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3 years ago

During aerobic activity, if your heart rate is lower than the lower limit, you are ___________________.

1 answer:

soldier1979 [14.2K]3 years ago

4 0

Not pushing yourself hard enough is the answer since your heart rate doesn't even hit your lower minimum.
Pushing yourself to the limit is at your max heart rate.
Just at the right spot is at your max heart rate.
Pushing yourself too hard is above your max heart rate.

You might be interested in

A 60 year old person has a threshold of hearing of 95.0 dB for a sound with frequency f=10,000 Hz. By what factor must the inten

Nadusha1986 [10]

Answer:

The intensity increased by a factor of 158489

Explanation:

Given that,

Sound level = 95.0 dB

Sound level = 43.0 dB

Frequency = 10000 Hz

We need to calculate the ratio of sound intensity

Using formula of sound level

$sound\ level =10 log\dfrac{I}{I_{0}}$

Put the value into the formula

$95.0=10\ log\dfrac{I_{1}}{I_{0}}$ ...(I)

$43.0=10\ log\dfrac{I_{2}}{I_{0}}$ .....(II)

Subtracting these equations

$52.0=10\ log\dfrac{I_{1}}{I_{2}}$

$\log\dfrac{I_{1}}{I_{2}}=5.2$

Taking inverse log

$\dfrac{I_{1}}{I_{2}}=10^{5.2}$

$\dfrac{I_{1}}{I_{2}}=158489$

Hence, The intensity increased by a factor of 158489

4 0

2 years ago

Which of the following is not a natural resource? a. time b. water c. land d. air Please select the best answer from the choices

AleksAgata [21]

Answer: Option (a) is the correct answer.

Explanation:

Resources which are created by human beings are known as man-made resources.

For example, glass, rayon, nylon etc are all man-made resources.

Whereas resources which are naturally created are known as natural resources.

For example, wind, air, water etc are all natural resources.

Thus, we can conclude that out of the given options time is not a natural resource.

3 0

3 years ago

An unlabeled hierarchical diagram of various astronomical bodies is shown below. The labels A, B, C, and D can be used to repres

Paha777 [63]

A) The biggest astronomical object is the Universe, which contains billions of galaxies among which there is the Milky Way.
The Milky Way contains thousands of planetary systems, among which the Solar System.
The Solar System contains many <span>planets <span>(but only one star, the Sun)</span>,</span> among which there is Earth.
Therefore you can label:
A = Universe, B = Milky Way, C = Solar system, D = Earth

b) Given what we said before, you could label D also any other planet in the Solar System, therefore you can choose among Mercury, Venus, Mars, Jupiter, Saturn, Uranus, and Neptune.

8 0

3 years ago

Newtons first law stateS that object will move With a constant velocity if nothing acts on it. Does our every day experience con

bazaltina [42]

Answer:

Explanation:

4 0

2 years ago

FIGURE 2 shows a 1.5 kg block is hung by a light string which is wound around a smooth pulley of radius 20 cm. The moment of ine

Sindrei [870]

Answer:

At t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

Explanation:

First, we consider all the forces acting on the pulley.

There is only one force acting on the pulley, and that is due to the 1.5 kg mass attached to it.

Therefore, the torque on the pulley is

$\tau=Fd=mg\cdot R$

where m is the mass of the block, g is the acceleration due to gravity, and R is the radius of the pulley.

Now we also know that the torque is related to angular acceleration α by

$\tau=I\alpha$

therefore, equating this to the above equation gives

$mg\cdot R=I\alpha$

solving for alpha gives

$\alpha=\frac{mgR}{I}$

Now putting in m = 1.5 kg, g = 9.8 m/s^2, R = 20 cm = 0.20 m, and I = 2 kg m^2 gives

$\alpha=\frac{1.5\cdot9.8\cdot0.20}{2}$ $\boxed{\alpha=1.47s^{-2}}$

Now that we have the value of the angular acceleration in hand, we can use the kinematics equations for the rotational motion to find the angular velocity and the number of revolutions at t = 4.2 s.

The first kinematic equation we use is

$\theta=\theta_0+\omega_0t+\frac{1}{2}\alpha t^2$

since the pulley starts from rest ω0 = 0 and theta = 0; therefore, we have

$\theta=\frac{1}{2}\alpha t^2$

Therefore, ar t = 4.2 s, the above gives

$\theta=\frac{1}{2}(1.47)(4.2)^2$

$\boxed{\theta=12.97}$

So how many revolutions is this?

To find out we just divide by 2 pi:

$\#\text{rev}=\frac{\theta}{2\pi}=\frac{12.97}{2\pi}$ $\boxed{\#\text{rev}=2.06}$

Or about 2 revolutions.

Now to find the angular velocity at t = 4.2 s, we use another rotational kinematics equation:

$\omega^2=w^2_0+2\alpha(\Delta\theta)_{}$

Since the pulley starts from rest, ω0 = 0. The change in angle Δθ we calculated above is 12.97. The value of alpha we already know to be 1.47; therefore, the above becomes:

$\omega^2=0+2(1.47)(12.97)$ $w^2=38.12$ $\boxed{\omega=6.17.}$

Hence, the angular velocity at t = 4.2 w is 6. 17 rad / s

To summerise:

at t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

3 0

1 year ago