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3 years ago

Cardiovascular exercise can

Physics

2 answers:

sesenic [268]3 years ago

6 0

Increase your resting heart rate

UkoKoshka [18]3 years ago

3 0

Can <span>get your heart rate up and increases blood circulation throughout the body.</span>

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In comparison with other ocean basins, major sedimentary features such as continental rises and abyssal plains are relatively ra

Maksim231197 [3]

Answer:

Why are continental rises and abyssal plains relatively rare in the Pacific? This is because the extensive system of trenches along the active margins of the Pacific, trap much of the sediments flowing off the continents, preventing them from building the broad, flat abyssal plains typical of the Atlantic ocean basins.

3 0

3 years ago

The transverse standing wave on a string fixed at both ends is vibrating at its fundamental frequency of 250 Hz. What would be t

hodyreva [135]

Answer:

Explanation:

fundamental frequency, f = 250 Hz

Let T be the tension in the string and length of the string is l ans m be the mass of the string initially.

the formula for the frequency is given by

$f=\frac{1}{2l}\sqrt{\frac{Tl}{m}}$ .... (1)

Now the length is doubled ans the tension is four times but the mass remains same.

let the frequency is f'

$f'=\frac{1}{2\times 2l}\sqrt{\frac{4T\times 2l}{m}}$ .... (2)

Divide equation (2) by equation (1)

f' = √2 x f

f' = 1.414 x 250

f' = 353.5 Hz

7 0

3 years ago

The greater the of an object the more force is needed to cause acceleration

d1i1m1o1n [39]

the greater the <u>mass</u> of an object the more force is needed to cause acceleration

8 0

3 years ago

A frog leaps up from the ground and lands on a step 0.1 m above the ground 2 s later. We want to find the

mash [69]

Answer:

$\Delta x = v_0 t + \frac{1}{2}at^2$

Explanation:

To solve this problem, we can use the following suvat equation:

$\Delta x = v_0 t + \frac{1}{2}at^2$

where

$\Delta x$ is the vertical displacement of the frog

$v_0$ is the initial vertical velocity

t is the time

a is the acceleration

We have chosen this formula because apart from $v_0$ , all the other quantities are known. In fact:

$\Delta x =0.1 m$ is the vertical displacement

t = 2 s is the total time of flight

$a=g=-9.8 m/s^2$ is the acceleration due to gravity (negative because it is downward)

Therefore, solving for $v_0$ , we find the initial velocity of the frog:

$v_0 = \frac{\Delta x-\frac{1}{2}at^2}{t}=\frac{0.1-\frac{1}{2}(-9.8)(2)^2}{2}=9.85 m/s$

4 0

4 years ago

Consider a semicircular ring of radius R. Its linear mass density varies as lambda =lambda not sin theta. Locate its centre of m

bearhunter [10]

Answer:

(0, πR/4)

Explanation:

The linear mass density (mass per length) is λ = λ₀ sin θ.

A short segment of arc length is ds = R dθ.

The mass of this short length is:

dm = λ ds

dm = (λ₀ sin θ) (R dθ)

dm = R λ₀ sin θ dθ

The x coordinate of the center of mass is:

X = ∫ x dm / ∫ dm

X = ∫₀ᵖ (R cos θ) (R λ₀ sin θ dθ) / ∫₀ᵖ R λ₀ sin θ dθ

X = R ∫₀ᵖ sin θ cos θ dθ / ∫₀ᵖ sin θ dθ

X = R ∫₀ᵖ ½ sin 2θ dθ / ∫₀ᵖ sin θ dθ

X = ¼R ∫₀ᵖ 2 sin 2θ dθ / ∫₀ᵖ sin θ dθ

X = ¼R (-cos 2θ)|₀ᵖ / (-cos θ)|₀ᵖ

X = ¼R (-cos 2π − (-cos 0)) / (-cos π − (-cos 0))

X = ¼R (-1 + 1) / (1 + 1)

X = 0

The y coordinate of the center of mass is:

Y = ∫ y dm / ∫ dm

Y = ∫₀ᵖ (R sin θ) (R λ₀ sin θ dθ) / ∫₀ᵖ R λ₀ sin θ dθ

Y = R ∫₀ᵖ sin² θ dθ / ∫₀ᵖ sin θ dθ

Y = R ∫₀ᵖ ½ (1 − cos 2θ) dθ / ∫₀ᵖ sin θ dθ

Y = ½R ∫₀ᵖ (1 − cos 2θ) dθ / ∫₀ᵖ sin θ dθ

Y = ½R (θ − ½ sin 2θ)|₀ᵖ / (-cos θ)|₀ᵖ

Y = ½R [(π − ½ sin 2π) − (0 − ½ sin 0)] / (-cos π − (-cos 0))

Y = ½R (π − 0) / (1 + 1)

Y = ¼πR

4 0

3 years ago