The focal length of a magnifying glass is the distance between the focal point and optical centre of the magnifying glass.
<h3>Focal length</h3>
The focal length, f is the distance from a lens or mirror to the focal point, F.
This is the distance from a lens or mirror at which parallel light rays will meet for a converging lens or mirror or appear to diverge from for a diverging lens or mirror.
A magnifying glass is a converging lens which produces a enlarged, erect and virtual image when an object is placed between the focal point and optical centre.
A magnifying glass will bring to focus at a point sun rays which can cause the paper to catch fire if it is held in place for long.
This point at which the most concentrated ray of light is shining on the paper, is the focal point for that magnifying glass.
Therefore, the focal length of a magnifying glass is the distance between the focal point and optical centre of the magnifying glass.
Learn more about about focal length at: brainly.com/question/25779311
It's the rate that energy is used or supplied
Explanation:
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Answer
Given,
y(x, t) = (3.5 cm) cos(2.7 x − 92 t)
comparing the given equation with general equation
y(x,t) = A cos(k x - ω t)
A = 3.5 cm , k = 2.7 rad/m , ω = 92 rad/s
we know,
a) ω =2πf
f = 92/ 2π
f = 14.64 Hz
b) Wavelength of the wave
we now, k = 2π/λ
2π/λ = 2.7
λ = 2 π/2.7
λ = 2.33 m
c) Speed of wave
v = ν λ
v = 14.64 x 2.33
v = 34.11 m/s
Answer:
v=12.5 i + 12.5 j m/s
Explanation:
Given that
m₁=m₂ = m
m₃ = 2 m
Given that speed of the two pieces
u₁=- 25 j m/s
u₂ =- 25 i m/s
Lets take the speed of the third mass = v m/s
From linear momentum conservation
Pi= Pf
0 = m₁u₁+m₂u₂ + m₃ v
0 = -25 j m - 25 i m + 2 m v
2 v=25 j + 25 i m/s
v=12.5 i + 12.5 j m/s
Therefore the speed of the third mass will be v=12.5 i + 12.5 j m/s
