Two frictionless lab carts start from rest and are pushed along a level surface by a constant force. Students measure the magnit

Answer;

• Heat is generated
• Light is emitted
• Bubbles form

Explanation;

• Chemical reactions are characterized by the interaction of two or more substances, leading to the formation of a new substance, and are typically irreversible.
• Indicators of chemical reactions include gas production, release of energy in the form of light or heat, formation of precipitates, and color changes.

As the plane heads toward Halifax, the wind speed supports the flight path

resulting in an overall improved speed

Conversely, during the return trip, the wind will resist the plane's motion, decreasing the net speed

The total journey lasts 13 hours

of which 2 hours was dedicated to the mathematics discussion

Consequently, the total flight time is 13 - 2 = 11 hours

Now we apply the formula to calculate the time for traveling to Halifax

Time needed to return

Let’s look at the total time

By solving the derived quadratic equation

the plane's speed calculates to 550 mph

Answer:

ΔL = MmRgt / (2m + M)

Explanation:

The system starts from rest, so the change in angular momentum correlates directly to its final angular momentum.

ΔL = L − L₀

ΔL = Iω − 0

ΔL = ½ MR²ω

To determine the angular velocity ω, begin by drawing a free body diagram for both the pulley and the block.

For the block, two forces act: the weight force mg downward and tension force T upward.

For the pulley, three forces are present: weight force Mg down, a reaction force up, and tension force T downward.

For the sum of forces in the -y direction on the block:

∑F = ma

mg − T = ma

T = mg − ma

For the sum of torques on the pulley:

∑τ = Iα

TR = (½ MR²) (a/R)

T = ½ Ma

Substituting gives:

mg − ma = ½ Ma

2mg − 2ma = Ma

2mg = (2m + M) a

a = 2mg / (2m + M)

The angular acceleration of the pulley is:

αR = 2mg / (2m + M)

α = 2mg / (R (2m + M))

Finally, the angular velocity after time t is:

ω = αt + ω₀

ω = 2mg / (R (2m + M)) t + 0

ω = 2mgt / (R (2m + M))

Substituting into the previous equations gives:

ΔL = ½ MR² × 2mgt / (R (2m + M))

ΔL = MmRgt / (2m + M)

Answer:

a)

b)

c)

Explanation:

According to the problem, the distance from the building where the ball hits is 16m, and its final elevation exceeds the initial height by 8m.

With this information, we can compute the ball’s starting speed.

a) Let's first assess the horizontal trajectory.

(1)

This gives us our initial equation.

Next, we need to examine the vertical trajectory.

Utilizing in our first equation (1)

Now let’s solve for t.

The ball takes two seconds to reach the adjacent building, allowing us to compute its initial speed.

b) To determine the velocity magnitude just before impact, we must calculate both x and y components.

The computed velocity magnitude is:

c) The ball's angle is:

Answer:

H = 109.14 cm

Explanation:

Given,                                                            

Assume that the total energy equals 1 unit.                                

Energy remaining after the first collision = 0.78 x 1 unit

Balance after the first impact = 0.78 units

Remaining energy after the second impact = 0.78 ^2 units

Balance after the second impact = 0.6084 units

Remaining energy after the third impact = 0.78 ^3 units

Balance after the third impact = 0.475 units

The height reached after the third collision is equivalent to the remaining energy.

Let H denote the height achieved after three bounces.

0.475 (m g h) = m g H                  

H = 0.475 x h                                    

H = 0.475 x 2.3 m                          

H = 1.0914 m                      

H = 109.14 cm