Initially, mg = kx. K = mg/x = 700/0.5x10^-3 = 1400000N/m. From second condition, applying work-energy theorem, potential enery- elastic potential energy = change in kinetic energy. Now change in kinetic energy is 0 since initial and final velocities are 0m/s. Therefore, potential energy = elastic potential energy. mgh = (1/2) * k* x^2. x^2 = 2(mg)h/k = 2 x 700 x 1.3/ 1400000. x = 0.036m. Hope it's clear.

3 0

3 years ago

3. A model rocket is launched straight upward at 58.8 m/s.

SIZIF [17.4K]

Let us assume that rocket only runs in initial energy and not using its own to flying.

Also , let upward direction is +ve and downward direction is -ve .

Initial velocity , u = 58.8 m/s .

Acceleration due to gravity , $g=-9.8\ m/s^2$ .

Final velocity , v - = 0 m/s .

We know , by equation of motion .

$v^2-u^2=2gh\\\\2gh_{max}=0^2-58.8^2\\\\h_{max}=\dfrac{0^2-58.8^2}{-2\times 9.8}\\\\h_{max}= 176.4\ m$

Hence, this is the required solution .

8 0

3 years ago

A manufacturer of printed circuit boards has a design capacity of 1,000 boards per day. the effective capacity, however, is 700

Alekssandra [29.7K]

A manufacturer of printed circuit boards has a design capacity of 1,000 boards per day. the effective capacity, however, is 700 boards per day. recently the production facility has been producing 950 boards per day. The design capacity utilization is (950/100) *100 = 95 %

3 0

3 years ago