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2 years ago

To improve your cardiovascular fitness:

Physics

2 answers:

RoseWind [281]2 years ago

4 0

C.Perform push ups and curl-ups daily

matrenka [14]2 years ago

4 0

Answer:

A. Include a minimum of 20-30 minutes of aerobic activity 3-4 times per week

Explanation:

Your cardiovascular fitness, also called cardiovascular fitness, says a lot about your health and how much other conditions can affect it. Basically, cardiovascular fitness measures how well your body absorbs oxygen and distributes it to your muscles and organs during prolonged periods of exercise. The best way to improve your cardiovascular fitness is through exercise that forces and stimulates the heart to work harder. Aerobic exercise is the best alternative for this.

For this reason, we can say that the best answer to your question is the letter A.

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Two samples of water are mixed together.

horrorfan [7]

Let the cold water go up x degrees.

Let the hot water go down 100 - x degrees.

The formula for heat exchange is m*c*delta t

Givens

Ice

deltat = x

m = 0.50 kg

c = 4.18

Hot water

deltat = 100 - x

m = 1.5 kg

c = 4.18

Formula

The heat up = heat down

0.50 * c * x = 1.5 * c * (100 - x) Divide both sides by c

Solution

0.50 *x = 1.5*(100 - x) Remove the brackets.

0.5x = 150 - 1.5x Add 1.5x to both sides.

0.5x + 1.5x = 150 - 1.5x + 1.5x Combine like terms

2x = 150 Divide by 2

x = 75

Answer

6 0

3 years ago

Astronomers have observed a small, massive object at the center of our Milky Way Galaxy. A ring of material orbits this massive

Setler [38]

Answer:

The mass of the massive object at the center of the Milky Way galaxy is $3.44\times10^{37}\ Kg$

Explanation:

Given that,

Diameter = 10 light year

Orbital speed = 180 km/s

Suppose determine the mass of the massive object at the center of the Milky Way galaxy.

Take the distance of one light year to be 9.461×10¹⁵ m. I was able to get this it is 4.26×10³⁷ kg.

We need to calculate the radius of the orbit

Using formula of radius

$r=\dfrac{d}{2}$

$r=\dfrac{15\times9.461\times10^{15}}{2}$

$r=7.09\times10^{16}\ m$

We need to calculate the mass of the massive object at the center of the Milky Way galaxy

Using formula of mass

$M=\dfrac{v^2r}{G}$

Put the value into the formula

$M=\dfrac{(180\times10^3)^2\times7.09\times10^{16}}{6.67\times10^{-11}}$

$M=3.44\times10^{37}\ Kg$

Hence, The mass of the massive object at the center of the Milky Way galaxy is $3.44\times10^{37}\ Kg$

5 0

2 years ago

A ball initially at rest rolls down a hill with an acceleration of 3.3 m/s2. If it accelerates for 7.5 s, how far will it move?

goldfiish [28.3K]

<u>Answer:</u>

Ball will move 92.8125 meter along the cliff in 7.5 seconds.

<u>Explanation:</u>

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this case initial velocity = 0 m/s, acceleration = 3.3 $m/s^2$ , we need to calculate displacement when time = 7.5 seconds.

Substituting

$s=0*7.5+\frac{1}{2} *3.3*7.5^2\\ \\ =92.8125 meter$

So ball will move 92.8125 meter along the cliff in 7.5 seconds.

5 0

3 years ago

Two cars are initially separated by 2500 m and traveling towards each other. One car travels at 4.5 m/s and the second car trave

mylen [45]

Answer:

714.285s

Explanation:

use relative velocity

8-4.5 = 3.5m/s

x = 2500m

2500/3.5 = 714.285s = 700s (with sig figs)

7 0

2 years ago

Help!!!!

BARSIC [14]

Can we see the diagram? Thanks.

6 0

2 years ago