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ValentinkaMS [17]
3 years ago
6

To improve your cardiovascular fitness:

Physics
2 answers:
RoseWind [281]3 years ago
4 0
C.Perform push ups and curl-ups daily
matrenka [14]3 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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A 75.0-kg man standing on a bathroom scale in an elevator. Calculate the scale in N, reading if the elevator moves upward at a c
Salsk061 [2.6K]

The scale in N, reading if the elevator moves upward at a constant speed of 1.5 m/s^2 is 862.5 N.

weight of man = 75kg

speed of elevator, a = 1.5 m/ s^{2}

F - w = ma \\

F = w + ma

F = m ( a +g )

F = 75 ( 1.5 + 10 ) \\

F = 75 ( 11.5 )

F = 862.5 N

So, the scale reading in the elevator is greater than his 862.5 N weight. This indicates that the person is being propelled upward by the scale, which it must do in order to do so, with a force larger than his weight. According to what you experience in quickly accelerating or slowly moving elevators, it is obvious that the faster the elevator acceleration, the greater the scale reading.

Speed can be defines as the pace at which the position of an object changes in any direction.  Since speed simply has a direction and no magnitude, it is a scalar quantity.

Learn more about speed here:-

brainly.com/question/19127881

#SPJ4

8 0
2 years ago
Two masses —m1 and m2— are connected by light cables to the perimeters of two cylinders of radii r1 and r2 respectively, as show
Aleksandr [31]

Answer:

Part a)

Mass of m2 is given as

m_2 = \frac{20}{3} kg

Part b)

Angular acceleration is given as

\alpha = 1.96 rad/s^2

Part c)

Tension in the rope is given as

T = 176.6 N

Explanation:

Part a)

When m1 and m2 both connected to the cylinder then the system is at rest

so we can use torque balance here

m_1g r_1 = m_2 g r_2

20 g(0.5) = m_2 g(1.5)

10 = 1.5 m_2

m_2 = \frac{20}{3} kg

Part b)

When block m_2 is removed then system becomes unstable

so force equation of mass m1

m_1g - T = m_1 a

also we have

T r_1 = I\alpha

now we have

m_1g = \frac{I a}{r_1^2} + m_1 a

a = \frac{m_1g}{\frac{I}{r_1^2} + m_1}

a = \frac{20 (9.81)}{\frac{45}{0.5^2} + 20}

a = 0.981 m/s^2

so angular acceleration is given as

\alpha = \frac{a}{r_1}

\alpha = \frac{0.981}{0.5}

\alpha = 1.96 rad/s^2

Part c)

Tension in the rope is given as

T = \frac{I\alpha}{r_1}

T = \frac{45 (1.96)}{0.5}

T = 176.6 N

7 0
3 years ago
A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotat
Natali [406]

centripetal acceleration is given by formula

a_c = \omega^2*R

given that

a_c = 34.1 m/s^2

R  =  5.91 m

now we have

\omega^2 R = 34.1

\omega^2 * 5.91 = 34.1

\omega^2 = 5.77

\omega = 2.4 rad/s

so the ratationa frequency is given by

\omega = 2 \pi f

2.4 = 2 \pi f

f = \frac{2.4}{2\pi}

f = 0.38 Hz

7 0
3 years ago
A 100 kg bungee jumper leaps from a bridge. The bungee cord has an un-streched equilibrium length of 10 m, and a spring constant
nika2105 [10]

Answer:

11.78meters

Explanation:

Given data

Mass m = 100kg

Length of cord= 10m

Spring constant k= 35N/m

At the greatest vertical distance, the spring potential energy is equal to the gravitational potential energy

That is

Us=Ug

Us= 1/2kx^2

Ug= mgh

1/2kx^2= mgh

0.5*35*10^2= 100*9.81*h

0.5*35*100=981h

1750=981h

h= 1750/981

h= 1.78

Hence the bungee jumper will reach 1.78+10= 11.78meters below the surface of the bridge

6 0
3 years ago
Read 2 more answers
If you were to divide the presentday universe into cubes whose sides are 10 million lightyears long, each cube would contain, on
vovikov84 [41]

Answer:

Explanation:

density of galaxies would be \frac{1}{0.27^3} times higher which is equal to 50.81.

It means  in a cube that today contains one galaxy the size of the Milky Way, we would instead find 50.81 galaxies this size.

You can round this off to 52

8 0
3 years ago
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