Daria was swimming in a friend’s pool yesterday, when she saw that a fly had landed in the water about 5 feet away from her. She

Answer:

2.87 m

Explanation:

Given parameters:

Mass of the ball (m) = 60 g = 0.06 kg

Height of the tube (h) = 0.70 m

Force applied on the ball by compressed air (F) = 3.0 N

Initial velocity of the ball (u) = 0 m/s (Assumed)

Final velocity of the ball at the tube's exit (v) =?

Acceleration of the ball (a) =?

The ball's weight is derived from multiplying mass and gravity. Therefore,

Weight (W) =

Thus, the total force acting on the ball equals the net of upward force minus the weight.

Net force = Air force - Weight

According to Newton's second law, net force equals the mass multiplied by acceleration.

Acceleration (a) is calculated as 40.2 m/s².

Using the motion equation, we find:

Let’s denote the maximum height achieved as 'H'.

Next, we apply the principle of energy conservation from the pipe's peak to the maximum height.

A decrease in kinetic energy equals an increase in potential energy.

Substituting the values, we solve for 'H', yielding:

Hence, the ball ascends to a height of 2.87 m above the top of the tube.

Answer:

Explanation:

To approach this problem, we need to understand two key concepts.

First, the gravitational force on an object in orbit equals its mass multiplied by centripetal acceleration.

Secondly, Newton's law of universal gravitation defines the force between two masses: Fg = mMG/r², where Fg denotes gravitational force, m and M signify the masses, G represents the gravitational constant, and r indicates the distance separating the two masses.

Thus:

Fg = m v²/r

mMG/r² = m v²/r

v² = MG/r

Potential energy for each planet is expressed as:

PE = mgr = m (MG/r²) r = mMG/r

Kinetic energy for each planet is computed as:

KE = 1/2 mv² = 1/2 m (MG/r) = 1/2 mMG/r

Total mechanical energy is calculated as:

ME = PE + KE = 3/2 mMG/r

Since both planets share the same mass, the only variable is their orbital radius. Consequently, Planet A, with a smaller radius, possesses greater potential, kinetic, and mechanical energy.

Answer:

155.38424 K

2.2721 kg/m³

Explanation:

= Reservoir pressure = 10 atm

= Reservoir temperature = 300 K

= Exit pressure = 1 atm

= Exit temperature

= Specific gas constant = 287 J/kgK

= Specific heat ratio = 1.4 for air

Assuming isentropic flow

Flow temperature at exit is 155.38424 K

Density at exit can be derived using the ideal gas equation

Flow density at exit measures 2.2721 kg/m³