¡Hola!
For this problem, first, lets recabe information:
v = 0 m/s
t = 3 s
g = 9,82 m/s²
v' = ?
d = ?
First, for calculate the final velocity:
v' = v + g * t
v' = 0 m/s + 9,82 m/s² * 3 s
v' = 29,46 m/s ![\leftarrow \boxed{\text{Velocity} }](https://tex.z-dn.net/?f=%5Cleftarrow%20%5Cboxed%7B%5Ctext%7BVelocity%7D%20%7D)
Now, for calculate how far did it fall:
d = v * t + g * t^2 / 2
Like v = 0 m/s, we can simplificate equation:
d = g * t^2 / 2
d = 9,82 m/s² * (3 s)^2 / 2
d = 9,82 m/s² * 9 s² / 2
d = 88,38 m / 2
d = 44.19 m ![\leftarrow \boxed{\text{Distance} }](https://tex.z-dn.net/?f=%5Cleftarrow%20%5Cboxed%7B%5Ctext%7BDistance%7D%20%7D)
¡Buena suerte!
I think it is force of acceleration but I'm not sure.. hope this helps though!
Answer:
![\hat{w}=\dfrac{\vec{5i+13j-k}}{13.92}](https://tex.z-dn.net/?f=%5Chat%7Bw%7D%3D%5Cdfrac%7B%5Cvec%7B5i%2B13j-k%7D%7D%7B13.92%7D)
Explanation:
It is given that,
![\vec{u}=2i-j-3k](https://tex.z-dn.net/?f=%5Cvec%7Bu%7D%3D2i-j-3k)
![\vec{v}=-3i+j-2k](https://tex.z-dn.net/?f=%5Cvec%7Bv%7D%3D-3i%2Bj-2k)
Taking the cross product of v and v such that,
![\vec{w}=u\times v](https://tex.z-dn.net/?f=%5Cvec%7Bw%7D%3Du%5Ctimes%20v)
![\vec{w}=(2i-j-3k)\times (-3i+j-2k)](https://tex.z-dn.net/?f=%5Cvec%7Bw%7D%3D%282i-j-3k%29%5Ctimes%20%28-3i%2Bj-2k%29)
![\vec{w}=5i+13j-k](https://tex.z-dn.net/?f=%5Cvec%7Bw%7D%3D5i%2B13j-k)
![|w|=\sqrt{5^2+13^2(-1)^2}](https://tex.z-dn.net/?f=%7Cw%7C%3D%5Csqrt%7B5%5E2%2B13%5E2%28-1%29%5E2%7D)
|w| = 13.92
Let
is the unit vector normal to the plane containing u and v. So,
![\hat{w}=\dfrac{\vec{w}}{|w|}](https://tex.z-dn.net/?f=%5Chat%7Bw%7D%3D%5Cdfrac%7B%5Cvec%7Bw%7D%7D%7B%7Cw%7C%7D)
![\hat{w}=\dfrac{\vec{5i+13j-k}}{13.92}](https://tex.z-dn.net/?f=%5Chat%7Bw%7D%3D%5Cdfrac%7B%5Cvec%7B5i%2B13j-k%7D%7D%7B13.92%7D)
Hence, this is the required solution.
The current flowing in each resistor of the circuit is 4 A.
<h3>
Equivalent resistance of the series resistors</h3>
The equivalent resistance of the series circuit is calculated as follows;
6 Ω and 4 Ω are in series = 10 Ω
5 Ω and 10Ω are in series = 15 Ω
<h3>Effective resistance of the circuit</h3>
![\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \\\\R = \frac{R_1R_2}{R_1 + R_2} \\\\R = \frac{10 \times 15}{10 + 15} \\\\R = 6 \ ohms](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BR%7D%20%3D%20%5Cfrac%7B1%7D%7BR_1%7D%20%2B%20%5Cfrac%7B1%7D%7BR_2%7D%20%5C%5C%5C%5CR%20%3D%20%5Cfrac%7BR_1R_2%7D%7BR_1%20%2B%20R_2%7D%20%5C%5C%5C%5CR%20%3D%20%5Cfrac%7B10%20%5Ctimes%2015%7D%7B10%20%2B%2015%7D%20%5C%5C%5C%5CR%20%3D%206%20%5C%20ohms)
<h3>Current flowing in the circuit</h3>
V = IR
I = V/R
I = 24/6
I = 4 A
Learn more about resistors in parallel here: brainly.com/question/15121871
A. The vibrations of the fields are perpendicular to the direction in which the wave moves.