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klemol [59]
2 years ago
15

It’s a sunny Saturday afternoon and you are walking around the lake by your house, enjoying the last few days of summer. The sid

ewalk surrounding the perimeter of the circular lake is crowded with walkers and runners. You then notice a runner approaching you wearing a T-shirt with writing on it. You read the first two lines, but are unable to read the third line before he passes. You wonder, ”Hmmm, if he continues around the lake, I bet I’ll see him again but I should anticipate the time when we’ll pass again.”
You look at your watch and it is 5:07pm. You estimate your walking speed at 3 m/s and the runner’s speed to be about 14 m/s. You also estimate that the diameter of the lake is about 2
miles. At what time should you expect to read the last line of the t-shirt?
Physics
1 answer:
Citrus2011 [14]2 years ago
7 0

The anticipated time when he will appear again is 5:17 pm

The given parameters;

your speed, V_a = 3 m/s

the runner's speed, V_b = 14 m/s

the diameter of the lake, d = 2 miles = 3218.69 meters

  • Let the anticipated time when he will appear again = t

The circumference of the lake is calculated as;

C = \pi d\\\\C = 3.142 \times 3218.69 = 10,113.12 \ m

Apply concept of relative velocity to determine the time, in which he will appear again.

By the time he appears again;

the distance you moved + distance he moved = circumference of the circle

V_at + V_bt = 10, 113.12\\\\(V_a + V_b)t = 10,113.12\\\\(3 + 14) t = 10,113.12\\\\17t = 10,113.12\\\\t = \frac{10,113.12}{17} \\\\t = 594.89 \ s = 9.92 \ \min \ \approx 10 \ \min

t\  \approx \ \ 5:17 \ pm

Thus, the anticipated time when he will appear again is 5:17 pm

Learn more here: brainly.com/question/20662992

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Answer:

Velocity of the two balls after collision: 60\; \rm m \cdot s^{-1}.

100\; \rm J of kinetic energy would be lost.

Explanation:

<h3>Velocity</h3>

Because the question asked about energy, convert all units to standard units to keep the calculation simple:

  • Mass of the first ball: 100\; \rm g = 0.1\; \rm kg.
  • Mass of the second ball: 400\; \rm g = 0.4 \; \rm kg.

The two balls stick to each other after the collision. In other words, this collision is a perfectly inelastic collision. Kinetic energy will not be conserved. The velocity of the two balls after the collision can only be found using the conservation of momentum.

Assume that the system of the two balls is isolated. Thus, the sum of the momentum of the two balls will stay the same before and after the collision.

The momentum of an object of mass m and velocity v is: p = m \cdot v.

Momentum of the two balls before collision:

  • First ball: p = m \cdot v = 0.1\; \rm kg \times 100\; \rm m \cdot s^{-1} = 10\; \rm kg \cdot m \cdot s^{-1}.
  • Second ball: p = m \cdot v = 0.4\; \rm kg \times 50\; \rm m \cdot s^{-1} = 20\; \rm kg \cdot m \cdot s^{-1}.
  • Sum: 10\; \rm kg \cdot m \cdot s^{-1} + 20 \; \rm kg \cdot m \cdot s^{-1} = 30 \; \rm kg \cdot m \cdot s^{-1} given that the two balls are moving in the same direction.

Based on the assumptions, the sum of the momentum of the two balls after collision should also be 30\; \rm kg \cdot m \cdot s^{-1}. The mass of the two balls, combined, is 0.1\; \rm kg + 0.4\; \rm kg = 0.5\; \rm kg. Let the velocity of the two balls after the collision v\; \rm m \cdot s^{-1}. (There's only one velocity because the collision had sticked the two balls to each other.)

  • Momentum after the collision from p = m \cdot v: (0.5\, v)\; \rm kg \cdot m \cdot s^{-1.
  • Momentum after the collision from the conservation of momentum: 30\; \rm kg \cdot m \cdot s^{-1}.

These two values are supposed to describe the same quantity: the sum of the momentum of the two balls after the collision. They should be equal to each other. That gives the equation about v:

0.5\, v = 30.

v = 60.

In other words, the velocity of the two balls right after the collision should be 60\; \rm m \cdot s^{-1}.

<h3>Kinetic Energy</h3>

The kinetic energy of an object of mass m and velocity v is \displaystyle \frac{1}{2}\, m \cdot v^{2}.

Kinetic energy before the collision:

  • First ball: \displaystyle \frac{1}{2} \, m \cdot v^2 = \frac{1}{2}\times 0.1\; \rm kg \times \left(100\; \rm m \cdot s^{-1}\right)^{2} = 500\; \rm J.
  • Second ball: \displaystyle \frac{1}{2} \, m \cdot v^2 = \frac{1}{2}\times 0.4\; \rm kg \times \left(50\; \rm m \cdot s^{-1}\right)^{2} = 500\; \rm J.
  • Sum: 500\; \rm J + 500\; \rm J = 1000\; \rm J.

The two balls stick to each other after the collision. Therefore, consider them as a single object when calculating the sum of their kinetic energies.

  • Mass of the two balls, combined: 0.5\; \rm kg.
  • Velocity of the two balls right after the collision: 60\; \rm m\cdot s^{-1}.

Sum of the kinetic energies of the two balls right after the collision:

\displaystyle \frac{1}{2} \, m \cdot v^{2} = \frac{1}{2}\times 0.5\; \rm kg \times \left(60\; \rm m \cdot s^{-1}\right)^2 = 900\; \rm J.

Therefore, 1000\; \rm J - 900\; \rm J = 100\; \rm J of kinetic energy would be lost during this collision.

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3 years ago
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