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2 years ago

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The 2004 landings of the Mars rovers Spirit and Opportunity involved many stages, resulting in each probe having zero vertical v

elocity about 12 m above the surface of Mars. Determine
(a) the time required for the final free-fall descent of the probes
(b) the vertical velocity at impact. The mass of Mars is 6 . 419 × 10 23 kg , and its radius is 3 . 397 × 10 6 m

1 answer:

Scrat [10]2 years ago

7 0

Answer:

2.54334 seconds

9.4364 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of Mars = $6.419\times 10^{23}\ kg$

R = Radius of Mars = $3.397\times 10^6\ m$

The acceleration due to gravity of a planet is given by

$g=\frac{GM}{R^2}\\\Rightarrow g=\frac{6.67\times 10^{-11}\times 6.419\times 10^{23}}{(3.397\times 10^6)^2}\\\Rightarrow g=3.71024\ m/s^2$

Equation of motion

$s=ut+\frac{1}{2}at^2\\\Rightarrow 12=0t+\frac{1}{2}\times 3.71024\times t^2\\\Rightarrow t=\sqrt{\frac{12\times 2}{3.71024}}\\\Rightarrow t=2.54334\ s$

The time taken to fall 12 m in Mars is 2.54334 seconds

$v=u+at\\\Rightarrow v=0+3.71024\times 2.54334\\\Rightarrow v=9.4364\ m/s$

The vertical velocity at impact is 9.4364 m/s

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Answer:

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Explanation:

<u>LC Circuit</u>

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

$q, q_1, q_2$ = charge of the capacitor in any time $t, t_1, t_2$

$q_o$ = initial charge of the capacitor

$\omega$ =angular frequency of the circuit

$i, i_1, i_2$ = current through the circuit in any time $t, t_1, t_2$

The charge in an LC circuit is given by

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The current is the derivative of the charge

$\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).$

We are given

$q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}$

It means that

$q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]$

$i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]$

From eq 1:

$\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}$

From eq 2:

$\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}$

Squaring and adding the last two equations, and knowing that

$sin^2x+cos^2x=1$

$\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1$

Operating

$\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2$

Solving for $q_o$

$\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}$

Now we know the value of $q_0$ , we repeat the procedure of eq 1 and eq 2, but now at the second time $t_2$ , and solve for $i_2$

$\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2$

Solving for $i_2$

$\displaystyle i_2=w\sqrt{q_o^2-q_2^2}$

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

$\displaystyle f=\frac{1}{4\pi}\ KHz$

$w=2\pi f=500\ rad/s$

$\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}$

$q_0=0.01166\ c$

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$\displaystyle i_2=500\sqrt{0.01166^2-.006^2}$

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brick wall, it'll be even harder for any math to find its way
in there, and you'll miss out on a lot of the fun.

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