3. A 3.0 × 10−3 kg copper penny drops a distance of 50.0 m to the ground. If 65 percent of the initial potential energy goes int

a) 19440 km/h²; b) 10 seconds.

Answer:

17 m/s south

Explanation:

Mass of the dog = 10 kg

Mass of skateboard = 2 kg

v = Combined velocity = 2 m/s

Velocity of the dog = 1 m/s

Velocity of skateboard

In this scenario, linear momentum is conserved

The speed of the skateboard post-dog jump will be 17 m/s south, considering north as the positive direction

The tension exerted in the cable amounts to T = 16653.32 N. Parameters: Cross section area A = 1.3 m² Drag coefficient CD = 1.2 Velocity V = 4.3 m/s The angle formed by the cable with the horizontal is 30 degrees. Density defined as follows: The drag force FD is determined by the equation: FD = (1/2) * ρ * V² * A * CD Calculating the drag force yields 14422.2 N acting opposite to motion. Given the cable's angle of 30 degrees with horizontal, the horizontal component contributes to the drag force calculation: T * cos(30) = F_D Thus, T = 16653.32 N.

Response:

216000 W or 216 kW

Details:

Power: This refers to the rate at which energy is utilized or consumed. The standard unit for power is the Watt (W).

In general,

Power = Energy/time

P = E/t........................ Equation 1.

However,

E = 1/2mv²..................... Equation 2

with m representing mass, and v indicating velocity.

Substituting equation 2 into equation 1,

P = 1/2mv²/t...................... Equation 3

Assuming flow rate (Q) = m/t,

Q = m/t................ Equation 4

Insert equation 4 into equation 3,

P = Qv²/2........................ Equation 5

Where Q stands for flow rate, v is velocity, and P denotes power.

Given: Q = 120 kg/s, v = 60 m/s

Plug these values into equation 5,

P = 120(60)²/2

P = 60(60)²

P = 60×3600

P = 216000 W.

Thus, the potential power generation of the water jet is 216000 W or 216 kW.