Answer: The bottom of the ladder is moving at 3.464ft/sec
Explanation:
The question defines a right angle triangle. Therefore using pythagorean
h^2 + l^2 = 10^2 = 100 ...eq1
dh/dt = -2ft/sec
dl/ dt = ?
Taking derivatives of time in eq 1 on both sides
2hdh/dt + 2ldl/dt = 0 ....eq2
Putting l = 5ft in eq2
h^ + 5^2 = 100
h^2 = 25 = 100
h Sqrt(75)
h = 8.66 ft
Put h = 8.66ft in eq2
2 × 8.66 × (-2) + 2 ×5 dl/dt
dl/dt = 17.32 / 5
dl/dt = 3.464ft/sec
Answer:
A driver.
Explanation:
Using a driver while at least 350 yds away is better than using a iron, because it will be a waste of the par 4 as it is not as powerful as the driver.
We will apply the Newton's second Law so the we will be able to find the acceleration.
F (tot) = ma
a = F(tot) / m
a = 32.0 N / 65.0 kg = 0.492 m/s^2
Approximately 0.492 m/s^2 is her initial acceleration if she is initially stationary and wearing steel-bladed skates.
