Answer:
![9.93\times 10^{19}Hz](https://tex.z-dn.net/?f=9.93%5Ctimes%2010%5E%7B19%7DHz)
Explanation:
Speed of light is the product of its wavelength and frequency, expressed as
S=fw
Where s represent speed, f is frequency while w is wavelength
Making f the subject of the formula then
f=s/w
Substituting 2.99x10^8 m/s for s and 3.012x10^-12 m for w then
![f=\frac {2.99\times 10^{8}}{3.012\times 10^{-12}}=9.926958831341\times 10^{19}\\f\approx 9.93\times 10^{19}Hz](https://tex.z-dn.net/?f=f%3D%5Cfrac%20%7B2.99%5Ctimes%2010%5E%7B8%7D%7D%7B3.012%5Ctimes%2010%5E%7B-12%7D%7D%3D9.926958831341%5Ctimes%2010%5E%7B19%7D%5C%5Cf%5Capprox%209.93%5Ctimes%2010%5E%7B19%7DHz)
Therefore, the frequency equals to ![9.93\times 10^{19}Hz](https://tex.z-dn.net/?f=9.93%5Ctimes%2010%5E%7B19%7DHz)
The velocity of B after elastic collision is 3.45m/s
This type of collision is an elastic collision and we can use a formula to solve this problem.
<h3>Elastic Collision</h3>
![v_2 = \frac{2m_1u_1}{m_1+m_2} - \frac{m_1 - m_2}{m_1 + m_2}u_2](https://tex.z-dn.net/?f=v_2%20%3D%20%5Cfrac%7B2m_1u_1%7D%7Bm_1%2Bm_2%7D%20-%20%5Cfrac%7Bm_1%20-%20m_2%7D%7Bm_1%20%2B%20m_2%7Du_2)
The data given are;
- m1 = 281kg
- u1 = 2.82m/s
- m2 = 209kg
- u2 = -1.72m/s
- v1 = ?
Let's substitute the values into the equation.
![v_1 = \frac{2*281*2.82}{281+209} -\frac{281-209}{281+209}(-1.72)\\v_1 = 3.45m/s](https://tex.z-dn.net/?f=v_1%20%3D%20%5Cfrac%7B2%2A281%2A2.82%7D%7B281%2B209%7D%20-%5Cfrac%7B281-209%7D%7B281%2B209%7D%28-1.72%29%5C%5Cv_1%20%3D%203.45m%2Fs)
From the calculation above, the final velocity of the car B after elastic collision is 3.45m/s.
Learn more about elastic collision here;
brainly.com/question/7694106
Recall this kinematic equation:
a = ![\frac{Vi+Vf}{Δt}](https://tex.z-dn.net/?f=%5Cfrac%7BVi%2BVf%7D%7B%CE%94t%7D)
This equation gives the acceleration of the object assuming it IS constant (the velocity changes at a uniform rate).
a is the acceleration.
Vi is the initial velocity.
Vf is the final velocity.
Δt is the amount of elapsed time.
Given values:
Vi = 0 m/s (the car starts at rest).
Vf = 25 m/s.
Δt = 10 s
Substitute the terms in the equation with the given values and solve for a:
a = ![\frac{0+25}{10}](https://tex.z-dn.net/?f=%5Cfrac%7B0%2B25%7D%7B10%7D)
<h3>a = 2.5 m/s²</h3>
The velocity is a vectorial quantity, whereas speed is a scalar quantity, meaning it depends on the direction!
As such, the velocity is changing because the direction is changing.
Answer:
![\phi_B = 0.216 T m^2](https://tex.z-dn.net/?f=%5Cphi_B%20%3D%200.216%20T%20m%5E2)
Explanation:
As we know that the length of the conductor is given as
![L = 2 m](https://tex.z-dn.net/?f=L%20%3D%202%20m)
now if it is converted into a square then we have
![L = 4a](https://tex.z-dn.net/?f=L%20%3D%204a)
![a = \frac{L}{4} = 0.5 m](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7BL%7D%7B4%7D%20%3D%200.5%20m)
now the are of the loop will be
![A = a^2 = 0.5(0.5) = 0.25 m^2](https://tex.z-dn.net/?f=A%20%3D%20a%5E2%20%3D%200.5%280.5%29%20%3D%200.25%20m%5E2)
now the magnetic flux is defined as
![\phi_B = BAcos\theta](https://tex.z-dn.net/?f=%5Cphi_B%20%3D%20BAcos%5Ctheta)
here we know
B = 1.0 T
![\theta = 30.0^o](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2030.0%5Eo)
![\phi_B = (1.0 T)(0.25 m^2)(cos30)](https://tex.z-dn.net/?f=%5Cphi_B%20%3D%20%281.0%20T%29%280.25%20m%5E2%29%28cos30%29)
![\phi_B = 0.216 T m^2](https://tex.z-dn.net/?f=%5Cphi_B%20%3D%200.216%20T%20m%5E2)