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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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Benjamin Franklin has convinced his hapless assistant Mike Piepan to participate in an experiment on electiricty. Ben has set up

ad-work [718]

Answer:

Mike will receive an electric shock

Explanation:

Human body is a conductor of electricity. When lightning strikes rod and give it negative charge, the rod will dissipate its charge as soon as it comes in contact with earth via conduting material. Mike will receive a severe electric shock as negatice charge pass through his body to other rod and to the ground.His body will feel numb. He may also get unconscious.

5 0

3 years ago

I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi

Masteriza [31]

Answer:

After it has traveled through 1 cm : $I(1 \ cm) = 0.5488 I_0$

After it has traveled through 2 cm : $I(2 \ cm) = 0.3012 I_0$

After it has traveled through 1 cm : $od( 1\ cm) = 0.2606$

After it has traveled through 2 cm : $od( 2\ cm) = 0.5211$

Explanation:

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient $\mu$ the formula is:

$I(x) = I_0 e^{-\mu x}$

where I is the intensity of the beam, $I_0$ is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

$I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = 0.5488 \ I_0$

After travelling 2 cm:

$I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = 0.3012 \ I_0$

The optical density od is given by:

$od(x) = - log_{10} ( \frac{I(x)}{I_0} )$ .

So, after travelling 1 cm:

$od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )$

$od( 1\ cm) = - log_{10} ( 0.5488 )$

$od( 1\ cm) = - ( - 0.2606)$

$od( 1\ cm) = 0.2606$

After travelling 2 cm:

$od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )$

$od( 2\ cm) = - log_{10} ( 0.3012 )$

$od( 2\ cm) = - ( - 0.5211)$

$od( 2\ cm) = 0.5211$

3 0

3 years ago

ffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff

mixer [17]

Answer:

fffffffffffffffffffffffffffff

5 0

3 years ago

Read 2 more answers

The formation of a cell plate is a characteristic of

RUDIKE [14]

The formation of a cell plate is a characteristic of cytokinesis in terrestrial plants.

4 0

3 years ago

A 1.0-kg block and a 2.0-kg block are pressed together on a horizontal frictionless surface with a compressed very light spring

egoroff_w [7]

Answer:

4. both blocks will both have the same amount of kinetic energy.

Explanation:

When the blocks are released free from the compression force, the spring exerts equal and opposite force on each block but the block with heavier (double) mass will attain slower ( half ) speed as compared to the lighter block according to the law of inertia. This works in synchronization to energy conservation.

Spring force is given as:

$F=k.\Delta x$

where: $\Delta x=$ length of compression in the spring

<u>We know kinetic energy is given by:</u>

$KE=\frac{1}{2} m.v^2$

Hence the kinetic energy of both the blocks is equal when they are released to move free.

8 0

3 years ago