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3 years ago

It's not important to monitor your heart rate during moderate intensity activities. True False

Physics

2 answers:

Mama L [17]3 years ago

5 0

I think it is False

hope this helps :3

8_murik_8 [283]3 years ago

5 0

Answer:

FALSE

Explanation:

Monitoring heart rate is important in all physical activities. Some heart problems are silent and may manifest during increased physical effort, albeit moderate. Therefore, the practice of exercises should be authorized by a cardiologist and you should monitor your heart rate whenever possible. This is a simple matter now, as there are applications and wristbands that measure heart rate.

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A wave pulse travels along a string at a speed of 230 cm/s . Note that parts a - d are independent and refer to changes made to

Oksanka [162]

a) The speed of the wave will increase by a factor $\sqrt{2}$ .

b) The speed of the wave will halve

c) The speed of the wave will double

Explanation:

The speed of a standing wave on a string is given by

$v=\sqrt{\frac{T}{m/L}}$

where

T is the tension in the string

m is the mass of the string

L is the length of the string

In this part of the problem, the tension in the string is doubled, so that the new tension is

T' = 2T

Substituting into the equation, we find the new speed of the wave in the string:

$v'=\sqrt{\frac{T'}{m/L}}=\sqrt{\frac{2T}{m/L}}=\sqrt{2}\sqrt{\frac{T}{m/L}}=\sqrt{2}v$

So, the speed will increase by a factor $\sqrt{2}$ .

We can solve also this part by referring to the formula

$v=\sqrt{\frac{T}{m/L}}$

where

T is the tension

m is the mass

L is the length

In this case, the string mass is quadrupled, so the new mass is:

m' = 4m

Substituting into the equation, we find what happens to the speed of the wave:

$v'=\sqrt{\frac{T}{m'/L}}=\sqrt{\frac{T}{4m/L}}=\frac{1}{\sqrt{4}}\sqrt{\frac{T}{m/L}}=\frac{1}{2}v$

So, the speed of the wave will halve.

Again, we can solve this part by referring to the same equation

$v=\sqrt{\frac{T}{m/L}}$

where

T is the tension

m is the mass

L is the length

In this case, the length of the string is quadrupled, so the new length is:

L' = 4L

Substituting into the equation, we find that the new speed is:

$v'=\sqrt{\frac{T}{m/L'}}=\sqrt{\frac{T}{m/(4L)}}=\sqrt{4}\sqrt{\frac{T}{m/L}}=2v$

So, the speed of the wave will double.

Learn more about waves:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

3 0

4 years ago

An accelerating voltage of 2.42 103 V is applied to an electron gun, producing a beam of electrons originally traveling horizont

Pavlova-9 [17]

Answer:

Part (a) The magnitude of the deflection of electron beam on the screen due to the Earth's gravitational field is 5.97* $10^{-16}$ m.

Part (b) The magnitude of the deflection of electron beam on the screen due to the vertical component of the Earth's magnetic field is $6.19* 10^-3$ m