The Professor's centripetal acceleration is 0.044 m/s²
Centripetal acceleration is the acceleration of an object moving in circular motion. It is usually directed towards the center of the rotation.
It is given by:
a = v²/r
where v is the velocity and r is the radius.
Given that the radius (r) = 4 m, velocity (v) = 0.419 m/s, hence:
a = v²/r = 0.419²/4 = 0.044 m/s²
The Professor's centripetal acceleration is 0.044 m/s²
Find out more at: brainly.com/question/6082363
To solve this problem it is necessary to apply the continuity equations in the fluid and the kinematic equation for the description of the displacement, velocity and acceleration.
By definition the movement of the Fluid under the terms of Speed, acceleration and displacement is,
Where,
Velocity in each state
g= Gravity
h = Height
Our values are given as,
Replacing at the kinetic equation to find 
we have,
Applying the concepts of continuity,
We need to find A_2 then,
So the cross sectional area of the water stream at a point 0.11 m below the faucet is
Therefore the cross-sectional area of the water stream at a point 0.11 m below the faucet is 
Answer:
#_photons = 30 photons / s
Explanation:
Let's start by finding the energy of a photon of light, let's use the Planck relation
E = h f
the speed of light is related to wavelength and frequency
c = λ f
we substitute
E = h c /λ
E₀ = 6.63 10⁻³⁴ 3 10⁸/500 10⁻⁹
E₀ = 3.978 10⁻¹⁹ J
now let's use a direct proportion rule. If the energy of a photon is Eo, how many fornes has an energy E = 1.2 10⁻¹⁷ J in a second
#_photons = 1 photon (E / Eo)
#_photons = 1 1.2 10⁻¹⁷ /3.978 10⁻¹⁹
#_photons = 3.0 10¹
#_photons = 30 photons / s
Answer:
Any incident ray traveling parallel to the principal axis of a converging lens will refract through the lens and travel through the focal point on the opposite side of the lens. ... These rays of light will refract when they enter the lens and refract when they leave the lens.
Hope this helps...
KE = 1/ 2 * 1252 * 144
as KE = 1/2 * m * v ^2
= 90144 J
