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mario62 [17]
2 years ago
6

Recall that the essential matrix E describing the stereo geometry of two calibrated cameras is a function only of the relative r

otation and translational offset between the two cameras. Compute the essential matrix for the following camera configurations.
The Point Grey "Bumblebee" stereo camera is a "simple stereo" system having two cameras side-by-side with their optic axes parallel to each other. The baseline distance between the two cameras is 12 cm.
Compute the values of the 3x3 essential matrix E for this stereo system.

Physics
1 answer:
Mashcka [7]2 years ago
7 0

Answer:

The essential matrix in this case is given by      

         E=[T_{x}]R=\left[\begin{array}{ccc}0&0&0\\0&0&d\\0&-d&0\end{array}\right]

the values of the 3x3 essential matrix E for this stereo system

E=[T_{x}]R=\left[\begin{array}{ccc}0&0&0\\0&0&0.12\\0&-0.12&0\end{array}\right]

Explanation:

The explanation is shown on the first uploaded image

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The capacitors in each circuit are fully charged before the switch is closed. Rank, from longest to shortest, the length of time
IgorLugansk [536]

Answer:

C has the greatest capacitance and the lightest load. It will provide current to the load for the longest.

A & E are the same

B is next, and finally

D has the smallest equivalent capacitance and the heaviest load. it will run out of juice the quickest.

Curious that the pictures are out of alphabetic order A B C E D.

Explanation:

6 0
2 years ago
The position of a particle moving along the x-axis varies with time according to x(t) = 5.0t^2 − 4.0t^3 m. Find (a) the velocity
KengaRu [80]
<h2>Answer:</h2>

(a) v(t) = [10.0t - 12.0t²] m/s  and a(t) = [10.0 - 24.0t ] m/s² respectively

(b) -28.0m/s and -38.0m/s² respectively

(c) 0.83s

(d) 0.83s

(e) x(t)  = 1.1573 m           [where t = 0.83s]

<h2>Explanation:</h2>

The position equation is given by;

x(t) = 5.0t² - 4.0t³ m           --------------------(i)

(a) Since velocity is the time rate of change of position, the velocity, v(t), of the particle as a function of time is calculated by finding the derivative of equation (i) as follows;

v(t) = dx(t) / dt = \frac{dx}{dt} = \frac{d}{dt} [ 5.0t² - 4.0t³ ]

v(t) = 10.0t - 12.0t²     --------------------------------(ii)

Therefore, the velocity as a function of time is v(t) = 10.0t - 12.0t² m/s

Also, since acceleration is the time rate of change of velocity, the acceleration, a(t), of the particle as a function of time is calculated by finding the derivative of equation (ii) as follows;

a(t) = dx(t) / dt = \frac{dv}{dt} =  \frac{d}{dt} [ 10.0t - 12.0t² ]

a(t) = 10.0 - 24.0t             --------------------------------(iii)

Therefore, the acceleration as a function of time is a(t) = 10.0 - 24.0t m/s²

(b) To calculate the velocity at time t = 2.0s, substitute the value of t = 2.0 into equation (ii) as follows;

=> v(t) =  10.0t - 12.0t²

=> v(2.0) = 10.0(2) - 12.0(2)²

=> v(2.0) = 20.0 - 48.0

=> v(2.0) = -28.0m/s

Also, to calculate the acceleration at time t = 2.0s, substitute the value of t = 2.0 into equation (iii) as follows;

=> a(t) = 10.0 - 24.0t

=> a(2.0) = 10.0 - 24.0(2)

=> a(2.0) = 10.0 - 48.0

=> a(2.0) = -38.0 m/s²

Therefore, the velocity and acceleration at t = 2.0s are respectively -28.0m/s and -38.0m/s²

(c) The time at which the position is maximum is the time at which there is no change in position or the change in position is zero. i.e dx / dt = 0. It also means the time at which the velocity is zero. (since velocity is dx / dt)

Therefore, substitute v = 0 into equation (ii) and solve for t as follows;

=> v(t) = 10.0t - 12.0t²

=> 0 = 10.0t - 12.0t²

=> 0 = ( 10.0 - 12.0t ) t

=> t = 0            or             10.0 - 12.0t = 0

=> t = 0            or             10.0 = 12.0t

=> t = 0            or             t = 10.0 / 12.0

=> t = 0            or             t = 0.83s

At t=0 or t = 0.83s, the position of the particle will be maximum.

To get the more correct answer, substitute t = 0 and t = 0.83 into equation (i) as follows;

<em>Substitute t = 0 into equation (i)</em>

x(t) = 5.0(0)² - 4.0(0)³ = 0

At t = 0; x = 0

<em>Substitute t = 0.83s into equation (i)</em>

x(t) = 5.0(0.83)² - 4.0(0.83)³

x(t) = 5.0(0.6889) - 4.0(0.5718)

x(t) = 3.4445 - 2.2872

x(t)  = 1.1573 m

At t = 0.83; x = 1.1573 m

Therefore, since the value of x at t = 0.83s is 1.1573m is greater than the value of x at t = 0 which is 0m, then the time at which the position is at maximum is 0.83s

(d) The velocity will be zero when the position is maximum. That means that, it will take the same time calculated in (c) above for the velocity to be zero. i.e t = 0.83s

(e) The maximum position function is found when t = 0.83s as shown in (c) above;

Substitute t = 0.83s into equation (i)

x(t) = 5.0(0.83)² - 4.0(0.83)³

x(t) = 5.0(0.6889) - 4.0(0.5718)

x(t) = 3.4445 - 2.2872

x(t)  = 1.1573 m            [where t = 0.83s]

8 0
2 years ago
The point on the graph that lies on the y-axis (vertical axis) is called the y-intercept. What does the y-intercept tell you abo
jekas [21]

Answer:

The starting position of the runner.

Explanation:

When you look at the graph, you can see that the first point on the graph is twenty on the y-axis.

The runner starts at twenty, and ends at thirty.

Therefore, the runner starts at twenty on the y-axis, so it's the starting position of the runner.

7 0
2 years ago
A pendulum is made of a small sphere of mass 0.250 kg attached to a lightweight string 1.20 m in length. As the pendulum swings
forsale [732]

Answer:v=2 m/s

Explanation:

Given

Length of string L=1.2 m

mass of pendulum m=0.25 kg

maximum inclination with vertical \theta =34

vertical Rise of Pendulum from its mean position  is given by

\Delta h=L(1-\cos \theta )

Conserving Energy at top and bottom point

Potential Energy of sphere is converted into kinetic energy of sphere

mgL(1-\cos \theta )=\frac{mv^2}{2}

v=\sqrt{2gL(1-\cos \theta )}

v=\sqrt{2\times 9.8\times 1.2(1-\cos 34)}

v=\sqrt{4.021}

v=2 m/s

5 0
2 years ago
A certain alarm clock ticks four times each second, with each tick representing half a period. The balance wheel consists of a t
Semenov [28]

Answer:

a. I=2.77x10^{-8} kg*m^2

b. K=4.37 x10^{-6} N*m

Explanation:

The inertia can be find using

a.

I = m*r^2

m = 0.95 g * \frac{1 kg}{1000g}=9.5x10^{-4} kg

r=0.54 cm * \frac{1m}{100cm} =5.4x10^{-3}m

I = 9.5x10^{-4}kg*(5.4x10^{-3}m)^2

I=2.77x10^{-8} kg*m^2

now to find the torsion constant can use knowing the period of the balance

b.

T=0.5 s

T=2\pi *\sqrt{\frac{I}{K}}

Solve to K'

K = \frac{4\pi^2* I}{T^2}=\frac{4\pi^2*2.7702 kg*m^2}{(0.5s)^2}

K=4.37 x10^{-6} N*m

3 0
2 years ago
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