Answer:
It's 1.0000042 times longer in summer than in winter. It represents a 1.6 centimeters difference between seasons.
Explanation:
The linear coefficient of thermal expansion for steel is about 
. From the equation of linear thermal expansion, we have:
Taking the winter day as the initial, and the summer day as the final, we can take the relationship between them:
![L_{summer}=L_{winter}[1+(1.2*10^{-7}\°C^{-1})(30\°C+5\°C)]\\\\L_{summer}=(1.0000042)L_{winter}](https://tex.z-dn.net/?f=L_%7Bsummer%7D%3DL_%7Bwinter%7D%5B1%2B%281.2%2A10%5E%7B-7%7D%5C%C2%B0C%5E%7B-1%7D%29%2830%5C%C2%B0C%2B5%5C%C2%B0C%29%5D%5C%5C%5C%5CL_%7Bsummer%7D%3D%281.0000042%29L_%7Bwinter%7D)
It means that the bridge is 1.0000042 times longer in summer than in winter. If we multiply it by the length of the bridge, we obtain that the difference is of about 1.6 centimeters between the two seasons.
Answer:
The distance the log has moved by the time Ernie reaches Bur is 1.33 m.
Explanation:
give information:
The log is 3.0 m long and has mass 20.0 kg.
Burt has mass 30.0 kg; Ernie has mass 40.0 kg
Ernie has mass 40.0 kg.
to find the distance, first, we have to calculate the center of mass
X = ∑ m x /∑m
= (20 x (3/2)) + (30 x 0) + (40 x 3)/ (20+30+40)
= 150/90
= 5/3
when Ernie walk, the center of the mass is
X = (70 x 0) + (20 x (3/2))/(70 + 20)
= 30/90
= 1/3
the distance of log moved = 5/3 - 4/3 = 1.33 m
The solution for this problem is:
r = [(2.90 + 0.0900t²) i - 0.0150t³ j] m/s²
this is for t in seconds and r in meters
v = dr/dt = [0.180t i - 0.0450t² j] m/s²
tan(-36.0º) = -0.0450t² / 0.180t
0.7265 = 0.25t
t = 2.91 s is the velocity vector of the insect
Calculate its average speed in meters per second
Answer:
5.77 m/s
Explanation:
Speed= Distance/Time
Distance= 40+ half of 40= 40+20= 60 m
Time= 8.8+1.6=10.4 s
Average speed= 60/10.4=5.769230769 m/s
Approximately, the average speed is 5.77 m/s
Answer:
A - 0 N
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