24.03.2021

Squid use jet propulsion for rapid escapes. A squid pulls water into its body and then rapidly ejects the water backward to propel itself forward. A 1.5 kg squid (not including water mass) can accelerate at \(20 \mathrm{~m} / \mathrm{s}^{2}\) by ejecting 0.15 kg of water. What is the magnitude of the thrust force on the squid? Express your answer with the appropriate units. \(F=\) \(\square\) Units Submit Previous Answers Request Answer

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To solve the problem of finding the magnitude of the thrust force on the squid, we need to use the concepts of Newton's Second Law of Motion and the information provided about the squid's mass and acceleration. ### Step-by-Step Solution 1. **Identify the Given Information**: - Mass of the squid \( m = 1.5 \, \text{kg} \) - Acceleration of the squid \( a = 20 \, \text{m/s}^2 \) - Mass of the water being ejected \( m_{\text{water}} = 0.15 \, \text{kg} \) 2. **Understand Thrust Force**: - The thrust force (\( F \)) is what the squid generates to propel itself forward by ejecting water backward. - According to Newton's Second Law, the net force acting on an object is equal to the mass of the object times its acceleration: \[ F_{\text{net}} = m \cdot a \] 3. **Calculate the Force on the Squid**: - We can substitute the mass and acceleration of the squid into this formula: \[ F_{\text{net}} = 1.5 \, \text{kg} \times 20 \, \text{m/s}^2 \] 4. **Perform the Calculation**: - Multiply the two values: \[ F_{\text{net}} = 1.5 \times 20 = 30 \, \text{N} \] - Here, \( N \) stands for Newtons, the unit of force. 5. **Consider the Ejection of Water**: - The thrust force also needs to account for the change in momentum due to the water being ejected. However, for simplicity in this scenario, the problem seems to focus directly on the thrust force generated by the squid to achieve the given acceleration. 6. **Final Answer**: - Therefore, the magnitude of the thrust force on the squid is: \[ F = 30 \, \text{N} \] ### Conclusion The thrust force that allows the squid to propel itself forward by ejecting water is **30 Newtons**. ### Final Answer Format: \[ F = 30 \, \text{N} \]
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To determine the magnitude of the force on the water being ejected by the squid, we can apply Newton’s second law of motion. This law states: \[ F = m \cdot a \] where: - \( F \) is the force, - \( m \) is the mass, - \( a \) is the acceleration. In this scenario, the mass \( m \) refers to the mass of the water being ejected, and the acceleration \( a \) is the rate at which the squid accelerates the water backward. ### Given Data: - Mass of water ejected, \( m = 0.15 \, \text{kg} \) - Acceleration of the squid (hence the acceleration of the water), \( a = 20 \, \text{m/s}^2 \) ### Step-by-Step Calculation: 1. **Calculate the Force**: We will use the values provided in the formula \( F = m \cdot a \). \[ F = 0.15 \, \text{kg} \cdot 20 \, \text{m/s}^2 \] 2. **Perform the Multiplication**: \[ F = 0.15 \cdot 20 \] \[ F = 3.0 \, \text{N} \] ### Conclusion: The magnitude of the force on the water being ejected is \[ F = 3.0 \, \text{N} \] ### Final Answer: \[ F = 3.0 \] Units: Newtons (N) So, when you write your answer, it should look like this: \[ F = \text{3.0} \quad \text{N} \] This analysis shows that the squid uses a force of 3.0 Newtons to eject water, helping it to rapidly escape from danger by moving forward quickly!
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a. FTh = 30 N

b. Fw = 30 N

c. a = 200 m/s2

Explanation:

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Squid use jet propulsion for rapid escapes. A squid pulls water into its body and then rapidly eject
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The idea of profit and loss in grammar is known as a noun phrase. It functions as the subject or the object of a sentence, depending on the context.

In the sentence "Profit and loss are important concepts in business", the noun phrase "profit and loss" functions as the subject. It identifies what is important.

In the sentence "The company experienced significant profit and loss last year", the noun phrase "profit and loss" functions as the object. It receives the action of the verb "experienced" and specifies what the company experienced.

Noun phrases like "profit and loss" can be made up of multiple nouns or noun equivalents and can function in various ways within a sentence.
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The type of language used in the question "Are there any reasons why this idea would not work?" is speculative.
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To solve this problem, first we need to understand that the volume of punch stays the same (2 gallons) before and after Gracie makes her adjustments. Therefore, we can create an equation using the information we have: the original amount of juice, how much juice Gracie adds, and the final amount of juice in the punch.

Let X = the amount of punch that Gracie pours out (and consequently, the volume of 100% juice she adds).

The initial punch is comprised of 50% (or 0.5) juice, therefore, the volume of the juice in the initial punch would be 0.5 * 2 = 1 gallon.

If Gracie pours out X gallons of punch, she also takes out 0.5 * X gallons of juice (because the original punch is 50% juice).

The final punch which is 2 gallons is comprised of 65% (or 0.65) juice, therefore, the volume of the juice in the final punch would be 0.65 * 2 = 1.3 gallons.

According to the problem, the final volume of juice (1.3 gallons) is equal to the initial volume of juice (1 gallon) minus the juice Gracie removed (0.5 * X) plus what she replaced it with (X gallons of 100% juice).

Therefore, we can write the equation:

1 - 0.5*X + X = 1.3

Solving this equation, we collect like terms and get:

0.5*X = 0.3
Which simplifies to:
X = 0.3/0.5 = 0.6 gallons

So, Gracie poured out 0.6 gallons of her original punch mixture and replaced it with pure 100% juice to achieve the 2 gallons of punch that contains 65% juice.


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To solve this problem, you need to know that the external angle of a triangle is equal to the sum of the two non-adjacent angles.

Here, ∠MLK is the external angle of ∆ZML.

So m∠MLK = m∠ZLK + m∠MLZ, which translates into:

130 = (x + 86) + (x + 66)

Combine the constants on the right hand side:

130 = 2x + 152

To isolate 2x, subtract 152 from both sides:

130 - 152 = 2x

-22 = 2x

Now divide each side by 2 to solve for x:

x = -22 / 2

x = -11

Now plug x = -11 into m∠ZLK = x + 86 to find m∠ZLK:

m∠ZLK = -11 + 86

m∠ZLK = 75

Therefore, m∠ZLK = 75°.
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To determine the pressure exerted by the trapped air in the mercury barometer, we can use the concept of equilibrium.

First, let's understand the setup. The mercury barometer consists of a long, vertical tube filled with mercury, inverted in a dish containing mercury. The top of the tube is sealed, creating a vacuum above the mercury column. The height of the mercury column indicates the atmospheric pressure.

When there is trapped air inside the tube, it exerts a pressure on the mercury column, balancing the atmospheric pressure.

To solve the problem, we need to compare the height of the mercury column in the barometer to that of the aneroid barometer, which reads 75 cm Hg. This means that the pressure exerted by the trapped air is equivalent to the pressure of a mercury column measuring 75 cm.

To calculate the pressure exerted by the trapped air, we use the equation for hydrostatic pressure:

Pressure = Density × Gravity × Height

In this case, the density will be that of mercury (which we'll denote as ρ) and the height of the mercury column will be 75 cm.

The equation becomes:

Pressure = ρ × g × Height

Now, let's substitute the values for density and gravity:

Pressure = (13.6 g/cm³) × (9.8 m/s²) × (75 cm)

Since the density is given in grams per cubic centimeter (g/cm³) and gravity is given in meters per second squared (m/s²), we need to convert the density to kilograms per cubic meter (kg/m³) to maintain consistency in the units. 1 g/cm³ is equal to 1000 kg/m³.

Substituting the values and performing the necessary conversions, we get:

Pressure = (13.6 × 1000) kg/m³ × 9.8 m/s² × (0.75 m)

Pressure = 127,680 kg/(m·s²) × 0.75 m

Pressure = 95,760 Pa

Therefore, the pressure exerted by the trapped air in the mercury barometer is approximately 95,760 Pascal (Pa).

Note: Pascal (Pa) is the unit of pressure in the International System of Units (SI).
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B. to help keep track of the developments in the meeting and any future action items to encourage all the participants contribute during the meeting.

Taking minutes during a meeting is important for several reasons:

1. Documentation: Minutes serve as a written record of what was discussed, decisions made, and actions planned during a meeting. They provide a historical reference that can be used for future reference, clarification, or legal purposes. Having an accurate record helps avoid any misunderstandings or disagreements about what was agreed upon.

2. Accountability: By assigning someone to take minutes, it ensures that there is a designated individual responsible for capturing the key points and outcomes of the meeting. This brings accountability and helps ensure that nothing important is missed or overlooked.

3. Tracking progress: Minutes help keep track of the developments and progress made during the meeting. They outline action items, responsibilities, and deadlines, making it easier to follow up on tasks and evaluate the progress at subsequent meetings. Without minutes, it can be challenging to remember all the details and actions agreed upon, leading to potential inefficiencies and missed opportunities.

4. Participation: Knowing that minutes will be recorded, participants are more likely to actively contribute during the meeting. They recognize that their ideas, suggestions, and feedback will be documented and taken into consideration. This promotes active engagement and collaboration among participants, leading to more productive meetings.

While option A and D may be part of the leader's responsibilities, they are not the primary reasons for taking minutes. Minutes primarily serve to keep a written record and facilitate the tracking of developments and action items during the meeting.

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