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3 years ago

What is the third basic component of an exercise program that includes cardio and flexibility?

Physics

2 answers:

marusya05 [52]3 years ago

3 0

Answer:

je3ui3ndiewncxihebcrebcrebdhcbrhdbcvihrbde

Explanation:

Tresset [83]3 years ago

3 0

Answer:

muscle development

Explanation:

Im a gym trainer

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Which of the following are likely to form a covalent bond? 1.two gold atom 2. two neon atoms 3. hydrgen and oxygen atoms 4. magn

Gwar [14]

Number 3.hydrogen and oxygen atoms because covalent bonds have both hydrogen and oxygen in them

4 0

3 years ago

What is the gravitational force that two objects would feel if they are 3.5 meters apart? Object 1 has a mass of 10x10^5 kg and

polet [3.4K]

Answer:

1.63 N

Explanation:

F = GMm/r^2

= (6.67x10^-11)(10x10^5)(3x10^5) / 3.5^2

= 1.63 N ( 3 sig. fig.)

6 0

3 years ago

A Parachutist with a camera, with descending at a speed of 12.5m/s, releases, the camera at an altitude of 64.3m. What is the ma

jonny [76]

Given :

Initial velocity, u = 12.5 m/s.

Height of camera, h = 64.3 m.

Acceleration due to gravity, g = 9.8 m/s².

To Find :

How long does it take the camera to reach the ground.

Solution :

By equation of motion :

$h = ut+\dfrac{gt^2}{2}$

Putting all given values, we get :

$12.5t+\dfrac{9.8t^2}{2}=64.3\\\\4.9t^2+12.5t=64.3$

t = 2.56 and t = −5.116.

Since, time cannot be negative.

t = 2.56 s.

Therefore, time taken is 2.56 s.

Hence, this is the required solution.

7 0

3 years ago

A stone is thrown horizontally from 2.4m above the ground at 35m/s. The wall is 14m away and 1m high.At what height the stone wi

KIM [24]

The stone reaches the wall at a height of 1.62 m.

The stone lands at a point 24.5 m from the point of projection.

The stone is projected horizontally with a velocity u at a height h from the ground. The wall is located at a distance x from the point of projection. The stone takes a time t to reach the wall and in the same time the stone falls a vertical distance y.

The horizontal distance x is traveled with a constant velocity u.

$x=ut$

Calculate the time taken t.

$t=\frac{x}{u} \\ =\frac{14m}{35 m/s} \\ =0.40s$

The stone's initial vertical velocity is zero. It falls through a distance y in the time t under the action of acceleration due to gravity g.

$y=\frac{1}{2} gt^2\\ \frac{1}{2} (9.81m/s^2)(0.40s)^2\\ =0.784m$

The height h₁ of the stone above the ground when it reaches the wall is given by,

$h_1=h-y\\ =(2.4m)-(0.784m)\\ =1.616m=1.62m$

When the stone reaches the wall, its height from the ground is 1.62m.

The stone thus crosses over the wall, since the height of the wall is 1 m. It reaches the ground at a distance R from the point of projection. If the time taken by the stone to reach the ground is t₁, then,

$h=\frac{1}{2} gt_1^2$