Answer:
24.71 mm
Explanation:
Distance is proportional to focal length, so
d∝f
which means
Magnification of first lens
and
Similarly, magnification of second lens
and
From the above equations we get
and
which means,
and
So, we get
∴ Focal length should this camera's lens is 24.71 mm
Answer: 1 m/s
Explanation:
We have an object whose position is given by a vector, where the components X and Y are identified by the unit vectors
and
(where each unit vector is defined to have a magnitude of exactly one):
On the other hand, velocity is defined as the variation of the position in time:
This means we have to derive :
This is the velocity vector
And when the velocity vector is:
This is the velocity vector at 2 seconds
However, the solution is not complete yet, we have to find the module of this velocity vector, which is the speed :
Finally:
This is the speed of the object at 2 seconds
Answer:
the expected distance is 4.32 m
Explanation:
given data
half life time = 1.8 × s
speed = 0.8 c = 0.8 × 3 ×
to find out
expected distance over
solution
we know c is speed of light in air is 3 × m/s
we calculate expected distance by given formula that is
expected distance = half life time × speed .........1
put here all these value
expected distance = half life time × speed
expected distance = 1.8 × × 0.8 × 3 ×
expected distance = 4.32
so the expected distance is 4.32 m
Answer:
x = 0.0685 m
Explanation:
In this exercise we can use the relationship between work and energy conservation
W = ΔEm
Where the work is
W = F x
The energy can be found in two points
Initial. Just when the block with its spring spring touches the other spring
Em₀ = K = ½ m v²
Final. When the system is at rest
=
= ½ k₁ x² + ½ k₂ x²
We can find strength with Newton's second law
∑ F = F - fr
Axis y
N- W = 0
N = W
The friction force has the equation
fr = μ N
fr = μ W
The job
W = (F – μ W) x
We substitute in the equation
(F - μ W) x = ½ m v² - (½ k₁ x² + ½ k₂ x²)
½ x² (k₁ + k₂) + (F - μ W) x - ½ m v² = 0
We substitute values and solve
½ x² (20 + 40) + (15 -0.2 2) x - ½ (2/32) 6² = 0
x² 30 + 14.4 x - 1,125 = 0
x² + 0.48 x - 0.0375 = 0
We solve the second degree equation
x = [-0.48 ±√(0.48 2 + 4 0.0375)] / 2
x = [-0.48 ± 0.617] / 2
x₁ = 0.0685 m
x₂ = -0.549 m
The first result results from compression of the spring and the second torque elongation.
The result of the problem is x = 0.0685 m