A world-class sprinter can burst out of the blocks to essentially top speed (of about 11.5 m/s) in the first 15.0 m of the race.
1 answer:
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Answer:
By using the acceleration formula,
Answer:
82.76m
Explanation:
In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.
You use the formula for the circumference of the steel ring:
(1)
C: circumference of the ring = 40000 km = 4*10^7m (you assume the circumference is the length of the steel tube)
you solve for r in the equation (1):
Next, you use the following formula to calculate the change in the length of the tube, when its temperature increases by 1°C:
(2)
L: final length of the tube = ?
Lo: initial length of the tube = 4*10^7m
ΔT = change in the temperature of the steel tube = 1°C
α: thermal coefficient expansion of steel = 13*10^-6 /°C
You replace the values of the parameters in the equation (2):
With the new length of the tube, you can calculate the radius of a ring formed with the tube. You again solve the equation (1) for r:
Finally, you compare both r and r' radius:
r' - r = 6,366,280.484m - 6,366,197.724m = 82.76m
Hence, the distance to the ring from the ground is 82.76m
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.