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Olin [163]
3 years ago
5

Raul wants to increase his upper body and cardiovascular strength. He plans to go to the gym and use his backyard swimming pool

to exercise and attain his goals. He plans to lift weights and do calisthenics on Mondays, Wednesdays and Fridays, and swim laps in the pool on Tuesdays, Thursdays and Saturdays with a rest on Sunday. In a paragraph below, explain how Raul can maximize his training by applying the seven principles of exercise to his workout routine. Be sure to mention all seven principles and circle or highlight them in your paragraph.
Physics
1 answer:
Mkey [24]3 years ago
4 0

Answer: An exercise, activity, or workout that targets a specific training need Progression - Gradually increasing your skill level and ability over time Overload -Working your body harder than usual to improve how it functions Reversibility - Losing what you have gained by taking time off Tedium - getting bored by doing the same thing

Explanation:

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3 years ago
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As it travels through a crystal, a light wave is described by the function E(x,t)=Acos[(1.57×107)x−(2.93×1015)t]. In this expres
Drupady [299]

Answer:

Speed, v=1.86\times 10^8\ m/s

Explanation:

It is given that,

A light wave is described by the following function as :

E(x,t)=A\ cos[(1.57\times 10^7)x-(2.93\times 10^{15})t].....(1)

The general equation of wave is given by :

E=Acos(kx-\omega t)........(2)

On comparing equation (1) and (2)

k=(1.57\times 10^7)

\dfrac{2\pi}{\lambda}=(1.57\times 10^7)

\lambda=\dfrac{2\pi}{(1.57\times 10^7)}

Wavelength, \lambda=4.002\times 10^{-7}\ m

\omega=(2.93\times 10^{15})

\dfrac{2\pi}{T}=(2.93\times 10^{15})

\dfrac{1}{T}=\dfrac{(2.93\times 10^{15})}{2\pi}

Frequency, f=4.66\times 10^{14}\ Hz

Let v is the speed of the light wave. It is given by :

v=f\times \lambda

v=4.66\times 10^{14}\ Hz\times 4.002\times 10^{-7}\ m

v=1.86\times 10^8\ m/s

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7 0
3 years ago
A plane leaves the airport in Galisteo and flies 170 km at 68.0° east of north; then it changes direction to fly 230 km at 36.0°
luda_lava [24]

Answer:

The direction will be 84.86^\circ and the distance 250.75km.

Explanation:

Let's say A is the displacement vector which represents the first 170km and B the one for the next 230km. Then the components of these vector will be:

A_x=170cos(68^{\circ})\\ A_y=170sin(68^\circ)\\\\B_x=230cos(-36^\circ)\\B_y=230sin(-36^\circ)

The vector which point from the origin to the final position of the plane will be R=A+B. We sum components on <em>x </em>and <em>y </em>independetly (vector property):

R_x=A_x+B_x=170cos(68^{\circ})+230cos(-36^\circ)=63.68km+186.07km=249.75km

R_y=A_y+B_y=170sin(68^\circ)+230sin(-36^\circ)=157.62km-135.19km=22.43km

If \theta is the direction of R then:

tan(\theta )=\frac{R_x}{R_y} ⇒ \theta = arctan(\frac{R_x}{R_y}) ⇒ \theta = 84.86^\circ.

The distance will be given by the magnitud of the vector R:

R=\sqrt{R_x^2 + R_y^2} ⇒ R=\sqrt{R_x^2 + R_y^2} = 250.75.

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Answer:

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Explanation:

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