First recall the equation that relates frequency to wavelength:
v = fw
Note that the v is the speed of light, a constant. Now plug in the information we know!
(3×10^8) = (6.67 × 10^14) w
Hit the numbers on the calculator and you'll get the wavelength, w. If you comment your answer I'll check it for you. :)
Answer:Dr. Hess' most significant contribution to the plate tectonic theory began in 1945 when he was the commander of the U.S.S. ... In the paper Hess described how hot magma would rise from under the crust at the Great Global Rift. When the magma cooled, it would expand and push the tectonic plates apart.
Explanation:
Initial Velocity formulas are used to find the initial velocity of the moving body if some of the terms are given. Initial velocity can be formulated in a unit of a meter per second i.e. ms^{-1}. ms−1.
Answer:
8 m/s²
Explanation:
Centripetal acceleration formula:
- a = v²/r
- where v = velocity and r = radius
As you can see, the mass of the yo-yo is not relevant to this problem.
We are given that the velocity of the yo-yo is 4 m/s.
The radius of the circle formed by the yo-yo on a 2-meter long string is 2 meters.
Substitute these known values into the equation.
- a = (4)²/2
- a = 16/2
- a = 8
The centripetal acceleration of the yo-yo is 8 m/s².
Answer:
Explanation:
1.
vf = vi + a(t)
vi = 0 m/s
a = 17/13
= 1.3 m/s^2
2.
F = M * a
= 15 * 1.3
= 19.62 N
3.
Normal force, Fn = m * g
= 15 * 9.8 m/s²
= 147 N.
Then find the maximum force of friction, knowing that μs = 0.76
Ff = Fn × μs
= 147 * 0.76
= 111.83 N
Maximum acceleration,
Ff = m × a
111.83 = 15 * a
a = 7.4556 m/s²
4.
In order to find the acceleration for the box, you need to know the net force of the box moving in the x direction and the frictional force, and you will end up with the Force of the vehicle minus the Frictional force (Ff) between the box and the vehicle, resulting your net force in the x direction.
F = m*a and Ff = μ * N
m*a = μk * N
m*a = μk * m * g
a = μk * g
a = 0.61 * 9.81
5.98 m/s^2 for the acceleration of the box on top of the vehicle.
5.
Assuming the box has re settled and is no longer sliding when braking begins and the surface remains horizontal, the maximum negative acceleration will again be
a = -7.4556 m/s².