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2 years 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]2 years 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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A point charge q1=−4.00nc is at the point x=0.600 meters, y=0.800 meters, and a second point charge q2=+6.00nc is at the point x

Katen [24]

electric field between mid point of two charges is given by

$E = E_1 + E_2$

here we have

$E_1 = \frac{kq_1}{r^2}$

$E_1 = \frac{9*10^9 * 4 * 10^{-9}}{0.4^2}$

$E_1 = 225 N/c$

now similarly for other charge

$E_2 = \frac{kq_2}{r^2}$

$E_2 = \frac{9*10^9 * 6 * 10^{-9}}{0.4^2}$

$E_2 = 337.5 N/C$

so the net electric field will be given as

$E = 225 + 337.5$

$E = 565.5 N/C$

8 0

2 years ago

Read 2 more answers

Steam at 700 bar and 600 oC is withdrawn from a steam line and adiabatically expanded to 10 bar at a rate of 2 kg/min. What is t

Setler79 [48]

Answer:

Final temperature of the steam = 304.29 K = 31.14°C

Rate Entropy generation of the process should be 0 because no heat is transferred into or out of the system. And ΔS = Q/T = 0 J/K.

But 0.01 J/k.s was obtained though, which is approximately 0.

Explanation:

For an adiabatic system, the Pressure and temperature are related thus

P¹⁻ʸ Tʸ = constant

where γ = ratio of specific heats. For steam, γ = 1.33

P₁¹⁻ʸ T₁ʸ = P₂¹⁻ʸ T₂ʸ

P₁ = 700 bar

P₂ = 10 bar

T₁ = 600°C = 873.15 K

T₂ = ?

(700⁻⁰•³³)(873.15¹•³³) = (10⁻⁰•³³)(T₂¹•³³)

T₂ = 304.29 K = 31.14°C

b) Rate Entropy generation of the process should be 0 because no heat is transferred into or out of the system. And ΔS = Q/T

To prove this

Entropy of the process

dQ - dW = dU

dQ = dU + dW

dU = mCv dT

dW = PdV

dQ = TdS

TdS = mCv dT + PdV

dS = (mCv dT/T) + (PdV/T) =

PV = mRT; P/T = mR/V

dS = (mCv dT/T) + (mRdV/V)

On integrating, we obtain

ΔS = mCv In (T₂/T₁) + mR In (V₂/V₁)

To obtain V₁ and V₂, we use PV = mRT

V/m = specific volume

Pv = RT

R for steam = 461.52 J/kg.K

For V₁

P = 700 bar = 700 × 10⁵ Pa, T = 873.15 K

v₁ = (461.52 × 873.15)/(700 × 10⁵) = 0.00570 m³/kg

For V₂

P = 10 bar = 10 × 10⁵ Pa, T = 304.29 K

v₂ = (461.52 × 304.29)/(10 × 10⁵) = 0.143 m³/kg

ΔS = mCv In (T₂/T₁) + mR In (V₂/V₁)

m = 2 kg/min = 0.0333 kg/s, Cv = 1410.8 J/kg.K

ΔS = [0.03333 × 1410.8 × In (304.29/873.15)] + (0.03333 × 461.52 × In (0.143/0.00570)

ΔS = - 49.56 + 49.57 = 0.01 J/K.s ≈ 0 J/K.s

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2 years ago

An astronaut hits a golf ball of mass m on the Moon, where there is no atmosphere and the acceleration due to gravity is g 6 , w

sasho [114]

Answer:

The magnitude of the average force exerted by the club on the ball during contact = mv/t

Explanation:

Impulse exerted on the ball = Momentum of the ball = mass * velocity = m*v

As we know,

m*v = Integration of F.dt with limits 0 to T

Ft = mv

F = mv/t

The magnitude of the average force exerted by the club on the ball during contact = mv/t

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2 years ago

This table shows Wayne’s weight on four different planets. Planet Wayne’s weight (pounds) Mars 53 Neptune 159 Venus 128 Jupiter

oee [108]

The correct order is (in decreasing order of gravity strength)
Jupiter - Neptune - Venus - Mars

In fact, Wayne's weight on each planet is given by
$W=mg$
where m is Wayne's mass, which is a constant value, and g is the gravity strength at the surface of the planet. Therefore, the Wayne's weight W on each planet is directly proportional to the gravity strength of that planet: so the planet with the strongest gravity is the one where Wayne's weight is the greatest (Jupiter, 333 pounds), followed by Neptune (159), Venus (128) and Mars (53).

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2 years ago

Ariel dropped a golf ball from her second story window. The ball starts from rest and hits the sidewalk 3.5 s later with a veloc

Aleks [24]

Answer:

By using the acceleration formula,

$a = \frac{v - u}{t}$

$a = \frac{14.7 - 0}{3.5}$

$a = 4.2m \: s ^{ - 2}$

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2 years ago