If my cylinder of air lasts 60 minutes while I am at the surface breathing normally, assuming all else is the same, how long wil

Answer:

A) denotes the resultant velocity of cart B post-collision.

B)

C)

D)

E)

F) Yes, kinetic energy remains conserved in this situation because both colliding bodies have identical mass.

G) Yes, momentum is conserved in every elastic collision.

Explanation:

Given:

• mass of car A,

• mass of car B,

• initial velocity of car A,

• final velocity of car A,

A)

The question mentions the cars experience an elastic collision:

By applying momentum conservation principles:

denotes the resulting velocity of cart B after collision.

B)

Initial kinetic energy of cart A:

C)

Initial kinetic energy of cart A:

D)

The final kinetic energy of cart A:

E)

The final kinetic energy of cart B:

F)

Yes, kinetic energy is conserved in this case due to both masses being identical in the collision.

G)

Indeed, momentum is consistently conserved in elastic collisions.

Response:

The ball remained airborne for 3.896 seconds

Explanation:

Given that

g = 9.8 m/s², representing gravitational acceleration,

If the angle of launch is 45°, the horizontal range will be maximized.

Both horizontal and vertical launch velocities are equal, each equating to

v_h  =  v cos θ

v_h  =  27 × cos 45°

         = 19.09 m/s.

The duration to reach maximum height is half of the flight time.

v = u + at   ∵ v = 0 (at maximum height)

19.09 - 9.8 t₁ = 0

t₁ = 1.948 s

The total time in the air equals twice the time to reach maximum height

2 t₁ = 3.896 s

The horizontal distance covered is

D = v × t

D = 3.896×19.09

   = 74.375 m

The ball was in the air for 3.896 seconds

Answer:

Explanation:

The distance between the electrodes is denoted as d.

The kinetic energy of the electron is represented as Ek when the electrodes are positioned at a distance of "d" apart.

Our goal is to determine the kinetic energy when they are separated by a distance of d/3.

K.E = ½mv²

It’s important to note that the mass remains constant; only velocity varies.

Additionally,

K.E = Work done by the electron

K.E = F × d

K.E = W = ma × d

Assuming constant acceleration

Hence, m and a are fixed,

therefore,

K.E is directly related to d

Thus, as d increases, K.E increases, and conversely, when d decreases, K.E decreases.

Consequently,

K.E_1 / d_1 = K.E_2 / d_2

With K.E_1 equating to E_k

and d_1 being d

while d_2 is represented as d/3

This leads to K.E_2 = K.E_1 / d_1 × d_2

Thus, K.E_2 = E_k × ⅓d / d

Finally,

K.E_2 = ⅓E_k

Therefore, the resultant kinetic energy is one third of the original E_k