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3 years ago

The heart rate that indicates your fitness level is the

Physics

1 answer:

MrRa [10]3 years ago

8 0

The heart rate that indicates your fitness level is the Resting Heart Rate.

Explanation:

The statement from Welch says that when we practise more and do an exercise harder, then the heart rate at rest would be low. This is the main reason for considering resting heart rates as a good indicator to one’s fitness level. The normal adult heart rate at rest is 60 to 100 bpm (beat per minute).

In general, a lower heart rate means a calmer heart rate and good heart health. For example, for a well-trained athlete, a normal heart rate can be closer to 40 bpm. Our resting rhythm, the measure of how often the heart beats every minute at rest.

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How are the electric field lines around a positive charge affected when a second positive charge is near it?

Lesechka [4]

Answer: The field lines bend away from the second positive charge

Explanation: opposite attracts, same repulse

4 0

3 years ago

Read 2 more answers

A girl weighing 50 kgf wears sandals of pencil heel of area of cross section 1 cm^2, stands on the floor.An elephant weighing 20

Klio2033 [76]

Answer:

$\boxed{{\boxed{\blue{ 12.5}}}}$

Explanation:

Given, for girl : Weight or force;

$\rm \: F_1 = 50 \: kgf$

Area of both heels;

$\rm \: A_1 = \; 2 ×1 \; cm^2 = 2 \: cm^2$

$\rm \: Pressure \: P_1 = \cfrac{F_1}{ A_1 } = \dfrac{50 \: kgf}{2 \: cm {}^{2} } = 25 \: kgf \: cm {}^{ - 1}$

For elephant, Weight = Force $\rm F_2$ = 2000 kg•f

Area of 4 feet;

$\rm \: A_2 = \; 4 \times 250 \; cm^2 = 1000 \: cm^2$

$\rm \: Pressure \: P_2 = {F_2}/{A_2} \; = \cfrac{2 \cancel{0 00 }\: kgf}{1 \cancel{000} \: cm^2} = 2 \: kgf \: { \:cm}^{- 1}$

Now;

$\rm = \dfrac{Pressure \: Exerted \: by \: the \: Girl}{Pressure \: exerted \: by \: the \: elephant}$

$= \rm \: P_1/P_2$

$\implies \rm\cfrac{25 \: kgf \: \: cm {}^{ - 2} }{2 \: kgf \: cm {}^{ - 2} } = \rm\cfrac{25 \: \cancel{kgf \: \: cm {}^{ - 2}} }{2 \: \cancel{ kgf \: cm {}^{ - 2}} } = \boxed{12.5}$

Thus, the girl's pointed heel sandals exert 12.5 times more pressure P than the pressure P exerted by the elephant.

I aspire this helps!

3 0

2 years ago

A construction worker is carrying a load of 40 kg over his head and is walking at a constant velocity if he travels a distance o

Anastasy [175]

Answer:

W = 0

Explanation:

We are given with, a construction worker is carrying a load of 40 kg over his head and is walking at a constant velocity. He travels a distance of 50 m.

The work done by an object is given by :

$W=Fd$

F = ma

So,

$W=mad$

m is mass

a is acceleration

d is displacement

The worker is moving with constant velocity, its acceleration will be 0. So, the work done by the worker is 0.

8 0

3 years ago

A force of 45 newtons is applied on an object, moving it 12 meters away in the same direction as the force. What is the magnitud

NARA [144]

<h2>Answer: 540 J</h2>

Explanation:

The Work $W$ done by a Force $F$ refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path.

Now, when the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:

$W=(F)(d)$ (1)

In this case both (the force and the distance in the path) are parallel (this means they are in the same direction), so the work $W$ performed is the product of the force exerted to push the box $F=45N$ by the distance traveled $d=12m$ .

Hence:

$W=(45N)(12m)$ (2)

$W=540Nm=540J$

6 0

3 years ago

Calculate the binding energy per nucleon of the helium nucleus 52he. express your answer in millions of electron volts to four s

Vitek1552 [10]

The main formula is given by Eb/nucleon = Eb/ mass of nucleid
as for 52He, the mass is 5
so by applying Einstein's formula Eb=DmC², Eb=binding energy
52He-----------> 2 x 11p + 3 x10n is the equation bilan
so Dm=2 mp + (5-2)mn-mnucleus, mp=mass of proton=1.67 10^-27 kg
mn=mass of neutron=1.67 10^-27 kg
 m nucleus= 5
Dm= 2x1.67 10^-27 kg+ 3x1.67 10^-27 kg-5= - 4.9 J
Eb= - 4.9 J x c²= -4.9 x 9 .10^16= - 45 10^16 J
so the answer is Eb /nucleon = Eb/5= -9.10^16 J, but 1eV=1.6 . 10^-19 J
so -9.10^16 J/ 1.6 10^-19= -5.625 10^35 eV

the final answer is Eb /nucleon = -5.625 x10^35 eV

8 0

3 years ago