The answer is: none of the above.
Explanation:
When light reflects from a surface, the frequency, wavelength, and speed do not change. They remain the same.
Answer with Explanation:
We are given that
![F=-8\hat{i}+6\hat{j}](https://tex.z-dn.net/?f=F%3D-8%5Chat%7Bi%7D%2B6%5Chat%7Bj%7D)
![r=3\hat{i}+4\hat{j}](https://tex.z-dn.net/?f=r%3D3%5Chat%7Bi%7D%2B4%5Chat%7Bj%7D)
a.We have to find the torque on the particle about the origin.
We know that
Torque=![\tau=r\times F=\begin{vmatrix}i&j&k\\3&4&0\\-8&6&0\end{vmatrix}](https://tex.z-dn.net/?f=%5Ctau%3Dr%5Ctimes%20F%3D%5Cbegin%7Bvmatrix%7Di%26j%26k%5C%5C3%264%260%5C%5C-8%266%260%5Cend%7Bvmatrix%7D)
By using the formula
![\tau=50\hat{k}](https://tex.z-dn.net/?f=%5Ctau%3D50%5Chat%7Bk%7D)
b.![\mid \tau\mid =\mid F\mid \mid r\mid sin\theta](https://tex.z-dn.net/?f=%5Cmid%20%5Ctau%5Cmid%20%3D%5Cmid%20F%5Cmid%20%5Cmid%20r%5Cmid%20sin%5Ctheta)
![\mid F\mid=\sqrt{(-8)^2+(6)^2}=10](https://tex.z-dn.net/?f=%5Cmid%20F%5Cmid%3D%5Csqrt%7B%28-8%29%5E2%2B%286%29%5E2%7D%3D10)
![\mid r\mid=\sqrt{3^2+4^2}=5](https://tex.z-dn.net/?f=%5Cmid%20r%5Cmid%3D%5Csqrt%7B3%5E2%2B4%5E2%7D%3D5)
![\mid \tau\mid=\sqrt{(-50)^2}=50](https://tex.z-dn.net/?f=%5Cmid%20%5Ctau%5Cmid%3D%5Csqrt%7B%28-50%29%5E2%7D%3D50)
Substitute the values then we get
![50=10\times 5 sin\theta](https://tex.z-dn.net/?f=50%3D10%5Ctimes%205%20sin%5Ctheta)
![sin\theta=\frac{50}{50}=1](https://tex.z-dn.net/?f=sin%5Ctheta%3D%5Cfrac%7B50%7D%7B50%7D%3D1)
![sin\theta=sin90^{\circ}](https://tex.z-dn.net/?f=sin%5Ctheta%3Dsin90%5E%7B%5Ccirc%7D)
Because ![sin90^{\circ}=1](https://tex.z-dn.net/?f=sin90%5E%7B%5Ccirc%7D%3D1)
![\theta=90^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D90%5E%7B%5Ccirc%7D)
Answer:
<em>v=40 m/s south</em>
Explanation:
<u>Momentum
</u>
It's a physical magnitude that measures the product of the mass by the velocity of a particle. Its units in the International System is kg.m/s and the formula is
![p=m.v](https://tex.z-dn.net/?f=p%3Dm.v)
Where m is the mass and v the velocity of the particle. If we wanted to solve for v, we have
![\displaystyle v=\frac{p}{m}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v%3D%5Cfrac%7Bp%7D%7Bm%7D)
The baseball has a momentum of 6.0 kg.m/s south and mass of 0.15kg, thus
![\displaystyle v=\frac{6}{0.15}=40\ m/s](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v%3D%5Cfrac%7B6%7D%7B0.15%7D%3D40%5C%20m%2Fs)
The velocity is directed to the south
ThIs is the same type of problem
find out the time value
3 = 1/2*a*T^2
6/10 = t^2
t = 0.77 seconds
and the distance is given 5 m
thus speed ,= distance/time
speed = 5/0.77
= 6.45 m/s
The resistance of a given conductor depends on its electrical resistivity (
), its length(L) and its cross-sectional area (A), as follows:
![R=\frac{\rho L}{A}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7B%5Crho%20L%7D%7BA%7D)
In this case, we have
,
and
. So, the total resistance of the wire with length of 138m is:
![R'=\frac{\rho' L'}{A'}\\R'=\frac{\rho 138L}{A}\\R'=138\frac{\rho L}{A}\\R'=138R\\R'=138(0.24\Omega)\\R'=33.12\Omega](https://tex.z-dn.net/?f=R%27%3D%5Cfrac%7B%5Crho%27%20L%27%7D%7BA%27%7D%5C%5CR%27%3D%5Cfrac%7B%5Crho%20138L%7D%7BA%7D%5C%5CR%27%3D138%5Cfrac%7B%5Crho%20L%7D%7BA%7D%5C%5CR%27%3D138R%5C%5CR%27%3D138%280.24%5COmega%29%5C%5CR%27%3D33.12%5COmega)