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3 years ago

An exercise program that lacks specific goals is failing what fitness principle?

Physics

2 answers:

rusak2 [61]3 years ago

8 0

An exercise program that lacks specific goals is failing into the specificity principle of fitness.

Further Explanation:

The fitness program helps in achieving good health. To start the fitness program we have to make fitness goals, create a healthy diet. Physical fitness is the capability of a body to perform daily activity.

Specificity is occurring in a specific reaction that occurs as a result of the exercise program. Specificity increases the long-term adaptations to occur with progressive overload. Specificity—a focus on particular movements, muscles, velocity and energy systems —drives the gains you make in training and practice. If an individual doesn't intentionally train for something, it may not occur.

The specificity principle states that these physiologic or metabolic changes are specific to the cardiorespiratory, neurologic and muscular responses that are entailed by the exercise activity. SAID (Specific Adaptations to Imposed Demands) is a common term of specificity principle. Physical fitness helps us to prevent heart disease, lung cancer, stroke, and type II diabetes. Progress of a fitness program is very slow it takes time. It is more difficult to maintain a program for a long time with the same enthusiasm. We can set the goal and check whether we are progressing or not.

Learn more:

Learn more about carbohydrate monomer <u>brainly.com/question/6947177 </u>
Learn more about core muscle stabilization <u>brainly.com/question/1231927 </u>
Learn more about energy storage <u>brainly.com/question/523624 </u>

Answer Details:

Grade: High School

Subject: Health

Chapter: Fitness Program

Keywords:

Physical fitness, healthy diet, heart disease, cancer, fitness program, goal, specificity, evaluation, fitness program, adaptation, cancer, stroke, response.

lbvjy [14]3 years ago

6 0

The answer is progression

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Answer:

hope this helps!

Explanation:

Volume of the air bubble, V1=1.0cm3=1.0×10−6m3

Bubble rises to height, d=40m

Temperature at a depth of 40 m, T1=12oC=285K

Temperature at the surface of the lake, T2=35oC=308K

The pressure on the surface of the lake: P2=1atm=1×1.103×105Pa

The pressure at the depth of 40 m: P1=1atm+dρg

Where,

ρ is the density of water =103kg/m3

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8 0

2 years ago

A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Co

Tems11 [23]

(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

<h3>Center mass of the airplane</h3>

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>

The wheels under the wind do not pass through the center line.
The position of the front wheel is constant and it is zero mark (origin).
The rear wheels are at 21.7 m mark

Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

<h3>Mass of the plane at the position of the rear wheels</h3>

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

$X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}$

$18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg$

<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

$W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN$

<h3>Mass of the plane at the position of the front wheel</h3>

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

<h3>Force exerted by the ground on the front wheel</h3>

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

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7 0

2 years ago

An object travels with a constant speed in a circular path. The net force on the object is

Pepsi [2]

Answer:

toward the center

Explanation:

Before answering, let's remind the first two Newton Laws:

1) An object at rest tends to stay at rest and an object moving at constant velocity tends to continue its motion at constant velocity, unless acted upon a net force

2) An object acted upon a net force F experiences an acceleration a according to the equation

$F=ma$

where m is the mass of the object.

In this problem, we have an object travelling at constant speed in a circular path. The fact that the trajectory of the object is circular means that the direction of motion of the object is constantly changing: this means that its velocity is changing, so it has an acceleration. And therefore, a net force is acting on it. The force that keeps the object travelling in the circular path is called centripetal force, and it is directed towards the center of the circle (because it prevents the object from continuing its motion straight away).

So, the correct answer is

toward the center

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3 years ago

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