<span>An imaginary line perpendicular to a reflecting surface is called "a normal" (principle line)
So, Your Answer would be Option B
Hope this helps!</span>
Answer:
For vector u, x component = 10.558 and y component =12.808
unit vector = 0.636 i+ 0.7716 j
For vector v, x component = 23.6316 and y component = -6.464
unit vector = 0.9645 i-0.2638 j
Explanation:
Let the vector u has magnitude 16.6
u makes an angle of 50.5° from x axis
So ![u_x=ucos\Theta =16.6\times cos50.5=10.558](https://tex.z-dn.net/?f=u_x%3Ducos%5CTheta%20%3D16.6%5Ctimes%20cos50.5%3D10.558)
Vertical component ![u_y=usin\Theta =16.6\times sin50.5=12.808](https://tex.z-dn.net/?f=u_y%3Dusin%5CTheta%20%3D16.6%5Ctimes%20sin50.5%3D12.808)
So vector u will be u = 10.558 i+12.808 j
Unit vector ![u=\frac{10.558i+12.808j}{\sqrt{10.558^2+12.808^2}}=0.636i+0.7716j](https://tex.z-dn.net/?f=u%3D%5Cfrac%7B10.558i%2B12.808j%7D%7B%5Csqrt%7B10.558%5E2%2B12.808%5E2%7D%7D%3D0.636i%2B0.7716j)
Now in second case let vector v has a magnitude of 24.5
Making an angle with -15.3° from x axis
So horizontal component ![v_x=vcos\Theta =24.5\times cos(-15.3)=23.6316](https://tex.z-dn.net/?f=v_x%3Dvcos%5CTheta%20%3D24.5%5Ctimes%20cos%28-15.3%29%3D23.6316)
Vertical component ![v_y=vsin\Theta =24.5\times sin(-15.3)=-6.464](https://tex.z-dn.net/?f=v_y%3Dvsin%5CTheta%20%3D24.5%5Ctimes%20sin%28-15.3%29%3D-6.464)
So vector v will be 23.6316 i - 6.464 j
Unit vector of v ![=\frac{23.6316i-6.464}{\sqrt{23.6316^2+6.464^2}}=0.9645i-0.2638j](https://tex.z-dn.net/?f=%3D%5Cfrac%7B23.6316i-6.464%7D%7B%5Csqrt%7B23.6316%5E2%2B6.464%5E2%7D%7D%3D0.9645i-0.2638j)
Answer:
0.572 Hz
Explanation:
given,
length of simple pendulum, l = 0.76 m
mass of the bob, m = 365 g = 0.365 Kg
angle made from the vertical, = 12°
frequency, f = ?
![f = \dfrac{1}{2\pi}\sqrt{\dfrac{g}{L}}](https://tex.z-dn.net/?f=f%20%3D%20%5Cdfrac%7B1%7D%7B2%5Cpi%7D%5Csqrt%7B%5Cdfrac%7Bg%7D%7BL%7D%7D)
![f = \dfrac{1}{2\pi}\sqrt{\dfrac{9.8}{0.76}}](https://tex.z-dn.net/?f=f%20%3D%20%5Cdfrac%7B1%7D%7B2%5Cpi%7D%5Csqrt%7B%5Cdfrac%7B9.8%7D%7B0.76%7D%7D)
![f = \dfrac{1}{2\pi}\times 3.59](https://tex.z-dn.net/?f=f%20%3D%20%5Cdfrac%7B1%7D%7B2%5Cpi%7D%5Ctimes%203.59)
f = 0.572 Hz
The frequency at which pendulum vibrates is equal to 0.572 Hz
Answer:
a) v₂ = 4.2 m/s
b) v₂ = 5 m/s
Explanation:
a)
We will use the law of conservation of momentum here:
![m_1u_1+m_2u_2=m_1v_1+m_2v_2](https://tex.z-dn.net/?f=m_1u_1%2Bm_2u_2%3Dm_1v_1%2Bm_2v_2)
where,
m₁ = m₂ = mass of bowling pin = 1.8 kg
u₁ = speed of first pin before collsion = 5 m/s
u₂ = speed of second pin before collsion = 0 m/s
v₁ = speed of first pin after collsion = 0.8 m/s
v₂ = speed of second after before collsion = ?
Therefore,
![(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0.8\ m/s)+(1.8\ kg)(v_2)\\v_2 = 5\ m/s - 0.8\ m/s](https://tex.z-dn.net/?f=%281.8%5C%20kg%29%285%5C%20m%2Fs%29%2B%281.8%5C%20kg%29%280%5C%20m%2Fs%29%3D%281.8%5C%20kg%29%280.8%5C%20m%2Fs%29%2B%281.8%5C%20kg%29%28v_2%29%5C%5Cv_2%20%3D%205%5C%20m%2Fs%20-%200.8%5C%20m%2Fs)
<u>v₂ = 4.2 m/s</u>
<u></u>
b)
We will use the law of conservation of momentum here:
![m_1u_1+m_2u_2=m_1v_1+m_2v_2](https://tex.z-dn.net/?f=m_1u_1%2Bm_2u_2%3Dm_1v_1%2Bm_2v_2)
where,
m₁ = m₂ = mass of bowling pin = 1.8 kg
u₁ = speed of first pin before collsion = 5 m/s
u₂ = speed of second pin before collsion = 0 m/s
v₁ = speed of first pin after collsion = 0 m/s
v₂ = speed of second after before collsion = ?
Therefore,
![(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0\ m/s)+(1.8\ kg)(v_2)](https://tex.z-dn.net/?f=%281.8%5C%20kg%29%285%5C%20m%2Fs%29%2B%281.8%5C%20kg%29%280%5C%20m%2Fs%29%3D%281.8%5C%20kg%29%280%5C%20m%2Fs%29%2B%281.8%5C%20kg%29%28v_2%29)
<u>v₂ = 5 m/s</u>