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(II) According to a simplified model of a mammalian heart, at each pulse approximately 20 \(g\) of blood is accelerated from 0.25 m/s to 0.35 m/s during a period of 0.10 s. What is the magnitude of the force exerted by the heart muscle?

Short Answer

Expert verified
The force exerted by the heart muscle is 0.02 N.

Step by step solution

01

Understand the Problem

We are given the mass of blood, its initial and final velocities, and the time duration over which the acceleration occurs. We need to calculate the force exerted by the heart.
02

Identify the Relevant Formula

Force can be calculated using Newton's second law of motion, which states that force is the product of mass and acceleration. The formula is: \( F = m \cdot a \).
03

Calculate Acceleration

First, we need to calculate the acceleration using the formula: \( a = \frac{{v_f - v_i}}{t} \), where \( v_f = 0.35 \ m/s \), \( v_i = 0.25 \ m/s \), and \( t = 0.10 \ s \). Plug in these values to get the acceleration: \( a = \frac{0.35 - 0.25}{0.10} = 1 \ m/s^2 \).
04

Convert Mass to Kilograms

The mass of the blood is given as 20 grams, which needs to be converted to kilograms for consistency in units. 1 gram = 0.001 kilograms, so 20 grams is 0.02 kilograms.
05

Calculate the Force

Now that we have both mass and acceleration, we can calculate force: \( F = m \cdot a = 0.02 \ kg \cdot 1 \ m/s^2 = 0.02 \ N \).

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Key Concepts

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

Newton's second law of motion
Newton's second law establishes a relationship between the motion of an object and the forces acting upon it. It states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This can be captured in the equation: \( F = m \cdot a \). Here, \( F \) represents force, \( m \) denotes mass, and \( a \) symbolizes acceleration.

This law implies that if you apply a greater force on an object, it will accelerate more. Conversely, a heavier object will accelerate less than a lighter one if the same force is applied. Think of pushing a child's tricycle versus a loaded truck; the tricycle moves easily because it has much less mass. Knowing this helps solve physics problems where you need to compute the force when mass and acceleration are given.
Understanding acceleration
Acceleration refers to how quickly an object changes its velocity. In simpler terms, it's the rate at which something speeds up or slows down. It's important to note that acceleration is a vector quantity, meaning it has both magnitude and direction.

The formula for acceleration is \( a = \frac{{v_f - v_i}}{t} \), where:
  • \( v_f \) is the final velocity
  • \( v_i \) is the initial velocity
  • \( t \) is the time it takes to change from \( v_i \) to \( v_f \)
This formula allows us to calculate the change in velocity per unit time. For instance, if a car goes from 0 to 60 km/h in 10 seconds, it accelerates more quickly than a bicycle going from 0 to 15 km/h in the same amount of time. Understanding this formula is crucial when solving problems that involve force calculation.
Mass conversion in physics
In many physics problems, especially when dealing with forces and accelerations, it's essential to ensure that the units used are consistent. Mass, in particular, needs to be in kilograms when using the metric system.

To convert mass from grams to kilograms, you multiply by 0.001 (since 1 gram = 0.001 kilograms). For example, a mass of 20 grams would convert to 0.02 kilograms. This conversion is critical in solving problems where you're applying formulas like Newton's second law.

Paying attention to units can make the difference between a correct answer and a wrong one. By consistently converting units in your calculations, you ensure accuracy and align with standard scientific conventions.
Solving physics problems
Physics problems often revolve around understanding fundamental principles and then applying them systematically. Here’s a simplified approach to tackling these challenges:
  • First, understand the problem by identifying what's given and what's needed.
  • Second, decide which physics principles or formulas are relevant. In many cases, like force calculations, Newton's second law could be the key.
  • Next, calculate the necessary quantities step-by-step, ensuring unit consistency. Acceleration, for example, may need to be computed before you can find force.
  • Lastly, perform the final calculations and double-check your work for accuracy.
This structured approach can help you streamline the process of solving physics problems effectively. Practice with varied examples to become familiar with different scenarios and calculations.

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Most popular questions from this chapter

Two blocks made of different materials, connected by a thin cord, slide down a plane ramp inclined at an angle to the horizontal, Fig. 4-76 (block \(B\) is above block \(A\)). The masses of the blocks are \(m_A\) and \(m_B\) and the coefficients of friction are \(\mu_A\) and \(\mu_B\). If \(m_A=m_B=\) 5.0 kg and \(\mu_A=\) 0.20 and \(\mu_B=\) 0.30, determine (a) the acceleration of the blocks and (b) the tension in the cord, for an angle \(\theta =\) 32\(^\circ\).

(II) One 3.2-kg paint bucket is hanging by a massless cord from another 3.2-kg paint bucket, also hanging by a massless cord, as shown in Fig. 4-49. (\(a\)) If the buckets are at rest, what is the tension in each cord? (\(b\)) If the two buckets are pulled upward with an acceleration of 1.25 m/s\(^2\) by the upper cord, calculate the tension in each cord

(II) Drag-race tires in contact with an asphalt surface have a very high coefficient of static friction. Assuming a constant acceleration and no slipping of tires, estimate the coefficient of static friction needed for a drag racer to cover 1.0 km in 12 s, starting from rest.

A 65-kg ice skater coasts with no effort for 75 m until she stops. If the coefficient of kinetic friction between her skates and the ice is \(\mu_k=\) 0.10, how fast was she moving at the start of her coast?

A 72-kg water skier is being accelerated by a ski boat on a flat ("glassy") lake. The coefficient of kinetic friction between the skier's skis and the water surface is \(\mu_k=\) 0.25 (Fig. 4-74). (\(a\)) What is the skier's acceleration if the rope pulling the skier behind the boat applies a horizontal tension force of magnitude \(F_T=\) 240 N to the skier (\(\theta=0^\circ\))? (\(b\)) What is the skier's horizontal acceleration if the rope pulling the skier exerts a force of \(F_T=\) 240 N on the skier at an upward angle \(\theta= 12^\circ\)?(\(c\)) Explain why the skier's acceleration in part (\(b\)) is greater than that in part (\(a\)).

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