Question

Old Faithful geyser in Yellowstone National Park shoots water every hour to a height of 40.0 m. With what velocity does the water leave the ground? (Disregard air resistance. g 9.81 m/s) a. 7.00 m/s c. 28.0 m/s d. 14.0 m/s b. 19.8 m/s 30.) A moderate force will break an egg. However, an egg dropped on the road usually breaks, while one dropped on the grass usually does not break because for the egg dropped on the grass, a. the b. the change in momentum is greater. time interval for stopping is greater. the change in momentum is less the time interval for stopping is less. . c. d.

Answer 1

Answer:

Option C is the answer = 28.0m/s

Option B

Explanation:

Applying the principle of conservation of energy which takes into consideration the total energy of the system , i.e the summation of the kinetic energy, potential energy. the conservation of energy deals with total energy of a system that changes by the transferring of energy to the system or from the system.

from the question,

height of water h = 40m

g = acceleration due to gravity = 9.81m/seconds square

from the addition of the kinetic energy(energy due to motion) and potential energy(energy by virtue of the position of an object) = initial (Kinetic + potential) = final ( kinetic + potential)

(mgh + 1/2mv -square)initial =( mgh + 1/2 mv -square) final

neglecting or eliminating the mass for both the initial and final,

V-square = 2gh = 2 x 9.81 x 40 = 784.8 = 28.01m/s

Answer 2

Answer:

a) $v \approx 28.010\,\frac{m}{s}$, b) B. Time interval for stopping is greater.

Explanation:

a) The initial velocity is found with the help of the Principle of Energy Conservation:

$K = U_{g}$

$m\cdot g \cdot h = \frac{1}{2}\cdot m \cdot v^{2}$

$v = \sqrt{2\cdot g \cdot h}$

$v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (40\,m)}$

$v \approx 28.010\,\frac{m}{s}$

b) According to the Impact Theorem, the egg does not break on the grass since contact force is significantly lower and time interval is greater. Hence, the right answer is option B.