All categories

3 years ago

Why do stretching exercises increase flexibility more than cardio exercises?

1 answer:

5 0

Stretching exercises increase flexibility more than cardio exercises because stretching exercises stretches your muscles in around your body while cardio exercises only work out the muscles of your cardiovascular system. It will help you in endurance but not so much in flexibility.

You might be interested in

A dragster runs the quarter mile in 8.96 s. What is the car's velocity (in ft/s) at the finish line?

lara [203]

The first thing you should know to answer this question is the following conversion:
1mi = 5280feet
We have then that the speed is:
v = ((1/4) * (5280)) / (8.96)
v = 147.32 feet / s
Answer:
the car's velocity (in ft / s) at the finish line is 147.32 feet / s

7 0

3 years ago

Based on this passage, what is campylobacter?

alexdok [17]

I believe the correct answer from the choices listed above is option 4. Base from the passage given above, it is very clear that the Campylobacter is a bacteria used to test in identifying drug-resistant bacteria. Hope this answers the question. Have a nice day.

4 0

3 years ago

Read 2 more answers

A satellite that goes around the earth once every 24 hours iscalled a geosynchronous satellite. If a geosynchronoussatellite is

lesantik [10]

Answer:

35870474.30504 m

Explanation:

r = Distance from the surface

T = Time period = 24 h

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

m = Mass of the Earth = 5.98 × 10²⁴ kg

Radius of Earth = $6.38\times 10^6\ m$

The gravitational force will balance the centripetal force

$\dfrac{GMm}{R^2}=m\dfrac{v^2}{R}\\\Rightarrow v=\sqrt{\dfrac{GM}{R}}$

$T=\dfrac{2\pi r}{v}\\\Rightarrow T=\dfrac{2\pi r}{\sqrt{\dfrac{GM}{r}}}$

From Kepler's law we have relation

$T^2=\dfrac{4\pi^2r^3}{GM}\\\Rightarrow r^3=\dfrac{T^2GM}{4\pi^2}\\\Rightarrow r=\left(\dfrac{(24\times 3600)^2\times 6.67\times 10^{-11}\times 5.98\times 10^{24}}{4\pi^2}\right)^{\dfrac{1}{3}}\\\Rightarrow r=42250474.30504\ m$

Distance from the center of the Earth would be

$42250474.30504-6.38\times 10^6=\mathbf{35870474.30504\ m}$

8 0

3 years ago

1. (30 pts) Let x(t) = cos(πt/2) be a continuous-time signal,

posledela

Given that the function of the wave is f(x) = cos(π•t/2), we have;

a. The graph of the function is attached

b. 4 units of time

c. Even

d. 4.935 J/kg

e. 1.234 W/kg

<h3>How can the factors of the wave be found?</h3>

a. Please find attached the graph of the signal created with GeoGebra

b. The period of the signal, T = 2•π/(π/2) = 4

c. The signal is even, given that it is symmetrical about the y-axis

d. The energy of the signal is given by the formula;

$\frac{1}{2} \cdot \mu^{2} \cdot \omega ^{2} \cdot \: {a}^{2} \times \lambda$

Which gives;

E = 0.5 × 1.571² × 1² × 4 = 4.935 J/kg

e. The power of the wave is given by the formula;

E = 0.5 × 1.571² × 1² × 4 × 0.25 = 1.234 W/kg

Learn more about waves here:

brainly.com/question/14015797

8 0

2 years ago

A football is place kicked with a velocity having a vertical component of 12 m/s and a horizontal component of 6 m/s. Find the r

SSSSS [86.1K]

The velocity is given by:

V = √(Vx²+Vy²)

V = velocity, Vx = horizontal velocity, Vy = vertical velocity

Given values:

Vx = 6m/s, Vy = 12m/s

Plug in and solve for V:

V = √(6²+12²)

V = 13.42m/s

Now find the direction:

θ = tan⁻¹(Vy/Vx)

θ = angle of velocity off horizontal, Vy = vertical velocity, Vx = horizontal velocity

Given values:

Vx = 6m/s, Vy = 12m/s

Plug in and solve for θ:

θ = tan⁻¹(12/6)

θ = 63.4°

The resultant velocity is 13.42m/s at an angle of 63.4° off the horizontal.

6 0

3 years ago