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vitfil [10]
4 years ago
6

How does exercise affect your heart rate??

Physics
2 answers:
mart [117]4 years ago
8 0
Exercise increases your heart rate, making your blood pump much faster to be able to accommodate your moving body.<span />
Setler [38]4 years ago
6 0
It makes your heart rate go up example like if you were running your heart rate would go up then when you stopped and took a break it’d go down slowly.
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According to the law of conservation of energy, which changes would
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3 years ago
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How much potential energy foes a 5kg mass have when its 2 meters above the ground?(Hint :PE=m*g*h)​
frez [133]

Answer:

<h2>98 J</h2>

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 9.8 m/s²

From the question we have

PE = 5 × 9.8 × 2

We have the final answer as

<h3>98 J</h3>

Hope this helps you

5 0
3 years ago
A cardboard box sits on top of a concrete sidewalk where the coefficient of friction between the surfaces is 0.3. The mass of th
blagie [28]
Hope these r correct:
Normal Force≈ 68.5
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Accel. ≈ 58.5

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6 0
2 years ago
The long handle on a rake is a _____.
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6 0
3 years ago
The speaker at the concert has the sound intensity level of 100 dB if we listen from the distance 5 m.How far from the speaker d
Mademuasel [1]

Answer:

28.11m far from the speaker the intensity drops to 85 dB.

Explanation:

In the equation for the Decibel scale

  (1). \: \:\beta =10 log(\dfrac{I}{I_0})

The ratio of the intensities can be written as

$ \frac{I}{I_0} = \dfrac{\frac{P}{A} }{\frac{P}{A_0} } $

\dfrac{I}{I_0} = \dfrac{A_0}{A}.

And since

A = 4\pi r^2

and

A_0 = 4 \pi r_0^2,

\dfrac{A_0}{A} = \dfrac{4 \pi r_0^2}{4 \pi r^2}  = \dfrac{r_0^2}{r^2}

meaning

\dfrac{I}{I_0} = \dfrac{r_0^2}{r^2}.

Putting this into equation (1), we get:

\boxed{ (2).\: \: \beta = 10log(\dfrac{r_0^2}{r^2})}

Now, if the intensity is 100 dB when the distance is 5 meters, we have:

100dB=10 log(\dfrac{r_02}{(5m)^2})

10= log(\dfrac{r_0^2}{25})

by taking both sides to the exponent:  

10^{10}= \dfrac{r_0^2}{25}

r^2 = 25 *10^{10}\\r = 5 *10^5

Now equation (2) becomes

\beta = 10log(\dfrac{25*10^{10}}{r^2})

when the intensity level is 85 dB we have

85 = 10log(\dfrac{25*10^{10}}{r^2})

8.5 = log(\dfrac{25*10^{10}}{r^2})

take both sides to exponents and we get:

10^{8.5} =10^{ log(\dfrac{25*10^{10}}{r^2})}

10^{8.5} =\dfrac{25*10^{10}}{r^2}

r^2 = \dfrac{25*10^{10}}{10^{8.5}}

\boxed{r = 28.11m}

Thus, 28.11m far from the speaker the intensity drops to 85 dB.

7 0
3 years ago
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