When the object is at the focal point the angular magnification is 2.94.
Angular magnification:
The ratio of the angle subtended at the eye by the image formed by an optical instrument to that subtended at the eye by the object when not viewed through the instrument.
Here we have to find the angular magnification when the object is at the focal point.
Focal length = 6.00 cm
Formula to calculate angular magnification:
Angular magnification = 25/f
= 25/ 8.5
= 2.94
Therefore the angular magnification of this thin lens is 2.94
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Answer:
Artificial weight = 70.27 N = 15.80 lbs
Explanation:
The earth weight of the astronaut = 160 lbs = 711.72 N
The weight on earth = m × g(earth)
g(earth) = 9.8 m/s²
711.72 = m × 9.8
m = (711.72/9.8)
m = 72.62 kg
But at the space station, the space station rotates once every 70 s to create an artificial radial acceleration that creates a radial gravity pulling the objects on the space station towards the centre of that space station.
radial acceleration = α = (v²/r)
v = rw,
α = (rw)²/r
α = rw²
r = radius of rotation = 120 m
w = angular velocity = (2π/70) (it completes 1 rotation, 2π radians, in 70 s)
w = 0.0898 rad/s
α = 120 × (0.0898²)
α = 0.968 m/s²
Artificial weight = (mass of astronaut) × (Radial acceleration) = 72.62 × 0.968
Artificial weight = 70.27 N = 15.80 lbs
Hope this Helps!!!
Answer:
The correct solution is "1.2 m".
Explanation:
The given values are:
Wavelength of waves,
λ = 1.5 m
Speed of waves on surface,
V = 2 m/sec
Speed of waves in water,
V₁ = 1.6 m/sec
As we know,
⇒ 
or,
⇒ 
On substituting the values, we get
⇒ 
⇒ 
hence,
⇒ 
On substituting the values, we get
⇒ 
⇒ 
Answer:
0. 153m
Explanation:
displacement= velocity/ time
S = 2.3 m/s / 15s
S = 0. 153m
