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Anna [14]
3 years ago
5

Which of these is not a component of physical fitness

Physics
2 answers:
kherson [118]3 years ago
8 0
Cardiovascular endurance...I believe (I'm not sure)
Nostrana [21]3 years ago
5 0
Flexibility does not involve physical fitness itself. flexibility can help with physical fitness but being flexible is not being fit.
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If vector A ⃗  has components A x and A y and makes an angle θ with the +x axis, then
ziro4ka [17]
Then the tangent of angle-Θ is (Ay / Ax).
5 0
2 years ago
Wavelength of blue photons 495 nm, what is the frequency? and what is the energy?
inessss [21]

Answer:

1.F: About 6*10^14 Hz

2.E: About 4*10^ -19 J

Explanation:

Frequency: We knew that the speed of a wave is its wavelength(λ)* frequency(f, in Hz).  By the wave-particle duality we know we can calculate the frequency of light in the same way. So, c=495nm *f, f=c/495nm=> (299,792,458 m/s) / (4.95*10^-7 m)

=6.05*10^14 /s

Energy: The energy photon contains can be calculate by this formula-- E=hf

f is the frequency and h is Planck's constant which is about 6.62 ×10^-34 *m^2*kg/s (after dimensional analysis ) =6.62*10^ -34 J*s.

So, the energy of a blue photon is (6.05*10^14)*(6.62*10^-34)=40.051*10^-20=  4.051*10^-19 J

3 0
2 years ago
What is the magnitude of the force needed to keep a 60 newton rubber block moving across level,dry asphalt in a straight line at
pantera1 [17]

let Coefficients of Friction of Rubber on asphalt (dry) =0.7

F= Coefficients of Friction * normal force = 0.7 * 60 =42 N

so the net force of the rubber is zero, meaning it will travel at a constant speed.

When the rubber is travel at 2m/s, 42N is required to keep moving at constant speed

6 0
3 years ago
Read 2 more answers
A space station, in the form of a wheel 119 m in diameter, rotates to provide an "artificial gravity" of 2.20 m/s2 for persons w
Rina8888 [55]
<span>Well, since it's in the shape of a wheel and the person walks around the edge of it, they must have a centripetal acceleration. Since a=v^2/r you can solve for "v" using 2.20 as your "a" and 59.5 as your "r" (r=half of the diameter).
</span> a=v^2/r 
 v=(a*r)^(1/2)=((2.20)*(59.5))^(1/2)=<span> <span>11.44 m/s.
</span></span><span> After you get "v," plugged that into T=2 pi r/ v. This will give you the 1rev per sec.
</span> T=2 pi r/ v= T=(2)*(pi)*(59.5)/(11.44)= <span> <span>32.68 rev/s
</span></span> Use dimensional analysis to get rev per min (1rev / # sec) times (60 sec/min). 
 (32.68 rev/s)(60 s/min)=<span> <span>1960.74 rev/min
</span></span>
5 0
3 years ago
A physicist's right eye is presbyopic (i.e., farsighted). This eye can see clearly only beyond a distance of 97 cm, which makes
Ivan

Answer:

f = 19,877 cm   and  P = 5D

Explanation:

This is a lens focal length exercise, which must be solved with the optical constructor equation

        1 / f = 1 / p + 1 / q

where f is the focal length, p is the distance to the object and q is the distance to the image.

In this case the object is placed p = 25 cm from the eye, to be able to see it clearly the image must be at q = 97 cm from the eye

let's calculate

        1 / f = 1/97 + 1/25

        1 / f = 0.05

        f = 19,877 cm

the power of a lens is defined by the inverse of the focal length in meters

         P = 1 / f

         P = 1 / 19,877 10-2

         P = 5D

5 0
2 years ago
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