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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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Calculate the energy released in joules when one mole of polonium-214 decays according to the following equation21484 Po -->

GuDViN [60]

Answer:

ΔE = 8.77 × 10¹¹ J

Explanation:

given,

²¹⁴₈₄Po -----> ²¹⁰₈₂Pb + 42 He

Atomic masses: Pb-210 = 209.98284 amu

Po-214 = 213.99519 amu

He-4 = 4.00260 amu

1 kg = 6.022 × 10²⁶ amu;

NA = 6.022 × 10²³ mol⁻¹

c = 2.99792458 × 10⁸ m/s

energy of molecule using equation

ΔE = Δm c²

Δm is mass difference and c is speed of light

Δm = 209.98284 + 4.00260 - 213.99519

Δm = - 0.00975 amu

1 amu = 1.66 x 10⁻²⁷ kg

- 0.00975 amu = - 0.00975 x 1.66 x 10⁻²⁷ Kg

= -0.016185 x 10⁻²⁷ Kg

total mass = 6.022 × 10²³ x -0.016185 x 10⁻²⁷

= -0.097467 x 10⁻⁴ Kg

ΔE = -(0.097467 x 10⁻⁴) (3 x 10^8)²

ΔE = - 8.77 × 10¹¹

ΔE = 8.77 × 10¹¹ J

8 0

2 years ago

1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n

sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

$(1).\: x =v_0t$

$(2).\: y= 15m-\dfrac{1}{2}gt^2$

where $v_0$ is the initial velocity.

(a).

When the projectile hits the 50m mark, $y=0$ ; therefore,

$0=15-\dfrac{1}{2}gt^2$

solving for $t$ we get:

$t= 1.75s.$

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

$50m = v_0(1.75s)$

which gives

$\boxed{v_0 = 28.58m/s.}$

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

$v_x = 28.58m/s.$

the vertical component of the velocity is

$v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.$

which gives a speed $v$ of

$v = \sqrt{v_x^2+v_y^2}$

$\boxed{v =33.3m/s.}$

4 0

1 year ago

Two wires are stretched between two fixed supports and have the same length. One wire A there is a second-harmonic standing wave

lina2011 [118]

(a) Greater

The frequency of the nth-harmonic on a string is an integer multiple of the fundamental frequency, $f_1$ :

$f_n = n f_1$

So we have:

- On wire A, the second-harmonic has frequency of $f_2 = 660 Hz$ , so the fundamental frequency is:

$f_1 = \frac{f_2}{2}=\frac{660 Hz}{2}=330 Hz$

- On wire B, the third-harmonic has frequency of $f_3 = 660 Hz$ , so the fundamental frequency is

$f_1 = \frac{f_3}{3}=\frac{660 Hz}{3}=220 Hz$

So, the fundamental frequency of wire A is greater than the fundamental frequency of wire B.

(b) $f_1 = \frac{v}{2L}$

For standing waves on a string, the fundamental frequency is given by the formula:

$f_1 = \frac{v}{2L}$

where

v is the speed at which the waves travel back and forth on the wire

L is the length of the string

(c) Greater speed on wire A

We can solve the formula of the fundamental frequency for v, the speed of the wave:

$v=2Lf_1$

We know that the two wires have same length L. For wire A, $f_1 = 330 Hz$ , while for wave B, $f_B = 220 Hz$ , so we can write the ratio between the speeds of the waves in the two wires: