A ladder 25 feet long is leaning against a house. The base of the ladder is pulled away at a rate of 2 ft/sec.
a.) How fast is the top of the ladder moving down the wall when the base of the ladder is 12 feet from the wall?
Answer:
dy/dt = -1.094ft/sec
Explanation:
Given that:
dz/dt = 0,
dx/dt = 2,
dy/dt = ?
Hence, we have the following
Using Pythagoras theorem
We have 25ft as the hypotenuse, y as the opposite or height of wall, and x as the base of the triangle
X² + y² = z²,
12² + y² = 25²,
y² = 25² - 12²
y = √481
Therefore, we have the following:
2x dx/dt + 2y dy/dt,
= 2z dz/dt,
= 12 (2) √481 dy/dt,
= √481 dy/dt = -24,
= dy/dt = -1.094ft/sec
Therefore, final answer is -1.094ft/sec
The speed stops as it hits the girls hand but speed reduces as it reaches her hand and increases at the beginning of the movement
Answer: the element with the lowest electronegativity
Explanation:
Answer:
a)Counterclockwise
b)ω=2 rad/s
Explanation:
Given that
Object at = --2 m
r= - 2 m
Velocity pf object V = - 4 m/s
We know that
r= - 2 i m
V= - 4 j m/s
- 4 j= ω x (- 2 i)
ω should be in counter clockwise (k) to satisfy the above equation.
It means that the direction of object will in counter clockwise.
4 = ω x 2
ω=2 rad/s
So object velocity at 2 m will be 2 rad/s.
