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tensa zangetsu [6.8K]
3 years ago
10

A 52 kg pole vaulter running at 11 m/s vaults over the bar. Her speed when she is above the bar is 1.2 m/s. The acceleration of

gravity is 9.8 m/s 2 . Find her altitude as she crosses the bar. Neglect air resistance, as well as any energy absorbed by the pole. Answer in units of m.
Physics
1 answer:
kicyunya [14]3 years ago
5 0

Answer:

6.1 m

Explanation:

m = Mass of person = 52 kg

h = Altitude

v = Velocity

Kinetic energy of the person on the ground

\dfrac{1}{2}mv^2=\dfrac{1}{2}52\times 11^2\\ =3146\ J

Kinetic energy of the person at the top

\dfrac{1}{2}52\times 1.2^2\\ =37.44\ J

At the top the potential energy is given by

mgh=52\times 9.8h

Balancing the energy of the system

3146=37.44+52\times 9.8h\\\Rightarrow h=\dfrac{3146-37.44}{52\times 9.8}\\\Rightarrow h=6.1\ m

Her altitude is 6.1 m

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Describe in your own words, What is insulation?
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3 years ago
(a) Suppose that your measured weight at the equator is one-half your measured weight at the pole on a planet whose mass and dia
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Answer:

7160.2812 s or 1.988 hours

Explanation:

m = Mass of person

R = Radius of Earth = 6.37\times 10^{6}\ m

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\omega = Angular speed

Force at equator would be

F_e=m(g-\omega^2R)

Force at pole

F_p=mg

From the question

F_e=\dfrac{1}{2}F_p\\\Rightarrow m(g-\omega^2R)=\dfrac{1}{2}F_p\\\Rightarrow \omega=\sqrt{\dfrac{g}{2R}}

Time period is given by

T=\dfrac{2\pi}{\omega}\\\Rightarrow T=2\pi\sqrt{\dfrac{2R}{g}}\\\Rightarrow T=2\pi\sqrt{\dfrac{2\times 6.37\times 10^6}{9.81}}\\\Rightarrow T=7160.2812\ s=1.988\ hours

The rotational period of the planet is 7160.2812 s or 1.988 hours

5 0
3 years ago
you stand on a straight desert road at night and observe a vehicle approaching. this vehicle is equipped with two small headligh
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Answer:

The distance that you marginally able to discern that there are two headlights rather than a single light source is 6.084 km

Explanation:

Given:

d = distance = 0.679 m

λ = wavelength of the light = 537 nm = 537x10⁻⁹m

dp = pupil diameter = 4.81 mm = 0.00481 m

Question: What distance, in kilometers, are you marginally able to discern that there are two headlights rather than a single light source, dx = ?

For the separation of the peak from the central maximum it is:

sin\theta =\frac{\lambda }{d_{p} } =\frac{537x10^{-9} }{0.00481} =1.116x10^{-4}

In this case, the two small sources of the headlights have the same angle as the images that form inside the eye

d_{x} =\frac{d}{sin\theta } =\frac{0.679}{1.116x10^{-4} } =6.084x10^{3} m=6.084km

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4 years ago
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