A graph of the net force F exerted on an object as a function of x position is shown for the object of mass M as it travels a ho

The question Ellen is likely exploring is "In what way does distance influence the gravitational force acting on objects?"

Explanation:

It is all right, so what is it that they said didn't. Fiend. Diehard d fish

La force agissant pendant 9 s et la décélération pendant 12 - 9 = 3 s.

Distance totale parcourue = 990 m

vitesse initiale u = 0

Distance parcourue pendant l'accélération

s₁ = 1/2 a 9² où a est l'accélération

= 40.5 a

vitesse finale après 9 s

v = at = 9a

pendant la décélération

v² = u² - 5 x s₂

0 = (9a)² - 5 s₂

s₂ = 16.2 a²

Distance parcourue pendant la décélération = 16.2 a²

s₁ + s₂ = 990

40.5 a + 16.2 a² = 990

16.2 a² + 40.5 a - 990 = 0

a = 6.5

Answer:

The output power of the circuit is 3 Watts.

Given:

a loss in decibels = 3 dB

Input power = 6 Watts

To find:

What is the output power?

Formula used:

Output power = Input power × loss in ratio

Solution:

3 dB loss corresponds to a ratio of 0.5

Output power can be calculated as follows:

Output power = Input power × loss in ratio

Output power = 6 × 0.5

Output power = 3 Watts

Therefore, the output power of the circuit is 3 Watts.

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.