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alexandr402 [8]
3 years ago
8

Physical activity is the most important thing you can do to improve and maintain health-related physical fitness. Which list con

tains factors that contribute to physical fitness?
A.age, environment, heredity, motivation

B.age, environment, height, maturation

C.age, emotional state, heredity, motivation

D.age, environment, heredity, maturation
Physics
1 answer:
katovenus [111]3 years ago
8 0

Answer:

I guess it's d because age environment heredity Nd maturation r more obvious factors if we compre with only physical fitness

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A square coil of wire of side 3.95 cm is placed in a uniform magnetic field of magnitude 2.25 T directed into the page as in the
Len [333]

Answer:

Explanation:

In this case we shall calculate rate of change of flux in the coli to calculate induced emf .

Flux through the coil  = no of turns x area x magnetic field perpendicular to it

=34 x  2.25 x (3.95 )²x 10⁻⁴ Weber

= 1193.4  x 10⁻⁴Weber

Final flux through the coil after turn by 90°

= 1193.4 x 10⁻⁴ cos 90 ° =0

Change of flux

= 1193.4 x 10⁻⁴ weber.

Time taken = 0.335 s .

Average emf= Rate of change of flux

= change in flux / time

=1193.4 x 10⁻⁴ / .335

= 3562.4 x 10⁻⁴

356.24 x 10⁻³

=356.24 mV.

Current induced = emf induced / resistance

= 356.24/.780

= 456.71 mA.

8 0
3 years ago
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3 0
3 years ago
The lens-makers’ equation applies to a lens immersed in a liquid if n in the equation is replaced by n2/n1. Here n2 refers to th
pickupchik [31]

Answer:

a

The focal length of the lens in water is  f_{water} = 262.68 cm

b

The focal length of the mirror in water is  f =79.0cm

Explanation:

From the question we are told that

    The index of refraction of the lens material = n_2

    The index of refraction of the medium surrounding the lens = n_1

 

The lens maker's formula is mathematically represented as

            \frac{1}{f} = (n -1) [\frac{1}{R_1} - \frac{1}{R_2}  ]

Where f is the focal length

            n is the index of refraction

            R_1 and R_2 are the radius of curvature of sphere 1 and 2 of the lens

From the question When the lens in air  we have  

           \frac{1}{f_{air}} = (n-1) [\frac{1}{R_1} - \frac{1}{R_2}  ]

    When immersed in liquid the formula becomes

          \frac{1}{f_{water}} = [\frac{n_2}{n_1} - 1 ] [\frac{1}{R_1} - \frac{1}{R_2}  ]

The ratio of the focal length of the the two medium is mathematically evaluated as

           \frac{f_water}{f_{air}} = \frac{n_2 -1}{[\frac{n_2}{n_1} - 1] }

From the question

      f_{air }= 79.0 cm

       n_2 = 1.55

and the refractive index of water(material surrounding the lens) has a constant value of  n_1 = 1.33

         \frac{f_{water}}{79}  = \frac{1.55- 1}{\frac{1.55}{1.44}  -1}

           f_{water} = 262.68 cm

b

The focal length of a mirror is dependent on the concept of reflection which is not affected by medium around it.

   

7 0
3 years ago
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