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posledela
3 years ago
15

Michael Jordan, el célebre basquetbolista, ganó el torneo de clavadas de la NBA en 1988. Para lograr la hazaña saltó 1.35 metros

verticalmente ¿Cuánto tiempo permaneció en los
20 cm de la parte más alta de su salto? ¿y en los 20 cm de su parte más baja?
Physics
1 answer:
kozerog [31]3 years ago
8 0

(a) 0.40 s

First of all, let's find the initial speed at which Jordan jumps from the ground.

The maximum height is h = 1.35 m. We can use the following equation:

v^2-u^2=2gh

where

v = 0 is the velocity at the maximum height

u is the initial velocity

g=-9.8 m/s^2 is the acceleration of gravity

Solving for u,

u=\sqrt{-2gh}=\sqrt{-2(-9.8)(1.35)}=5.14 m/s

The time needed to reach the maximum height can now be found by using the equation

v=u+gt

Solving for t,

t=\frac{v-u}{g}=\frac{0-5.14}{-9.8}=0.52s

Now we can find the velocity at which Jordan reaches a point 20 cm below the maximum height, so at a height of

h' = 1.35 - 0.20 = 1.15 m

Using again the equation

v'^2-u^2=2gh'

we find

v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(1.15)}=1.97 m/s

And the corresponding time is

t'=\frac{v'-u}{g}=\frac{1.97-5.14}{-9.8}=0.32s

So the time to go from h' to h is

\Delta t = t-t'=0.52-0.32=0.20 s

And since we have also to take into account the fall down (after Jordan reached the maximum height), which is symmetrical, we have to multiply this time by 2 to get the total time of permanence in the highest 20 cm of motion:

\Delta t=2\cdot 0.20 = 0.40 s

(b) 0.08 s

This part is easier since we need to calculate only the velocity at a height of h' = 0.20 m:

v'^2-u^2=2gh'

v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(0.20)}=4.74 m/s

And the corresponding time is

t'=\frac{v'-u}{g}=\frac{4.74-5.14}{-9.8}=0.04s

So this is the time needed to go from h=0 to h=20 cm; again, we have to take into account the motion downwards, so we have to multiply this by 2:

\Delta t = 2\cdot 0.04 =0.08 s

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Answer:

44,000 Nm^2/C

Explanation:

The electric flux through a certain surface is given by (for a uniform field):

\Phi = EA cos \theta

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In this problem, we have:

E=1.1\cdot 10^4 N/C is the electric field

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A=L^2=(2.0)^2=4.0 m^2

\theta=0^{\circ}, since the electric field is perpendicular to the surface

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\Phi =(1.1\cdot 10^4)(4.0)(cos 0^{\circ})=44,000 Nm^2/C

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2 years ago
Determine the index of refraction of glass that is struck by unpolarized light at 53.8 degrees and resulting in a fully polarize
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Answer:

The refractive index of glass, \mu_{g} = 1.367

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To determine the refractive index of glass:

tan\theta_{P} = \frac{\mu_{g}}{\mu_{a}}                 (1)

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tan{53.8^{\circ}} = \frac{\mu_{g}}{1}

\mu_{g} = tan53.8^{\circ}

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3 years ago
A plastic rod of length d = 1.5 m lies along the x-axis with its midpoint at the origin. The rod carries a uniform linear charge
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Answer:

Explanation:

Let the plastic rod extends from - L to + L .

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It will create a field at point P on y -axis . Distance of point P

= √ x² + .15²

electric field at P due to small charged length

dE = k λ dx x  / (x² + .15² )

Its component along Y - axis

= dE cosθ where θ is angle between direction of field dE and y axis

= dE x .15 / √ x² + .15²

=  k λ dx  .15 / (x² + .15² )³/²

If we consider the same strip along the x axis at the same position  on negative x axis , same result will be found . It is to be noted that the component of field in perpendicular to y axis will cancel out each other . Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L

E = ∫  k λ  .15  / (x² + .15² )³/² dx

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6 0
3 years ago
A rigid, insulated tank whose volume is 10 L is initially evacuated. A pinhole leak develops and air from the surroundings at 1
balandron [24]

Answer:

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Explanation:

For point a:

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\frac{dU}{dt}= Q-Wm_ih_i-m_eh_e\\\\

W=0\\\\Q=0\\\\m_e=0

From the above equation:

\frac{dU}{dt}=0-0+m_ih_i-0\\\\\Delta U=\int^{2}_{1}m_ih_idt\\\\

because the rate of air entering the tank that is h_i constant.

\Delta U = h_i \int^{2}_{1} m_i dt \\\\= h_i(m_2 -m_1)\\\\m_2u_2-m_1u_2=h_i(M_2-m_1)\\\\

Since the tank was initially empty and the inlet is constant hence, m_2u-0=h_1(m_2-0)\\\\m_2u_2=h_1m_2\\\\u_2=h_1\\\\

Interpolate the enthalpy between T = 300 \ K \ and\ T=295\ K. The surrounding air  

temperature:

T_1= 25^{\circ}\ C\ (298.15 \ K)\\\\\frac{h_{300 \ K}-h_{295\ K}}{300-295}= \frac{h_{300 \ K}-h_{1}}{300-295.15}

Substituting the value from ideal gas:

\frac{300.19-295.17}{300-295}=\frac{300.19-h_{i}}{300-298.15}\\\\h_i= 298.332 \ \frac{kJ}{kg}\\\\Now,\\\\h_i=u_2\\\\u_2=h_i=298.33\ \frac{kJ}{kg}

Follow the ideal gas table.

The u_2= 298.33\ \frac{kJ}{kg} and between temperature T =410 \ K \ and\  T=240\ K.

Interpolate

\frac{420-410}{u_{240\ k} -u_{410\ k}}=\frac{420-T_2}{u_{420 k}-u_2}

Substitute values from the table.

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For point b:

Consider the ideal gas equation.  therefore, p is pressure, V is the volume, m is mass of gas. \bar{R} \ is\  \frac{R}{M} (M is the molar mass of the  gas that is 28.97 \ \frac{kg}{mol} and R is gas constant), and T is the temperature.

n=\frac{pV}{TR}\\\\

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For point c:

 Entropy is given by the following formula:

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mestny [16]

Answer:

Electric potential = 0.00054 V

Explanation:

We are given;

Charge; q = 3 pC = 3 × 10^(-12) C

Radius; r = 2 cm = 0.02 m

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V = kqr

Where;

K is a constant = 9 × 10^(9) N⋅m²/C².

Thus;

V = 9 × 10^(9) × 3 × 10^(-12) × 0.02

V = 0.00054 V

8 0
3 years ago
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