The surfaces between a 14 kg block, the 16 kg wedge and between the 16 kg wedge and the horizontal plane are smooth (without friction). the acceleration of gravity is 9.8 m/s^2. a block is released on the inclined plane (top side of the wedge).

what is the force f which must be exerted on the 16 kg block in order that the 14 kg block does not move up or down the plane?

If the wedge has an angle with the horizontal plane of α, the required force is F = 294N·tan(α)

Explanation:

To solve this problem you have to analyze the block and the wedge separately and like a single body. If there is no relative movement between them, that means that both have the same acceleration a*.

Knowing that we analyze the block first:

We can use a non-inertial system for the block. As seen in the image, the total acceleration in the NIS is 0 m/s². Therefore we applied the Newton equations:

Y) Ny - Wb = 0 ⇒ N cos(α) - 14kg·g=0  ⇒ N = 137.2N / cos(α)

X) F* - Nx = 0 ⇒ a*·14kg - 14kg·N sin(α) ⇒ a* = g·tan(α)

For the wedge, we can use an inertial system:

X) F - N´x = 16kg·a*  ⇒ F - 137.2N·tan(α) = 156.8N·tan(α)

F = 294N·tan(α)

1 - wave length

2 - amplitude

3 - crest

4 - trough

explanation:

duh.

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.