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1 year ago

You decide to work at a heart rate of 150 instead of 120. What area of F.I.T.T. did you change?

Physics

1 answer:

Rina8888 [55]1 year ago

8 0

Key concepts

Heart rate

Exercising

The heart

Cardiovascular system

Health

Introduction

As Valentine's Day approaches, we're increasingly confronted with "artistic" images of the heart. Real hearts hardly resemble to two-lobed shapes adorning cards and candy boxes this time of year. And the actual shape of the human heart is important for its function of supplying blood to the entire body. You have likely noticed that your heart beats more quickly when you exercise. But have you ever taken the time to observe how long it takes to return to its normal rate after you're done exercising? In this science activity you'll get to do some exercises to explore your own heart-rate recovery time.

Background

Your heart is continuously beating to keep blood circulating throughout your body. Its rate changes depending on your activity level; it is lower while you are asleep and at rest and higher while you exercise—to supply your muscles with enough freshly oxygenated blood to keep the functioning at a high level. Because your heart is also a muscle, exercise, in turn, helps keep it healthy. The American Heart Association recommends that a person does exercise that is vigorous enough to raise their heart rate to their target heart-rate zone—50 percent to 85 percent of their maximum heart rate, which is 220 beats per minute (bpm) minus their age for adults—for at least 30 minutes on most days, or about 150 minutes a week in total. So for a 20-year-old, the maximum heart rate would be 200 bpm, with a target heart-rate zone of 100 to 170 bpm. (For those 19 or younger, target zones can vary more than they do for adults.)

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Part a)

As we know that

$P_1V_1^{1.35} = P_2V_2^{1.35}$

here we know that

P1 = 20 bar

V1 = 0.5 m^3

V2 = 2.75 m^3

from above equation

$20* 0.5^{1.35} = P * (2.75)^{1.35}$

$P = 2 bar$

so final state pressure will be 2 bar

Part b)

now in order to find the work done

$W = \int PdV$

$W = \int \frac{c}{V^{1.35}}dV$

$W = c\frac{V^{-0.35}}{-0.35}$

$W = \frac{P_1V_1 - P_2V_2}{0.35}$

$W = \frac{20* 0.5 - 2 * 2.75}{0.35}* 10^5 = 12.86 * 10^5 J$

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1 year ago

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Lamar writes several equations trying to better understand potential energy. H = d with an arrow to the equation W = F d and P E

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Answer:

The gravitational potential energy equals the work needed to lift the object.

Explanation:

here we know that

$H = \vec d$

work done is given as

$W = \vec F . \vec d$

Potential energy is given as

$PE_g = mgh$

force due to gravity is given as

$\vec F_g = mg$

now here if we plug in the value of distance and force in the formula of work done then we will have

$W = (mg)(h)$

so here we got

$W = PE_g$

so we can concluded that

The gravitational potential energy equals the work needed to lift the object.

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A biophysics experiment uses a very sensitive magnetic field probe to determine the current associated with a nerve impulse trav

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Answer:

The peak current carried by the axon is 5.85 x 10⁻⁸ A

Explanation:

Given;

distance of the field from the axon, r = 1.3 mm

peak magnetic field strength, B = 9 x 10⁻¹² T

To determine the peak current carried by the axon, apply the following equation;

$B = \frac{\mu I}{2\pi r}$

where;

B is the peak magnetic field

r is the distance of the magnetic field from axon

μ is permeability of free space = 4π x 10⁻⁷

I is the peak current

Re-arrange the equation and solve for "I"

$B = \frac{\mu I}{2\pi r} \\\\I = \frac{B*2\pi r}{\mu} \\\\I = \frac{9*10^{-12}*2*\pi *1.3*10^{-3}}{4\pi *10^{-7}} \\\\I = 5.85 *10^{-8} \ A$

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1 year ago

A thin hoop with a mass of 5.0 kg rotates about a perpendicular axis through its center. A force F is exerted tangentially to th

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Given :

Thin hoop with a mass of 5.0 kg rotates about a perpendicular axis through its center.

A force F is exerted tangentially to the hoop. If the hoop’s radius is 2.0 m and it is rotating with an angular acceleration of 2.5 rad/s².

To Find :

The magnitude of F.

Solution :

Torque on hoop is given by :

$\tau =F\times R\\\\I\alpha = FR\\\\MR^2\alpha = FR\\\\F = MR\alpha$ ( Moment of Inertia of hoop is MR² )

Putting value of M, R and α in above equation, we get :

$F=5\times 2\times 2.5\ N\\\\F = 25 \ N$

Therefore, the magnitude of force F is 25 N.

Hence, this is the required solution.

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Often what one expects to see influences what is perceived in the surrounding environment. Please select the best answer from th

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Hello <span>Andijwiltbank
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Question: <span>Often what one expects to see influences what is perceived in the surrounding environment. True or False?

Answer: True

Reason: What we observe about the environment decides what we believe about it and how we react.

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