The energy of a single photon is given by:
where
E is the energy
h is the Planck constant
f is the frequency of the light
The light in our problem has a frequency
, so the energy of each photon of that light is:
We are given a bucket with pennies. We are asked to increase the gravitational potential energy. The gravitational potential energy is the energy that is related to the position of the object and is determined by the following formula:
This means that the gravitational potential energy depends on the height and the mass. Therefore, to increase the gravitational potential energy we must increase the height of the bucket. Therefore, we need to add more books. Therefore, the right option is B.
A special purpose nozzles that are often lowered through holes or other openings to the cellar of an occupancy are called cellar nozzles.
What is a nozzle:
A nozzle is a device in which steadily flowing fluid can be made to accelerate by a pressure drop along the duct in a cross-sectional area.
So when a fluid flows through a nozzle, its velocity increases continuously and pressure decreases continuously.
There are different types of tips like hollow cone, solid cone or flat fan.
The cellar nozzle (also called a Bresnan nozzle) is designed to be used under the surface the operator is standing on.
Hence,
The cellar nozzles are often lowered through holes or other opening to the cellar of an occupancy.
Learn more about nozzles here:
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Answer:
1) a) I₁ = 0.2941 kg m², b) I₂ = 0.2963 kg m², c) I_{total} = 0.5904 kg m²
3) α = 6.31 10⁶ rad / s²
Explanation:
1) The moment of inertia for bodies with high symmetry is tabulated, for a divo with an axis passing through its center is
I = ½ m r²
a) moment of inertia of the upper disk
I₁ = ½ m₁ r₁²
I₁ = ½ 1,468 0.633²
I₁ = 0.2941 kg m²
b) upper aluminum disc moment of inertia
I₂ = ½ m₂ r₂²
I₂ = ½ 1.479 0.633²
I₂ = 0.2963 kg m²
c) the moment of inertia is an additive scalar quantity therefore
I_{total} = I₁ + I₂
I_{total} = 0.2941 + 0.2963
I_{total} = 0.5904 kg m²
3) ask the value of the angular acceleration, that is, the second derivative of the angle with respect to time squared
indicate the angular velocity of the system w = 400 rev / s
Let's reduce the SI system
w = 400 rev / s (2π rad / rev) = 2513.27 rad / s
as the system is rotating we can calculate the centripetal acceleration
a = w² R
a = 2513.27² 0.633
a = 3.998 10⁶ m / s²
the linear and angular variable are related
a = α r
α = a / r
α = 3.998 10⁶ / 0.633
α = 6.31 10⁶ rad / s²