15) A 328-kg car moving at 19.1 m/s in the +x direction hits from behind a second car moving at 13 m/s in the same direction. If

Answer:

All observers are accurate.

Explanation:

This situation reflects a matter of reference frames regarding the book's motion as perceived by different observers.

From their distinct frames of reference, each observer's perspective is valid.

Observer A is in an inertial reference frame.

Observers capable of explaining the book's behavior and its relationship to the car through the interplay of forces and changes in velocity are classified as being in inertial reference frames.

Observer A's observations illustrate this, for she pointed out the relative motion between the book and the car, indicating her position in an inertial reference frame.

Likewise, observers in these inertial reference frames can elucidate object velocity changes based on the forces affecting them from other objects.

This is exemplified by observer B, who notes the car's force impacting the book's velocity.

Observer C occupies a non-inertial reference frame, as Newton's laws of motion do not apply. This scenario arises within non-inertial frames.

Answer:

1)

2)

Explanation:

Projectile Motion

When an object is projected near the surface of the Earth at an angle to the horizontal, it follows a trajectory known as a parabola. The only force acting on it (ignoring wind resistance) is gravity, affecting the vertical axis.

The height of a projectile can be calculated using

where represents the initial height from ground level, is the vertical component of the initial velocity, and t is the elapsed time.

The vertical speed component is expressed as

1) To proceed, we will determine the initial vertical velocity component since we lack sufficient data to calculate the absolute value of .

The peak height is attained when , which allows us to compute the time to reach that height.

Solving for

Thus, the maximum height reached is

We know this value is equal to 8 meters

Continuing with the calculations for

Substituting known values yields

2) At t=1.505 seconds, the ball is positioned above Julie’s head; we can calculate

T= 24.5 feet per second. That’s the speed it attains just before hitting the ground.

To respond to the previous question:
<span>Q = heat quantity (kJ) </span>
<span>cp = specific heat (kJ/kg.K) = 4.187 kJ/kgK </span>
<span>m = weight (kg) </span>
<span>dT = temperature change between hot and cold water (K). Note: dt in °C is identical to dt in Kelvin </span>

<span>Q = 100kg * (4.187 kJ/kgK) * 15 K </span>
<span>Q = 6,280.5 KJ = 6,280,500 J = 1,501,075.5 cal</span>