A diver explores a shallow reef off the coast of Belize. She initially swims d1 = 74.8 m north, makes a turn to the east and con

Answer:

a) The ball's velocity just prior to hitting the ground measures -6.3 m/s

b) The ball's velocity right after bouncing off the ground registers at 3.1 m/s

c) The average acceleration's magnitude is 470 m/s², and its direction is upward, forming a 90º angle with the ground.

Explanation:

To begin, let’s assess the time it takes for the ball to reach the floor:

The equation outlining the ball's position is:

y = y0 + v0 * t + 1/2 g * t²

Where:

y = position at given time t

y0 = initial position

v0 = initial velocity

t = time

g = acceleration triggered by gravity

We establish the ground as the reference origin.

a) Since the ball is released rather than thrown, the initial velocity v0 is 0. The direction of acceleration is downward, directed towards the origin; thus, “g” is treated as negative. When the ball contacts the ground, its position will be 0. Therefore:

0 = 2.0 m + 0 m/s *t - 1/2 * 9.8 m/s² * t²

-2.0 m = -4.9 m/s² * t²

t² = -2.0 m / - 4.9 m/s²

t = 0.64 s

The motion equation for a falling body is:

v = v0 + g * t      where "v" denotes the velocity

Since v0= 0:

v = g * t = -9.8 m/s² * 0.64 s = -6.3 m/s

b) The pebble's speed reaches 0 during its maximum height. To find the time taken for the pebble to achieve that height, we can use the velocity equation and then substitute that time in the position equation to derive the initial velocity:

v = v0 + g * t

0 = v0 + g * t

-v0/g = t

Replacing t in the position equation, knowing the maximum height is 1.5 m:

y = y0 + v0 * t + 1/2* g * t² y = 1.5 m y0 = 0 m t = -v0/g

1.5 m = v0 * (-v0/g) + 1/2 * g (-v0/g)²

1.5 m = - v0²/g - 1/2 * v0²/g

1.5 m = -3/2 v0²/g

1.5 m * (-2/3) * g = v0²

1.5 m * (-2/3) * (-9.8 m/s²) = v0²

v0 = 3.1 m/s

c) The average acceleration can be determined by:

a = final velocity - initial velocity / time

a = 3.1 m/s - (-6.3 m/s) / 0.02 s = 470 m/s²

The direction of the acceleration is upward, perpendicular to the ground.

The vector average acceleration will be:

a = (0, 470 m/s²) or (470 m/s² * cos 90º, 470 m/s² * sin 90º)

Complete Question

An aluminum "12 gauge" wire measures a diameter of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field E in the wire varies over time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is recorded in seconds.

At time 5 seconds, I = 1.2 A.

We need to find the charge Q traveling through a cross-section of the conductor from time 0 to time 5 seconds.

Answer:

The charge is  

Explanation:

The question indicates that

    The wire’s diameter is  

     The radius of the wire is  

     Aluminum's resistivity is

       The electric field variation is described as

The charge is effectively given by the equation

Where A is the area expressed as

 Thus,

Therefore

By substituting values

The question states that t =  5 seconds