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2 years ago

A ball is thrown horizontally from a 16 m -high building with a speed of 2.0 m/s .

Physics

1 answer:

BigorU [14]2 years ago

4 0

Answer:

3.61 m

Explanation:

First, find how long it takes to land.

Given (in the y direction):

Δy = 16 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

(16 m) = (0 m/s) t + ½ (9.8 m/s²) t²

t = 1.81 s

Given (in the x direction):

v₀ = 2.0 m/s

a = 0 m/s²

t = 1.81 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (2.0 m/s) (1.81 s) + ½ (0 m/s²) (1.81 s)²

Δx = 3.61 m

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A system of two objects has ΔKtot = 6 J and ΔUint = -5 J. Part A How much work is done by interaction forces? Express your answe

Elina [12.6K]

A) +5 J

B) +1 J

Explanation:

The internal forces (interaction forces) acting on a system do not change the mechanical energy (sum of potential and kinetic energy) of the system.

However, these forces are responsible for converting the energy from one form into another; the work done by these forces is equal to the amount of energy converted from one form into the other.

In this problem, we have:

$\Delta U=-5 J$ is the loss in potential energy of the system

$\Delta K=+6 J$ is the gain in kinetic energy of the system

By looking at these numbers, this means that the internal forces have converted 5 J of energy from potential energy into kinetic energy (while the additional +1 J missing is due to external forces, as explained in part B).

Therefore, the work done by internal forces is

W = +5 J

First of all, we calculate the change in mechanical energy of the system.

The mechanical energy of a system is the sum of its kinetic energy (K) and its potential energy (U):

$E=K+U$

So, the change in mechanical energy is equal to the sum of the changes of kinetic energy and the changes of potential energy:

$\Delta E= \Delta K + \Delta U$

In this problem:

$\Delta K=+6 J$

$\Delta U=-5 J$

So, the change in mechanical energy is:

$\Delta E=+6+(-5)=+1 J$

According to the work-energy theorem, the work done by external forces on a system is equal to the change in mechanical energy of the system: therefore in this case, the work done by external forces is

$W=\Delta E=+1 J$

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3 years ago

Darryl throws a basketball at the gym floor. The ball bounces once on the floor and comes to rest in his coach’s hands. At which

mina [271]

Answer:

when the ball is at rest in his coach's hands.

Explanation:

The forces on the basketball are balanced when the basketball is not experiencing any acceleration. This happens when the ball is in his coach's hand: in fact, at that moment the ball is at rest, so it means that its acceleration is zero. According to Newton's second law, this also mean that the net force on the basketball is zero, so the forces on the ball are balanced:

$F=ma$