Luke Autbeloe drops an approximately 5.0 kg object (weight = 50.0 N) off the roof of his house into the swimming pool below. Upon encountering the pool, the object encounters a 50.0 N upward resistance force (assumed to be constant). Use this description to answer the following questions. (Down is usually considered a negative direction) a. Which one of the velocity-time graphs best describes the motion of the object? Why?

Answers

Answer 1

Answer:

The motion of the object is best depicted by graph B. Therefore, option B is correct.

This is because the object is initially accelerating downwards due to the force of gravity. However, when it encounters the pool, the upward resistance force counteracts the force of gravity, and the object's velocity slows down. The object will eventually reach a terminal velocity, where the upward resistance force is equal to the force of gravity, and the object will no longer accelerate.

Graph A shows a constant velocity, which is not possible in this case. Graph C shows an initial acceleration downwards, followed by an acceleration upwards. This is also not possible, as the object cannot accelerate upwards while it is still in contact with the pool. Graph D shows a constant acceleration downwards.

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Answer 2

Answer:

Graph B represents correct graph for velocity time Explanation:Initially the object is dropped under the influence of constant force due to gravitySo here we can say its acceleration is constant due to gravityhence here the velocity will increase downwards due to gravity and it is given asnow when it enters into the pool an upward force of same magnitude will act on itso net force on it is so here we will have no acceleration and it will move with constant speed after that so here correct graph is Graph B

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Heat in a substance is an example of _____ energy?

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Heat in a substance is still an example of heator thermal energy. Heat or thermal energy is further defined as the amount ofavailable molecules of kinetic energy transferred in a specific amount of time.Heat is nothing if it is not transferred from one object to another. There aremany uses of heat energy. When we are cooking using our stoves, there is heatenergy. When we pour hot water onto our cup of coffee, there is heat energy.

Two cars collide at an intersection. Car A, with a mass of 1900 kg, is going from west to east, while car B, of mass 1500 kg, is going from north to south at 17.0 m\s. As a result of this collision, the two cars become enmeshed and move as one afterwards. In your role as an expert witness, you inspect the scene and determine that, after the collision, the enmeshed cars moved at an angle of 60.0degrees south of east from the point of impact.Part A WAS: How fast were the enmeshed cars moving just after the collision? I got 8.66 for velocity in part a which was CORRECT but i can't figure out PART B??...Part B:How fast was car A going just before the collision

Answers

Part A: The enmeshed cars were moving at a velocity of approximately 8.66 m/s just after the collision.

Part B: Car A was traveling at a velocity of approximately 8.55 m/s just before the collision.

How to compute the above velocities

To find the speed of car A just before the collision in Part B, you can use the principle of conservation of momentum.

The total momentum of the system before the collision should equal the total momentum after the collision. You already know the total momentum after the collision from Part A, and now you want to find the velocity of car A just before the collision.

Let's denote:

- v_A as the initial velocity of car A before the collision.

- v_B as the initial velocity of car B before the collision.

In Part A, you found that the enmeshed cars were moving at a velocity of 8.66 m/s at an angle of 60 degrees south of east. You can split this velocity into its eastward and southward components. The eastward component of this velocity is:

v_east = 8.66 m/s * cos(60 degrees)

Now, you can use the conservation of momentum to set up an equation:

Total initial momentum = Total final momentum

(mass_A * v_A) + (mass_B * v_B) = (mass_A + mass_B) * 8.66 m/s (the final velocity you found in Part A)

Plug in the known values:

(1900 kg * v_A) + (1500 kg * v_B) = (1900 kg + 1500 kg) * 8.66 m/s

Now, you can solve for v_A:

(1900 kg * v_A) + (1500 kg * v_B) = 3400 kg * 8.66 m/s

1900 kg * v_A = 3400 kg * 8.66 m/s - 1500 kg * v_B

v_A = (3400 kg * 8.66 m/s - 1500 kg * v_B) / 1900 kg

Now, plug in the values from Part A to find v_A:

v_A = (3400 kg * 8.66 m/s - 1500 kg * 8.66 m/s) / 1900 kg

v_A = (29244 kg*m/s - 12990 kg*m/s) / 1900 kg

v_A = 16254 kg*m/s / 1900 kg

v_A ≈ 8.55 m/s

So, car A was going at approximately 8.55 m/s just before the collision in Part B.

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Answers

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Answers

An object is thrown horizontally from the open
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s = 9.8*2 = 19.6 m

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