = 32 m/s, u = 0 m/s, and a= 1.6 .

   32 = 0 + 1.6 * t

    t = 20 seconds.

  Therefore, the locomotive attains a speed of 32 m/s after 20 seconds, and it passes the crossing in 2.4 seconds.

Thus, after clearing the crossing, it takes an additional 17.6 seconds to reach the speed of 32 m/s.

Answer

Given data:

height of the dam = 15 m

effective area for water flow = 2.3 x 10⁻³ m²

Applying the principle of energy conservation:

v = 17.15 m/s

water discharge

Q = A V

Q = 2.3 x 10⁻³ x 17.15

Q = 0.039 m³/s

Hopefully this clarifies things. The position remains unchanged over the span of 5 seconds, which indicates that it is at rest while time continues to pass. When the slope turns negative, it illustrates that the bear is moving backward, thus changing its direction.

Answer: Her velocity magnitude (v) relative to the shore is 5.70 km/h.

Explanation:

Let Q be the speed of the boat, and P be the speed of the river flow.

R represents the resultant velocity combining boat velocity and river current.

According to vector addition using the law of triangles:

From the diagram:

P = 3.5 km/h, Q = 4.5 km/h

Therefore, her velocity magnitude relative to the shore is 5.70 km/h.

For this issue, the answer is clarified as the system takes in energy (+). The surroundings contribute 84 KJ of work. Whenever a system is receiving work from its surroundings, the value will be positive. Therefore, it sums to 12.4 KJ + 4.2 = 16.6 KJ.