Great experiment ! Everybody should try it if they can get the equipment.
It demonstrates a lot of things that are very hard to explain in words.
I hope the students remembered to tilt the axis of the globe. If they didn't,
and instead kept it straight up and down, then each city had pretty much
the same amount of bulb-light all the way around, and there were no seasons.
If the axis of the globe was tilted, then City-D had the least variation in
seasons. City-D is only 2° from the equator, so the sun is more direct
there all year around than it is at any of the others.
The model bridge captures all the structural attributes of the real bridge, at a reduced scale.
Part a.
Note that volume is proportional to the cube of length. Therefore the actual bridge will have 100^3 = 10^6 times the mass of the model bridge.
Because the model bridge weighs 50 N, the real bridge weighs
(50 N)*10^6 = 50 MN.
Part b.
The model bridge matches the structural characteristics of the actual bridge.
Therefore the real bridge will not sag either.
Answer:
The final velocity of the runner at the end of the given time is 2.7 m/s.
Explanation:
Given;
initial velocity of the runner, u = 1.1 m/s
constant acceleration, a = 0.8 m/s²
time of motion, t = 2.0 s
The velocity of the runner at the end of the given time is calculate as;
where;
v is the final velocity of the runner at the end of the given time;
v = 1.1 + (0.8)(2)
v = 2.7 m/s
Therefore, the final velocity of the runner at the end of the given time is 2.7 m/s.
Answer:
(a) t = 1.14 s
(b) h = 0.82 m
(c) vf = 7.17 m/s
Explanation:
(b)
Considering the upward motion, we apply the third equation of motion:
where,
g = - 9.8 m/s² (-ve sign for upward motion)
h = max height reached = ?
vf = final speed = 0 m/s
vi = initial speed = 4 m/s
Therefore,
<u>h = 0.82 m</u>
Now, for the time in air during upward motion we use first equation of motion:
(c)
Now we will consider the downward motion and use the third equation of motion:
where,
h = total height = 0.82 m + 1.8 m = 2.62 m
vi = initial speed = 0 m/s
g = 9.8 m/s²
vf = final speed = ?
Therefore,
<u>vf = 7.17 m/s</u>
Now, for the time in air during downward motion we use the first equation of motion:
(a)
Total Time of Flight = t = t₁ + t₂
t = 0.41 s + 0.73 s
<u>t = 1.14 s</u>
Answer:
y_red / y_blue = 1.11
Explanation:
Let's use the constructor equation to find the image for each wavelength
1 /f = 1 /o + 1 /i
Where f is the focal length, or the distance to the object and i the distance to the image
Red light
1 / i = 1 / f - 1 / o
1 / i_red = 1 / f_red - 1 / o
1 / i_red = 1 / 19.57 - 1/30
1 / i_red = 1,776 10-2
i_red = 56.29 cm
Blue light
1 / i_blue = 1 / f_blue - 1 / o
1 / i_blue = 1 / 18.87 - 1/30
1 / i_blue = 1,966 10-2
i_blue = 50.863 cm
Now let's use the magnification ratio
m = y ’/ h = - i / o
y ’= - h i / o
Red Light
y_red ’= - 5 56.29 / 30
y_red ’= - 9.3816 cm
Light blue
y_blue ’= 5 50,863 / 30
y_blue ’= - 8.47716 cm
The ratio of the height of the two images is
y_red ’/ y_blue’ = 9.3816 / 8.47716
y_red / y_blue = 1,107
y_red / y_blue = 1.11
