(1) The image of an object placed further from the lens than the focal point will be upside down and smaller than the object.
(2) When light rays reflect, they bounce back.
(3) Images formed by a concave lens will look magnified.
(4) When light rays enter a different medium, they bend.
<h3>
1.0 Object placed further from the lens than the focal point</h3>
The image of an object placed further from the lens than the focal point will be diminished and inverted.
Thus, the correct answer will be "upside down and smaller than the object".
<h3>2.0 What is reflection of light?</h3>
The ability of light to bounce back when it strike a hard surface is known as refection.
<h3>3.0 Image formed by concave lens</h3>
A concave lens is diverging lens is usually virtual, erect and magnified.
<h3>4.0 Refraction of light</h3>
The change in speed of light when it travels from medium to another medium is known as refraction. Refraction is also, the ability of light to bend around obstacles.
Learn more about reflection and refraction of light here: brainly.com/question/1191238
Answer:
Atomic Size and Mass:
convert given density to kg/m^3 = 8900kg/m^3 2) convert to moles/m^3 (kg/m^3 * mol/kg) = 150847 mol/m^3 (not rounding in my actual calculations) 3) convert to atoms/m^3 (6.022^23 atoms/mol) = 9.084e28 atoms/m^3 4) take the cube root to get the number of atoms per meter, = 4495309334 atoms/m 5) take the reciprocal to get the diameter of an atom, = 2.2245e-10 m/atom 6) find the mass of one atom (kg/mol * mol/atoms) = 9.7974e-26 kg/atom Young's Modulus: Y=(F/A)/(dL/L) 1) F=mg = (45kg)(9.8N/kg) = 441 N 2) A = (0.0018m)^2 = 3.5344e-6 m^2 3) dL = 0.0016m 4) L = 2.44m 5) Y = 1.834e11 N/m^2 Interatomic Spring Stiffness: Ks,i = dY 1) From above, diameter of one atom = 2.2245e-10 m 2) From above, Y = 1.834e11 N/m^2 3) Ks,i = 40.799 N/m (not rounding in my actual calculations) Speed of Sound: v = ωd 1) ω = √(Ks,i / m,a) 2) From above, Ks,i = 40.799 N/m 3) From above, m,a = 9.7974e-26 kg 4) ω=2.0406e13 N/m*kg 5) From above, d=2.2245e-10 m 6) v=ωd = 4539 m/s (not rounding in actual calculations) Time Elapsed: 1) length sound traveled = L+dL = 2.44166 m 2) From above, speed of sound = 4539 m/s 3) T = (L+dL)/v = 0.000537505 s
Answer:
a = 3.27 m/s²
v = 2.56 m/s
Explanation:
given,
mass A = 1 kg
mass B = 2 kg
vertical distance between them = 1 m
a = 3.27 m/s²
The speed of the system at that moment is:
v² = u² + 2×a×s
v² = 0² + 2× 3.27 × 1
v ² = 6.54
v = 2.56 m/s
Hello,
Your answer to this problem is 400/3
Hope this helps!
The EM spectrum has no limits. Any frequency you can imagine
is the frequency of some electromagnetic radiation somewhere.
