The number of gallons of water in the tank at t=10 is
... W(10) = 160,000 -10(8000 -10) = 80100
The number of gallons of water in the tank at t=10.5 is
... W(10.5) = 160,000 -10.5(8000 -10.5) = 76110.25
The rate of change over the interval is
... (W(10.5) - W(10))/(10.5 - 10) = (76110.25 - 80100)/(0.5) = -7979.5
The average rate of change in the number of gallons of water in the tank over the interval is -7979.5 gal/min.
The sign is negative, so the amount of water is decreasing.
The digit in the hundred thousands place is the 2
Answer:
Step-by-step explanation:
Given
Required
The area of the shaded sector
This is calculated as:
This gives
Answer:
A. 5.16 s.
B. 5.66 s.
Step-by-step explanation:
A.
For a simple harmonic motion,
T = 2pi (sqrt * (l/g))
Given:
L1 = 3 cm
T1 = 4 s
L2 = 5 cm
T2 = ?
4 = 2pi*sqrt(3/g)
g = 7.4
At, L2,
T2 = 2pi*sqrt(5/7.4)
= 5.16 s.
B.
M1 = M1
M2 = 2*M1
For a simple harmonic motion,
T = 2pi (sqrt * (m/k))
4 = 2pi (sqrt * (M1/k))
M1/k = 0.405
Inputting the above values,
T2 = 2pi (sqrt * (2*M1/k))
= 2pi (sqrt * (2 * 0.405))
= 5.66 s.
