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taurus [48]
3 years ago
7

An average person can reach a maximum height of about 60 cm when jumping straight up from a crouched position. During the jump i

tself, the person's body from the knees up typically rises a distance of around 50 cm. To keep the calculations simple and yet get a reasonable result, assume that the entire body rises this much during the jump.
I need to answer this:
a) With what initial speed does the person leave the ground to reach a height of 60 cm?
and
b) In terms of this jumper's weight W , what force does the ground exert on him or her during the jump?
Physics
1 answer:
Iteru [2.4K]3 years ago
4 0

Answer:

Explanation:

Height attained by body = 50 cm

= .5 m

Initial velocity = u

v² = u² - 2gh

0 = u² - 2gh

u² = 2 x 9.8 x .5

u = 3.13 m /s

During the initial period , the muscle stretches by around 10 cm during which force by ground reacts on the body and gives acceleration to achieve velocity of 3.13 m/s from zero .

v² = u² + 2as

3.13² = 0 + 2 a x .10

a = 49  m/s²

reaction by ground R

Net force

R-mg = ma

R= m ( g +a )

= mg + ma

=W + (W/g) x a

W ( 1 + a / g )

= W ( 1 + 49 / 9.8 )

= 6W

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Smaller refractive power

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A the person ages, the ciliary muscles of the eyes weakens and as a result the lens contracts and the formation of the image takes place behind the retina instead of forming at the retina.

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Where do we hear loud sound, at antinode or node?​
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A charged particle having mass 6.64 x 10-27 kg (that of a helium atom) moving at 8.70 x 105 m/s perpendicular to a 1.30-T magnet
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Answer:

the charge of the particle is 2.47 x 10⁻¹⁹ C

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velocity of the particle, v = 8.7 x 10⁵ m/s

strength of the magnetic field, B = 1.3 T

radius of the circle, r = 18 mm = 1.8 x 10⁻³ m

The magnetic force experienced by the charge is calculated as;

F = ma = qvB

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a is the acceleration of the charge in the circular path

a = \frac{v^2}{r} \\\\ma = qvB\\\\q = \frac{ma}{vB} \\\\q = \frac{mv^2}{rvB} = \frac{mv}{rB} \\\\q = \frac{(6.64\times 10^{-27} ) \times (8.7\times 10^5)}{(1.8\times 10^{-2}) \times (1.3)} \\\\q = 2.47 \ \times 10^{-19} \ C

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A sample of metallic frewium weighs 185N on a spring scale in air. When immersed in pure water, the frewium pulls on the scale w
balu736 [363]

Wow !  This one could have some twists and turns in it.
Fasten your seat belt.  It's going to be a boompy ride.

-- The buoyant force is precisely the missing <em>30N</em> .

--  In order to calculate the density of the frewium sample, we need to know
its mass and its volume.  Then, density = mass/volume .

-- From the weight of the sample in air, we can closely calculate its mass.

   Weight = (mass) x (gravity)
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The buoyant force is equal to the weight of displaced water, and the
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The weight of the displaced water is 30N, and weight = (mass) (gravity).

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Finally, density of the frewium sample = (mass)/(volume)

      Density = (18,858 grams) / (3,058 cm³) = <em>6.167 gm/cm³</em> (rounded)

================================================

I'm thinking that this must  be the hard way to do it,
because I noticed that

       (weight in air) / (buoyant force) =  185N / 30N = <u>6.1666...</u>

So apparently . . .

        (density of a sample) / (density of water) =

                                  (weight of the sample in air) / (buoyant force in water) .

I never knew that, but it's a good factoid to keep in my tool-box.


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