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.

<u>Explanation:</u>

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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The current in some DC circuits decays according to the function I=I0e−t/τ, where I is the current at some point in time, I0 is

almond37 [142]

Answer: 1.95

Explanation:

You should start off from the decay formula and solve for τ:

$I = I_{0}e^{\frac{t}{\tau\\ } }$

$\frac{I}{I_{0}} = e^{\frac{-t}{\tau} }$

Apply inverse logarithmic function:

$ln(\frac{0.2 A}{1.2 A} ) = \frac{-t}{\tau}$

The final form will be:

$\tau=\frac{-3.5s}{ln(\frac{0.2A}{1.2A} )}$

Inputing values for I, IO, and t:

$\tau=\frac{-3.5S}{ln(\frac{0.2 A}{1.2 A} )} = 1.95$

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3 years ago

How are the spiral arms of the milky way detected?

Step2247 [10]

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3 years ago

A student is swimming south applying a force of 256 N. The water exerts a westward force of 104 N. If the student has a mass of

grigory [225]

Answer:

$a=2.9\ m/sec^2$

Explanation:

<u>Net Forces and Acceleration</u>

The second Newton's Law relates the net force $F_r$ acting on an object of mass m with the acceleration a it gets. Both the net force and the acceleration are vector and have the same direction because they are proportional to each other.

$\vec F_r=m\vec a$

According to the information given in the question, two forces are acting on the swimming student: One of 256 N pointing to the south and other to the west of 104 N. Since those forces are not aligned, we must add them like vectors as shown in the figure below.

The magnitude of the resulting force $F_r$ is computed as the hypotenuse of a right triangle

$|F_r|=\sqrt{256^2+104^2}$