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4 years ago

How does exercise affect your heart rate??

2 answers:

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

Svetach [21]

I think the answer is a that what i think

4 0

3 years ago

Read 2 more answers

How much potential energy foes a 5kg mass have when its 2 meters above the ground?(Hint :PE=m*g*h)

frez [133]

Answer:

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

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
Frictional Force≈ 20.5
Accel. ≈ 58.5

These are rounded

6 0

2 years ago

The long handle on a rake is a _____.

artcher [175]

From what i know it is c. it is a lever

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