The wrong type of lens-Microscope, concave
Explanation:
A microscope Basically uses t<u>wo convex lenses to magnify an object, or specimen.</u>
There are 2 lenses in a microscope
- <u>Object Lens:</u>The lens that is closer to the object
- <u>Eyepiece:</u>The lens that is closer to the eye
Both the object lens and the eyepiece, is a convex lens.
The conclusion is, medium Q is most likely a solid because solids have the highest density and sound waves travel fastest in high density media.
<h3>
Effect of density on speed of sound</h3>
Sound wave is mechanical wave that requires material medium for its propagation.
A high dense medium, is a medium with closely packed molecules. Since sound wave requires material medium for its propagation, it will travel faster in a high dense medium than a less dense medium.
Thus, the speed of sound increases as the density of the medium increases.
<h3>Speed of sound in the different media</h3>
The conclusion that can be made from the speed of sound in the different media is "Medium Q is most likely a solid because solids have the highest density and sound waves travel fastest in high density media".
Learn more about effect of density on speed of sound here: brainly.com/question/3323620
From the information given,
diameter of ornament = 8
radius = diameter/2 = 8/2
radius of curvature, r = 4
Recall,
focal length, f = radius of curvature/2 = 4/2
f = 2
Recall,
magnification = image d
Answer:
Power output = 96.506 watts
Explanation:
Drag coefficient (Cd) = 0.9
V = 7.3 m/s
Air density (ρ) = 1.225 kg/m^(3)
Area (A) = 0.45 m^2
Let's find the drag force ;
Fd=(1/2)(Cd)(ρ)(A)(v^(2))
So Fd = (1/2)(0.9)(1.225)(0.45)(7.3^(2)) = 13.22N
Drag power = Drag Force x Drag velocity.
Thus drag power, = 13.22 x 7.3 = 96.506 watts
Answer:
208 Joules
Explanation:
The radius of the circular path the charge moves, r = 26 m
The magnetic force acting on the charge particle, F = 16 N
Centripetal force, 
= m·v²/r
Kinetic energy, K.E. = (1/2)·m·v²
Where;
m = The mass of the charged particle
v = The velocity of the charged particle
r = The radius of the path of the charged particle
Whereby the magnetic force acting on the charge particle = The centripetal force, we have;
F = = m·v²/r = 16 N
(1/2) × r × = (1/2) × r × m·v²/r = (1/2)·m·v² = K.E.
∴ (1/2) × r × = (1/2) × 26 m × 16 N = = (1/2)·m·v² = K.E.
∴ 208 Joules = K.E.
The kinetic energy of an particle moving in the circular path, K.E. = 208 Joules.
