/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 77 Archerfish are tropical fish tha... [FREE SOLUTION] | 91Ó°ÊÓ

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Archerfish are tropical fish that hunt by shooting drops of water from their mouths at insects above the water's surface to knock them into the water, where the fish can eat them. \(\mathrm{A} 65 \mathrm{~g}\) fish at rest just on the water's surface can expel a \(0.30 \mathrm{~g}\) drop of water in a short burst of \(5.0 \mathrm{~ms}\). High-speed measurements show that the water has a speed of \(2.5 \mathrm{~m} / \mathrm{s}\) just after the archerfish expels it. What is the average force the fish exerts on the drop of water? A. \(0.00015 \mathrm{~N}\) B. \(0.00075 \mathrm{~N}\) C. \(0.075 \mathrm{~N}\) D. \(0.15 \mathrm{~N}\)

Short Answer

Expert verified
D. 0.15 N

Step by step solution

01

Understand the problem statement

The archerfish expels a drop of water weighing 0.30 g at a speed of 2.5 m/s in 5.0 ms. We need to find the average force exerted by the fish on the water drop.
02

Convert mass to kilograms

The mass of the water drop is given in grams. Convert it to kilograms: \(0.30 \text{ g} = 0.30 \times 10^{-3} \text{ kg}\).
03

Convert milliseconds to seconds

The time of expulsion is given in milliseconds. Convert it to seconds: \(5.0 \text{ ms} = 5.0 \times 10^{-3} \text{ s}\).
04

Use the impulse-momentum theorem

The impulse-momentum theorem states: \( F_{\text{avg}} \cdot \Delta t = m \cdot \Delta v \). Solve for \( F_{\text{avg}} \):\[ F_{\text{avg}} = \frac{m \cdot v}{\Delta t} \]Substitute the values: mass \(m = 0.30 \times 10^{-3} \text{ kg}\), velocity \(v = 2.5 \text{ m/s}\), and time \(\Delta t= 5.0 \times 10^{-3} \text{ s}\).
05

Calculate the average force

Substitute the known values into the formula:\[F_{\text{avg}} = \frac{0.30 \times 10^{-3} \text{ kg} \cdot 2.5 \text{ m/s} }{5.0 \times 10^{-3} \text{ s}}\]\[ F_{\text{avg}} = \frac{0.00075 \text{ kg} \cdot \text{m/s}}{0.005 \text{ s}} \]\[ F_{\text{avg}} = 0.15 \text{ N} \]
06

Verify the answer with the options provided

Compare the calculated average force, \(0.15 \text{ N}\), with the available options. The correct answer is D. \(0.15 \text{ N}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Archerfish Physics
The archerfish is known for its remarkable ability to hunt by shooting water droplets at insects, knocking them off perches into the water where they can be easily consumed. This fascinating behavior is not just a random action but rooted deeply in physics principles. When the fish expels water, it showcases how momentum and force come into play even in underwater environments.

By expelling water with precision and speed, an archerfish can compensate for the upward trajectory needed to reach its target. The physics of these fish involves understanding how forces exerted can change momentum over a short duration, allowing the fish to be efficient hunters.

Through studying the physics behind the archerfish, we can explore broader concepts of motion and force that apply to a wide variety of situations in the natural world.
Force Calculation
To calculate the force the archerfish exerts on the drop of water, we need to delve into the basic laws of physics dealing with force and motion. Force can be computed using the formula derived from the impulse-momentum theorem:

\[ F_{\text{avg}} = \frac{m \cdot v}{\Delta t} \]

where \( m \) is the mass of the water drop, \( v \) is its velocity after expulsion, and \( \Delta t \) is the time over which the force is applied.

For this exercise:
  • The mass of the drop is \( 0.30 \text{ g} = 0.30 \times 10^{-3} \text{ kg} \).
  • The velocity is \( 2.5 \text{ m/s} \).
  • The time of expulsion is \( 5.0 \text{ ms} = 5.0 \times 10^{-3} \text{ s} \).

Plug these values into the formula to find the average force. This simple calculation allows us to see how even a small fish can exert significant force in a short amount of time. Understanding these calculations forms the basis of how we interpret forces in practical scenarios.
Projectile Motion
Projectile motion involves the flight path of an object that is subject only to the force of gravity and initial thrust. The movements of the water droplet expelled by the archerfish are a great example of this concept in action—where it is given an initial velocity and then moves along a curved path influenced by gravity.

The main aspects of projectile motion discussed here involve:
  • Initial Speed: The speed at which the water droplet is expelled, \(2.5 \text{ m/s}\) as per this scenario.
  • Gravity's Influence: As soon as the droplet leaves the fish's mouth, gravity starts to affect its path.
  • Trajectory Path: The curved path followed by the droplet as it travels toward the target insect.

Grasping projectile motion opens up a broader understanding of how objects move through space and enhances our ability to solve various physics problems involving motion.
Physics Problem Solving
Physics problem-solving skills are crucial for understanding and applying complex concepts. To effectively solve physics problems, it's essential to break them down into clear, manageable steps. Our exercise illustrates a successful approach to physics problem solving by following these general steps:

  • Understand the Problem: Determine what is being asked and identify the known values and unknowns.
  • Convert Units: Ensure all units are consistent, e.g., converting grams to kilograms and milliseconds to seconds.
  • Apply Physics Principles: Use relevant physics laws, such as the impulse-momentum theorem, to set up equations.
  • Calculate: Perform the necessary calculations carefully to arrive at the correct answer.
  • Verify: Check the solution's feasibility given practical constraints and available answers.

Employing these strategies enables students to tackle not just theoretical problems but also real-world situations where physics provides insights and solutions.

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Most popular questions from this chapter

A \(15.0 \mathrm{~kg}\) block is attached to a very light horizontal spring of force constant \(500.0 \mathrm{~N} / \mathrm{m}\) and is resting on a frictionless horizontal table. (See Figure \(8.40 .\) ) Suddenly it is struck by a \(3.00 \mathrm{~kg}\) stone traveling horizontally at \(8.00 \mathrm{~m} / \mathrm{s}\) to the right, whereupon the stone rebounds at \(2.00 \mathrm{~m} / \mathrm{s}\) horizontally to the left. Find the maximum distance that the block will compress the spring after the collision. (Hint: Break this problem into two parts the collision and the behavior after the collision -and apply the appropriate conservation law to each part.)

Accident analysis. A \(1500 \mathrm{~kg}\) sedan goes through a wide intersection traveling from north to south when it is hit by a \(2200 \mathrm{~kg}\) SUV traveling from east to west. The two cars become enmeshed due to the impact and slide as one thereafter. On-the-scene measurements show that the coefficient of kinetic friction between the tires of these cars and the pavement is \(0.75,\) and the cars slide to a halt at a point \(5.39 \mathrm{~m}\) west and \(6.43 \mathrm{~m}\) south of the impact point. How fast was each car traveling just before the collision?

A \(750 \mathrm{~kg}\) car is stalled on an icy road during a snowstorm. A \(1000 \mathrm{~kg}\) car traveling eastbound at \(10 \mathrm{~m} / \mathrm{s}\) collides with the rear of the stalled car. After being hit, the \(750 \mathrm{~kg}\) car slides on the ice at \(4 \mathrm{~m} / \mathrm{s}\) in a direction \(30^{\circ}\) north of east. (a) What are the magnitude and direction of the velocity of the \(1000 \mathrm{~kg}\) car after the collision? (b) Calculate the ratio of the kinetic energy of the two cars just after the collision to that just before the collision. (You may ignore the effects of friction during the collision.)

In a volcanic eruption, a \(2400-\mathrm{kg}\) boulder is thrown vertically upward into the air. At its highest point, it suddenly explodes (due to trapped gases) into two fragments, one being three times the mass of the other. The lighter fragment starts out with only horizontal velocity and lands \(274 \mathrm{~m}\) directly north of the point of the explosion. Where will the other fragment land? Ignore any air resistance.

A stone with a mass of \(0.100 \mathrm{~kg}\) rests on a frictionless, horizontal surface. A bullet of mass \(2.50 \mathrm{~g}\) traveling horizontally at \(500 \mathrm{~m} / \mathrm{s}\) strikes the stone and rebounds horizontally at right angles to its original direction with a speed of \(300 \mathrm{~m} / \mathrm{s}\). (a) Compute the magnitude and direction of the velocity of the stone after it is struck. (b) Is the collision perfectly elastic?

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