A cleaver physics professor wants to create a situation where a block starts from rest at the top of a 31.0° inclined plane and
1 answer:
You might be interested in
Answer:
24.348mm
Explanation:
NB: I'll be attaching pictures so as to depict missing mathematical expressions or special characters which are not easily found on keyboards
K = d / €^n
Note : d represents the greek alphabet epsilion.
K = 345 / 0.02⁰.²² = 816mPa
The true strain based upon the stress of 414mPa =
€= (€/k)^1/n = (414/816)¹/⁰.²² = 0.04576
However the true relationship between true strain and length is given by
€ = ln(Li/Lo)
Making Li the subject of formula by rearranging,
Li = Lo.e^€
Li = 520e⁰.⁰⁴⁵⁷⁶
Li = 544.348mm
The amount of elongation can be calculated from
Change in L = Li - Lo = 544.348 - 520 change in L = 24.348mm.
A. 6.67 cm
The focal length of the lens can be found by using the lens equation:
where we have
f = focal length
s = 12 cm is the distance of the object from the lens
s' = 15 cm is the distance of the image from the lens
Solving the equation for f, we find
B. Converging
According to sign convention for lenses, we have:
- Converging (convex) lenses have focal length with positive sign
- Diverging (concave) lenses have focal length with negative sign
In this case, the focal length of the lens is positive, so the lens is a converging lens.
C. -1.25
The magnification of the lens is given by
where
s' = 15 cm is the distance of the image from the lens
s = 12 cm is the distance of the object from the lens
Substituting into the equation, we find
D. Real and inverted
The magnification equation can be also rewritten as
where
y' is the size of the image
y is the size of the object
Re-arranging it, we have
Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.
Also, the sign of s' tells us if the image is real of virtual. In fact:
- s' is positive: image is real
- s' is negative: image is virtual
In this case, s' is positive, so the image is real.
E. Virtual
In this case, the magnification is 5/9, so we have
which can be rewritten as
which means that s' has opposite sign than s: therefore, the image is virtual.
F. 12.0 cm
From the magnification equation, we can write
and then we can substitute it into the lens equation:
and we can solve for s:
G. -6.67 cm
Now the image distance can be directly found by using again the magnification equation:
And the sign of s' (negative) also tells us that the image is virtual.
H. -24.0 cm
In this case, the image is twice as tall as the object, so the magnification is
M = 2
and the distance of the image from the lens is
s' = -24 cm
The problem is asking us for the image distance: however, this is already given by the problem,
s' = -24 cm
so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.