All categories

3 years ago

What support system do you have in place to complete your scheduled exercises? How and why did you

Physics

1 answer:

Bingel [31]3 years ago

6 0

Answer:

It has to be important to you.

-It has to be something that you are in control of.

-It has to be attainable, or able to be achieved.

-It has to be clear and specific.

-It has to have a plan.

Wellness

Wellness refers to a healthy state of being in body, mind, and emotion.

Body wellness

means eating right, exercising, and keeping your bodily systems in good health.

Mental wellness

means feeling safe and secure in your environment and at peace with your decisions and actions.

Emotional wellness

means feeling your best and having the ability to deal with many different situations.

Wellness: Setting Goals

The first step to setting a goal is defining it. Keep in mind that not everything you wish for can be a goal. When it comes to physical activity, diet, and exercise, you may wish to run a marathon before the end of the school year. However, this goal could be difficult to achieve if school ends in just a few months and you are not already training. A more reasonable goal might be to be able to run a mile in the amount of time considered healthy for your age group.

Importance of Physical Education

Physical education explores the methods, types, and benefits of physical activity for the body. In addition to staying active, making healthy choices about diet and exercise is important to overall personal and long-term health. Long-term health passed down through generations can also promote healthy family lifestyles.

Fitness

is the ability of the body to function effectively and efficiently during physical activity and is the ability to move joints through an entire range of motion.

Physical exercise

is the act of improving one's body through specific intentional movements.

differences between exercise and fitness

while anyone can exercise, only someone who exercises regularly can be fit.

Why is it important to make healthy choices about your diet and exercise?

Choosing healthy behaviors when it comes to diet and exercise is important to personal health, long-term health, and promoting healthy family lifestyles.

five components of physical education:

-Cardiorespiratory endurance

-Muscular strength

-Muscular endurance

-Flexibility

-Body composition

Explanation:

Life is WORTH LIVING

You might be interested in

A parallel-plate capacitor is held at a potential difference of 250 V. A proton is fired toward a small hole in the negative pla

MissTica

Answer:

The speed of proton when it emerges through the hole in the positive plate is $2.05\times 10^5\ m/s$ .

Explanation:

Given that,

A parallel-plate capacitor is held at a potential difference of 250 V.

A A proton is fired toward a small hole in the negative plate with a speed of, $u=3\times 10^5\ m/s$

We need to find the speed when it emerges through the hole in the positive plate. It can be calculated using the conservation of energy as :

$qV=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2\\\\1.6\times10^{-19}\times250=\dfrac{1}{2}mv^2-\frac{1}{2}\cdot1.67\times10^{-27}\cdot(3\times10^{5})^{2}\\\\\dfrac{1}{2}mv^2=3.515\cdot10^{-17}\\\\v=\sqrt{\dfrac{3.515\cdot10^{-17}\cdot2}{1.67\times10^{-27}}}\\\\v=2.05\times 10^5\ m/s$

So, the speed of proton when it emerges through the hole in the positive plate is $2.05\times 10^5\ m/s$ .

5 0

3 years ago

Which energy does a car travelling 30 m/ph as it slows have:

Paladinen [302]

Answer:

c) kinetic energy

Explanation:

6 0

2 years ago

Read 2 more answers

liquid helium has a very low boiling point, 4.2 k, as well as a very low latent heat of vaporization, 2.00 104 j/kg. if energy i

aksik [14]

$4.80 \times 10^3 \text { seconds }$ long does it take to boil away 2.40 kg of the liquid.

Boiling point of He is $T=4.2 \mathrm{k}$

Latent heat of vapourization $L=2.00 \times 10^4 \mathrm{~J} / \mathrm{kg}$

Power of electrical heater $P=30 \mathrm{w}$

mass of liquid is $m=2.40 \mathrm{~kg}$

amount of heat required to boil

$\begin{aligned}&Q=m L \\&Q=2.40 \times 2 \times 10^4 \mathrm{~J} \\&Q=4.80 \times 10^4 \mathrm{~J}\end{aligned}$

Power $p=\frac{\text { work }}{\text { time }}=\frac{\text { Energy }}{\text { Time }}$

$\begin{aligned}P &=\frac{Q}{t} \\\text { tine } t &=\frac{Q}{P}=\frac{4.80 \times 10^4 \mathrm{~J}}{10} \\t &=4.80 \times 10^3 \text { seconds }\end{aligned}$

The heat or energy that is absorbed or released during a substance's phase shift is known as latent heat. It could go from a solid to a liquid or from a liquid to a gas, or vice versa. Enthalpy, a characteristic of heat, is connected to latent heat.

The heat that is used or lost as matter melts and transitions from a solid to a fluid form at a constant temperature is known as the latent heat of fusion.

Due to the fact that during softening the heat energy anticipated to transform the substance from solid to fluid at air pressure is the latent heat of fusion and that the temperature remains constant during the process, the "enthalpy" of fusion is a latent heat. The enthalpy change of any quantity of material during dissolution is known as the latent heat of fusion.

For learn more about Latent heat of vaporization, visit: brainly.com/question/14980744

#SPJ4

3 0

1 year ago

A ball is thrown at an angle of 38 degrees to the horizontal. What happens to the magnitude of the ball's vertical acceleration

Brut [27]

Answer:

<em>I belive it is 38 degrees</em>

<em />

Explanation:

7 0

2 years ago

A uniform disk with mass 35.2 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stati

Sergio [31]

Answer:

a) v = 1.01 m/s

b) a = 5.6 m/s²

Explanation:

If the disk is initially at rest, and it is applied a constant force tangential to the rim, we can apply the following expression (that resembles Newton's 2nd law, applying to rigid bodies instead of point masses) as follows:

$\tau = I * \alpha (1)$

Where τ is the external torque applied to the body, I is the rotational inertia of the body regarding the axis of rotation, and α is the angular acceleration as a consequence of the torque.
Since the force is applied tangentially to the rim of the disk, it's perpendicular to the radius, so the torque can be calculated simply as follows:
τ = F*r (2)
For a solid uniform disk, the rotational inertia regarding an axle passing through its center is just I = m*r²/2 (3).
Replacing (2) and (3) in (1), we can solve for α, as follows:

$\alpha = \frac{2*F}{m*r} = \frac{2*34.5N}{35.2kg*0.2m} = 9.8 rad/s2 (4)$

Since the angular acceleration is constant, we can use the following kinematic equation:

$\omega_{f}^{2} - \omega_{o}^{2} = 2*\Delta \theta * \alpha (5)$

Prior to solve it, we need to convert the angle rotated from revs to radians, as follows:

$0.2 rev*\frac{2*\pi rad}{1 rev} = 1.3 rad (6)$

Replacing (6) in (5), taking into account that ω₀ = 0 (due to the disk starts from rest), we can solve for ωf, as follows:

$\omega_{f} = \sqrt{2*\alpha *\Delta\theta} = \sqrt{2*1.3rad*9.8rad/s2} = 5.1 rad/sec (7)$

Now, we know that there exists a fixed relationship the tangential speed and the angular speed, as follows:

$v = \omega * r (8)$

where r is the radius of the circular movement. If we want to know the tangential speed of a point located on the rim of the disk, r becomes the radius of the disk, 0.200 m.
Replacing this value and (7) in (8), we get:

$v= 5.1 rad/sec* 0.2 m = 1.01 m/s (9)$

There exists a fixed relationship between the tangential and the angular acceleration in a circular movement, as follows:

$a_{t} = \alpha * r (9)$

where r is the radius of the circular movement. In this case the point is located on the rim of the disk, so r becomes the radius of the disk.
Replacing this value and (4), in (9), we get:

$a_{t} = 9.8 rad/s2 * 0.200 m = 1.96 m/s2 (10)$

Now, the resultant acceleration of a point of the rim, in magnitude, is the vector sum of the tangential acceleration and the radial acceleration.
The radial acceleration is just the centripetal acceleration, that can be expressed as follows:

$a_{c} = \omega^{2} * r (11)$

Since we are asked to get the acceleration after the disk has rotated 0.2 rev, and we have just got the value of the angular speed after rotating this same angle, we can replace (7) in (11).
Since the point is located on the rim of the disk, r becomes simply the radius of the disk,, 0.200 m.
Replacing this value and (7) in (11) we get:

$a_{c} = \omega^{2} * r = (5.1 rad/sec)^{2} * 0.200 m = 5.2 m/s2 (12)$

The magnitude of the resultant acceleration will be simply the vector sum of the tangential and the radial acceleration.
Since both are perpendicular each other, we can find the resultant acceleration applying the Pythagorean Theorem to both perpendicular components, as follows:

$a = \sqrt{a_{t} ^{2} + a_{c} ^{2} } = \sqrt{(1.96m/s2)^{2} +(5.2m/s2)^{2} } = 5.6 m/s2 (13)$

6 0

2 years ago