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4 years ago

A seated cable row is an example of which level of training in the NASM OPT model?

Physics

1 answer:

Vera_Pavlovna [14]4 years ago

8 0

Answer:

B. Strength

Explanation:

The OPT Model, or Optimum Performance Training Model, is a "fitness training system created by the NASM. The OPT Model is contructed with scientific evidence and principles that progresses an individual through five training phases: stabilization endurance, strength endurance, hypertrophy, maximal strength and power".

On the stabilization level we have the phase 1 called stabilization endurance.

For the level strength part we have 3 phases . Phase 2: Strength endurance , Phase 3: Hypertrophy, Phase 4: Maximal strength. And we can consider the case "A seated cable row" on this the level strength since we need to have some abilities to do this but not enough to stay on the power level since this one is the advancd level.

For the power level we have the last phase called power in order to mantain and conduct high training level programs.

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After the change of state occurs, the temperature _____________.

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4 years ago

A 77 turn, 10.0 cm diameter coil rotates at an angular velocity of 8.00 rad/s in a 1.18 T field, starting with the normal of the

Tasya [4]

Answer:

a) fem = 5.709 V, b) t = 0.196 s, c) t = 0.589 s, d) T = 0.785 s

Explanation:

This is an exercise in Faraday's law

fem= - N $\frac{d \Phi _B}{dt}$

fem = - N $\frac{d \ (B A cos \theta)}{dt}$

The magnetic field and the area are constant

fem = - N B A $\frac{d \ cos \ \theta}{dt}$

fem = - N B A (-sin θ) $\frac{d \theta}{dt}$

fem = N B (π d² / 4) sin θ w

fem= $\frac{\pi }{4}$ N B d² w sin θ

with this expression we can correspond the questions

a) the peak of the electromotive force

this hen the sine of the angle is 1

sin θ = 1

fem = $\frac{\pi }{4}$ 77 1.18 0.10² 8.0

fem = 5.709 V

b) as the system has a constant angular velocity, we can use the angular kinematics relations

θ = w₀ t

t = θ/w₀

Recall that the angles are in radians, so the angle for the maximum of the sine is

θ= π/2

t = $\frac{\pi }{2} \ \frac{1}{8}$

t = 0.196 s

c) for the electromotive force to be negative, the sine function of being

sin θ= -1

whereby

θ = 3π/ 2

t = $\frac{3\pi }{2} \ \frac{1}{8}$

t = 0.589 s

d) This electromotive force has values that change sinusoidally with an angular velocity of

w = 8 rad / s

angular velocity and period are related

w = 2π / T

T = 2π / w

T = 2π / 8

T = 0.785 s

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3 years ago

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4 years ago

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The moons of Mars, Phobos (Fear) and Deimos (Terror), are very close to the planet compared to Earth's Moon. Their orbital radii

tankabanditka [31]

Answer:

$0.2528$

Explanation:

To calculate the period we need the formula:

$T=\frac{2\pi r^{3/2}}{\sqrt{GM}}$

Where $r$ is the radius of the moon, $G$ is the universal constant of gravitation and $M$ is the mass of mars.

The period of Phobos:

$T_{p}=\frac{2\pi r_{p}^{3/2}}{\sqrt{GM}}$

The period of Deimos:

$T_{D}=\frac{2\pi r_{D}^{3/2}}{\sqrt{GM}}$

The ratio of the period of Phobos and Deimos:

$\frac{T_{p}}{T_{D}}=\frac{\frac{2\pi r_{p}^{3/2}}{\sqrt{GM}}}{\frac{2\pi r_{D}^{3/2}}{\sqrt{GM}}}$

$\frac{T_{p}}{T_{D}}=\frac{\sqrt{GM}2\pi r_{p}^{3/2}}{\sqrt{GM}2\pi r_{D}^{3/2}}$

Most terms get canceled and we have:

$\frac{T_{p}}{T_{D}}=\frac{r_{p}^{3/2}}{r_{D}^{3/2}}$

According to the problem

$r_{p}=9,378km\\r_{D}=23,459km$

so the ratio will be:

$\frac{T_{p}}{T_{D}}=\frac{(9,378)^{3/2}}{(23,459)^{3/2}}=\frac{908166.22}{3593058.125}=0.25275$ ≈ $0.2528$

the ratio of the period of revolution of Phobos to that of Deimos is 0.2528

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3 years ago

A ray of light strikes a smooth surface and is reflected. The angle of incidence is 35°. What can be predicted about the angle o

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4 years ago

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