An Aluminum bar is 2 m long at a temperature of 20 degrees Celsius. What will it be at 100 degrees Celsius?
2 answers:
Answer:
10 meters
Step-by-step explanation:
Let at 100 degrees Celsius, the aluminum bar is x m long.
We have been given that aluminum bar is 2 m long at a temperature of 20 degrees Celsius.
Thus, we have the equation
Solve the equation for x
Thus, at 100 degree Celsius, the aluminium bar is 10 meters long.
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1) Answer:
<u>Explanation:</u>
The equation of an ellipse is: ; where (h, k) is the center, "a" is the x-radius, and "b" is the y-radius.
Center Radius
x-axis: (10 + -10)/2 = 0 10 - 0 = 10
- h = 0 a = 10
y-axis: (2 + -2)/2 = 0 2 - 0 = 2
- k = 0 b = 2
Now, input the values into the equation:
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2) Answer:
<u>Explanation:</u>
Vertices are: (0, 1) and (0, -5) ------> x-values are the same, y = 1, -5
Covertices are: (-1, -2) and (1, -2) ----> y-values are the same, x = -1, 1
Center Radius
x-axis: (-1 + 1)/2 = 0 1 - 0 = 1
- h = 0 a = 1
y-axis: (1 + -5)/2 = -2 1 - (-2) = 3
- k = -2 b = 3
Now, input the values into the equation:
24 degrees per second is 360 degrees in 15 seconds (pretty fast for a Ferris wheel!).
a) The distance above the ground can be modeled using a cosine function.
.. h(t) = 7 -6cos(2πt/15)
b) h(t) = 10
.. 10 = 7 -6cos(2πt/15)
.. (10 -7)/-6 = cos(2πt/15)
.. 2πt/15 = π ±π/3 +2kπ
. t = 7.5*((1 ±1/3) +2k) . . . . . . . . seconds; k = any integer
You are 10 meters above the ground at t=5 seconds and every 15 seconds thereafter, as well as at t=10 seconds and every 15 seconds thereafter.
a) The distance above the ground can be modeled using a cosine function.
.. h(t) = 7 -6cos(2πt/15)
b) h(t) = 10
.. 10 = 7 -6cos(2πt/15)
.. (10 -7)/-6 = cos(2πt/15)
.. 2πt/15 = π ±π/3 +2kπ
. t = 7.5*((1 ±1/3) +2k) . . . . . . . . seconds; k = any integer
You are 10 meters above the ground at t=5 seconds and every 15 seconds thereafter, as well as at t=10 seconds and every 15 seconds thereafter.