A 1.10 kg block is attached to a spring with spring constant 13.5 n/m . while the block is sitting at rest, a student hits it wi
1 answer:
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The Image distance and Magnification of The Image will be 30 cm and 3.
<h3>What is focal length?</h3>The focal length of the lens, which is often expressed in millimeters, is the distance between the lens and the image sensor when the subject is in focus.
Given data;
Focal length,f=?
Image distance,v=?
Object distance,u= 10 cm
Magnification,m= 2.85
The focal length is half of the radius;
f=R/2
f=30 Cm/2
f= 15 Cm
The mirror equation is found as;
The magnification of the lens is found as;
Hence, the image distance and magnification of The image will be 30 cm and 3.
To learn more about the focal length refer;
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The final velocity is 5.87 m/s
<u>Explanation:</u>
Given-
mass, = 72 kg
speed, = 5.8 m/s
,
= 45 kg
,
= 12 m/s
Θ = 60°
Final velocity, v = ?
Applying the conservation of momentum:
X
+
X
= (
+
) v
72 X 5.8 + 45 X 12 X cos 60° = (72 + 45) v
v = 417.6 + 540 X
v = 417.6 +
v = 5.87 m/s
The final velocity is 5.87 m/s
Answer
given,
wavelength (λ)= 500 n m
thickness of film= 10⁻⁴ cm
refractive index = μ = 1.375
distance traveled is double which is equal to 2 x 10⁻⁴ cm
a) Number of wave
N = 2.91
N = 3
b) phase difference is equal to
Reflection from the first surface has a 180° (½λ) phase change.
There is no phase change for the 2nd surface reflection and there is no phase difference for the 2nd wave having traveled an exact whole number of waves.
net phase difference =
= 270°