Well-designed weight-training programs only target two or three body areas.a. True
b. False

Answers

Answer 1

Answer: False, well designed weight training programs actually target most of the muscles in the body. A good weight training program includes many compound exercises to activate multiple muscle groups and promote muscle hypertrophy. Some isolation exercises may also be included to target a specific muscle and help it grow.

Answer 2

Answer:

Well-designed weight-training programs do not only target two or three body areas.

The statement that well-designed weight-training programs only target two or three body areas is FALSE.

Well-designed weight-training programs aim to target different muscle groups in the body to promote overall strength and balanced development. These programs often include exercises that involve the chest, back, shoulders, arms, legs, and core.

For example, a weight-training program might include exercises like bench press (chest), rows (back), shoulder press (shoulders), bicep curls (arms), squats (legs), and planks (core).

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The ___ belt contains perhaps ten of thousands of objects orbiting within about 100 AU of the sun

Answers

That would probably be the asteroid belt as it is probably the only existing belt in our Solar System. And it fits the description of having objects orbiting as there are a lot of particles that can be found there.

Answer:

Kuiper

Explanation:

The region beyond the orbit of Neptune is known as Kuiper belt which contains ten of thousands of objects orbiting sun in long elliptical paths within 100 AU of the sun. Most of the objects in the Kuiper belt are composed of frozen gases. Comets which are icy bodies are found in the Kuiper belt. Kuiper belt is also home to dwarf planets like Pluto, Makemake and Haumea.

2. What happens when multiple forces act on an object?A. The object will always move.
B. The object will move if the forces are balanced.
C. The object will move if the forces are unbalanced.
D. The object will always stay in place.
3. An object moves in the direction of the __________ force.
A. net
B. balanced
C. strongest
D. weakest

Answers

2) The correct answer is option C i.e. The object will move if the forces are unbalanced. The objects move only on the application of unbalanced forces. 3. The correct answer is option A. i.e. net An object moves in the direction of the net force. The object moves in the direction of resultant or net force.

Answer:

C. The object will move if the forces are unbalanced.

A. net

Explanation:

just took the test swear to god that the answer

Your vessel's position should be plotted using bearings of____.

Answers

Your vessel's position should be plotted using bearings of fixed known objects on shore.

If the pressure in a gas container that is connected to an open-end U-tube manometer is 116 kPa and the pressure of the atmosphere at the open end of the tube is 752 mm Hg, the level of mercury in the tube will a. be 870 mm higher in the arm open to the atmosphere.
b. be 870 mm higher in the arm connected to the gas cylinder.
c. be 118 mm higher in the arm connected to the gas cylinder.
d. be 118 mm higher in the arm open to the atmosphere.

Answers

P = absolute pressure in gas container connected to open-end U-tube = 116 kPa = 116000 Pa

we know that, 1 atm = 101325 Pa and 1 atm = 760 mm of Hg , hence

P = 116000 Pa (1 atm/101325 Pa) (760 mm of Hg/1 atm)

P = (116000 x 760/101325) mm of Hg

P = 870.07 mm of Hg

P₀ = atmospheric pressure = 752 mm Hg

Level of mercury is given as

P' = P - P₀

inserting the values

P' = 870.07 - 752

P' = 118 mm

so correct choice is

d. be 118 mm higher in the arm open to the atmosphere.

Final answer:

The level of mercury in the tube would be 118 mm higher in the arm connected to the gas cylinder, as determined by converting kPa to mm Hg and calculating the pressure difference between the gas pressure and atmospheric pressure.

Explanation:

In order to solve this problem, we need to convert all units to the metric units system for consistency. First, let's convert the pressure of the atmosphere from mm Hg to kilopascals (kPa), knowing that one atmosphere which equals 760 mm Hg is approximately 101.3 kPa.

So, the atmospheric pressure is (752 mm Hg / 760 mm Hg) * 101.3 kPa ≈ 100.4 kPa.

The difference between the gas pressure and the atmospheric pressure gives the pressure of the mercury in the tube, since pressures are additive. So, the pressure difference is 116 kPa - 100.4 kPa = 15.6 kPa.

Finally, we convert this pressure in kPa back to mm Hg to get the height difference. By doing the inverse conversion, we get 15.6 kPa * (760 mm Hg / 101.3 kPa) ≈ 118 mm Hg. Therefore, the level of mercury in the tube will be 118 mm higher in the arm connected to the gas cylinder, corresponding to the option

c. be 118 mm higher in the arm connected to the gas cylinder.

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A shear force of 400 N is applied to one face of an aluminum cube with sides of 30 cm. What is the resulting relative displacement? (the shear modulus for aluminum is 2.5 X 1010N/m2).a) 4.4 X 10-8 m
b) 8.2 X 10-8 m
c) 1.9 X 10-8 m
d) 5.3 X 10-8 m