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3 years ago

What is the heart rate recorded a few minutes after completing a workout?

Physics

2 answers:

DochEvi [55]3 years ago

8 0

Correct answer should be C

Naily [24]3 years ago

5 0

The answer is D, recovery heart rate

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Pls answer fast !!!!!!!!!!!!!!!!!!!!!!!!

juin [17]

Answer:

tgffgg

Explanation:

8 0

3 years ago

Reading from left to right, describe the spacing between the dots in lines 1, 2, and 3.

Ierofanga [76]

Answer:

Line 1: the spacing is even the whole time

Line 2: the spacing increases over time

Line 3: the spacing decreases over time

hope this helps!

5 0

3 years ago

a 2.0 kg mass moving to the east at a speed of 4.0 m/s collides head-on in a perfectly inelastic collision with a stationary 2.0

cricket20 [7]

Answer:

12J

Explanation:

Kinetic Energy before collision = 1/2mv1^2 = 1/2×2×4^2 = 16J

Velocity after collision (v2) = m1v1/m1+m2 = 2×4/2+2 = 8/4 = 2m/s

Kinetic Energy after collision = 1/2mv2^2 = 1/2×2×2^2 = 4J

Kinetic Energy lost = 16J - 4J = 12J

3 0

3 years ago

Read 2 more answers

You've recently read about a chemical laser that generates a 20.0-cm-diameter, 22.0 MW laser beam. One day, after physics class,

Setler [38]

Answer:

0.3847 m/s

Explanation:

$I$ = Intensity = $\dfrac{P}{A}=\dfrac{P}{\pi r^2}$

d = Diameter = 20 cm

r = Radius = $\dfrac{d}{2}=\dfrac{20}{2}=10\ cm$

c = Speed of light = $3\times 10^8\ m/s$

s = Distance = 100 m

P = Power = 22 MW

Pressure due to the laser is given by

$P_r=\dfrac{I}{c}\\\Rightarrow P_r=\dfrac{P}{Ac}\\\Rightarrow P_r=\dfrac{P}{\pi r^2c}\\\Rightarrow P_r=\dfrac{22\times 10^{6}}{\pi 0.1^2\times 3\times 10^8}\\\Rightarrow P_r=2.33427\ N/m^2$

Force is given by

$F=P_rA\\\Rightarrow F=2.33427\times \pi 0.1^2\\\Rightarrow F=0.07333\ N$

Acceleration is given by

$a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{0.07333}{99}\\\Rightarrow a=0.00074\ m/s^2$

Speed of the block would be

$v=\sqrt{2as}\\\Rightarrow v=\sqrt{2\times 0.00074\times 100}\\\Rightarrow v=0.3847\ m/s$

The speed of the block is 0.3847 m/s

5 0

3 years ago

A rock is thrown downwards from the edge of the Grand Canyon. With

Olin [163]

Answer:

Vf = 28 m/s

Explanation:

In order to find the final velocity of the rock, we will use the 3rd equation of motion. The third equation of motion for vertical direction is written as follows:

2gh = Vf² - Vi²

where,

g = acceleration due to gravity = 9.8 m/s²

h = height dropped = 40 m

Vf = final velocity of the rock = ?

Vi = Initial Velocity of the rock = 0 m/s (since, rock was initially at rest)

Therefore,

(2)(9.8 m/s²)(40 m) = Vf² - (0 m/s)²

Vf = √(784 m²/s²)

3 0

4 years ago