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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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Phase change requires...

amm1812

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3 0

3 years ago

A seagull flies at a velocity of 9.00 m/s straight into the wind.

RideAnS [48]

a)If it takes the bird 18.0 minutes to fly 6 km away from the earth, the wind's speed will be 4 m/s.

b) The bird would need 7 minutes and 42 seconds to fly back 6 kilometers if he turned around and flew with the wind.

c)Compared to the 133.33 seconds it would take without the wind, the overall round-trip time is affected by the wind.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

A seagull flies at a velocity, $\rm v_{SA} = 9 \ m/sec$

The time the bird takes,t=18.0 min

The distance traveled relative to the earth = 6.00 km

The seagull's relative velocity with reference to the ground as;

$\rm v_{sg} = \frac{6.00 \times 10^3 \ m }{(20 min) \times \frac{60 s }{1 \ min}} \\\\ v_{sg}= 5.00 \ m/sec$

Air velocity with reference to the ground is;

$\rm v_{AG}= v_{SG}-v_{SA} \\\\ v_{AG} = 5.00 \ m/sec - 9.00 \ m/sec \\\\ v_{AG} = -4.00 \ m/sec$

If the bird turns around and flies with the wind, The time will he take to return 6.00 km is;

$\rm v_{SG}=v_{SA}+v_{AG} \\\\ v_{SG}=-900 \ m/sec +(-4.00 \ m/sec) \\\\ v_{SG}= -13.00 \ m/sec$

The time the bird takes;

$\rm t = \frac{x_{SG}}{v_{SG}} \\\\ t = \frac{6.00 \times 10^3 \ m }{13.00 \ m/sec } \\\\ t = 462 m/sec \\\\ t = 7 \ min \ and \ 42 \ sec$

c)\

The total round-trip time compared to what it would be with no wind. is;

$\rm t = 20 \ min( \frac{60 \ sec }{1 \ min} )+ 462 \ sec \\\\ t = 1200 \ sec +6 462 \ ec \\\\ t= 1662 \ sec$

The time for the round trip is;

$\rm t = \frac{12 \times 10^ 3 }{ 9 \ m/sec } \\\\ t = 1333.33 \ sec$

Hence the wind's speed, the time bird would need to fly back the total round-trip time will be 4 m/s, 7 minutes and 42 seconds and 1333.33 sec.

To learn more about the velocity, refer to the link: brainly.com/question/862972.

#SPJ1

4 0

2 years ago

Give two examples of vernier calliper

omeli [17]

Answer:

Example 1, if a vernier caliper output a measurement reading of 2.13 cm, this means that: The main scale contributes the main number(s) and one decimal place to the reading

E.g. 2. 1 cm, whereby 2 is the main number and 0.1 is the one decimal place number

Explanation:

plz mark as brainliest and hope it helps you

4 0

2 years ago

A raindrop of mass 0.5 * 10^-4 kg is falling verctically under the influence of gravity. The air drag on the raindrop is fdrag =

Elina [12.6K]

Answer:

The displacement of the air drop after 3 second is 18.27 m.

Explanation:

Mass of the rain drop = m = $0.5\times 10^{-4} kg$

Weight of the rain drop = W

Duration of time = t = 3 seconds

$W=m\times g$

Drag force on rain drop = $D=0.2\times 10^{-5} v^2$

$W=0.5\times 10^{-4} kg\time 10 m/s^2=0.5\times 10^{-3} N$

Motion of the rain drop:

$F=m\times a$

Net force on the rain drop , F= W - D

$W-D=m\times a$

$0.5\times 10^{-3} N-0.2\times 10^{-5} v^2=0.5\times 10^{-4} kg\times a$

$0.5\times 10^{-3} kg m/s^2-0.2\times 10^{-5} v^2=0.5\times 10^{-4} kg\times \frac{v}{t}$

$0.006v^2+0.05v-1.5=0$

v = 12.18 m/s

Initial velocity of the rain drop = u = 0 (since, it is starting from rest)

v=u+at (First equation of motion)

$12.18 m/s=0m/s+a\times 3 s$

$a=4.06 m/s^2$

$s=ut+\frac{1}{2}at^2$ (second equation of motion)

$s=0\times 3s+\frac{1}{2}\times 4.06m/s^2\times (3 s)^2$

s = 18.27 m

The displacement of the air drop after 3 second is 18.27 m.

6 0

2 years ago

The handle of a hammer is 20 cm long and the metal end pulls

artcher [175]

Answer:

Given the experimental setup shown below, where should you put object C, ... applies a force of 150 N. What is the mechanical advantage of the hammer? ... Explain why tin snips, designed for cutting metal, have long handles and short blades. ... advantage for scissors, a first-class lever, is the length of the handle divided ...

Explanation:

6 0

2 years ago