Calculate the partial pressure of ozone at 441 ppb if the atmospheric pressure is 0.67 atm.

The height is h = 17 10⁶ meters above the surface of Mars. To determine this, we apply Newton's second law according to the universal law of gravitation, represented by F = m a. The centripetal acceleration a is expressed as v² / r. Applying the gravitational force we have G m M / r² = m v² / r. Given that the speed of the object remains constant, we derive v from d / t, where d is the circumference and t is the orbital period. Substituting gives us d = 2π r and v = 2π r / T. Replacing these values leads to the equation G M / r² = (4π² r² / T) / r, so r³ = G M T² / 4π². Converting time into SI units, T = 24.66 h converts to 88776 seconds. Ultimately, the computed value of r is 2,045 10⁶ m, and after subtracting Mars’ radius of 3.39 10⁶ m, we find the height h to be 17 10⁶ m.

The elevator's acceleration is 0.422 m/s². To clarify the solution: By applying Newton's Law, the net forces in the motion's direction equal the mass multiplied by the acceleration. The forces comprise 460 N in the motion's direction and the person's weight acting in the opposite direction... The weight is determined by the mass and gravity's acceleration (W = mg). Here m = 45 kg and g = 9.8 m/s², leading to W = 441 N. With the scale indicating 460 N, we apply F - W = ma, yielding 19 = 45 a. Dividing both sides by 45 gives a = 0.422 m/s².

Explanation:

The term 'collision' refers to the interaction between two objects. There are two distinct types of collisions: elastic and inelastic.

In this scenario, two identical carts are heading towards each other at the same speed, resulting in a collision. In an inelastic collision, the momentum is conserved before and after the incident, but kinetic energy is lost.

After the event, both objects combine and move together at a single velocity.

The graph representing a perfectly inelastic collision is attached, illustrating that both carts move together at the same speed afterward.

Answer:

Explanation:

We start with the fact that

Initially, the spacecraft was at rest, u = 0

The final velocity of the rocket is given as v = 11 m/s

The distance that the rocket covers during acceleration is given as d = 213 m

We seek to determine the acceleration that the occupants of the spacecraft experience during launch, which is derived from the principles of kinematics. By applying the

third motion equation we can find the acceleration.

Thus, the acceleration felt by those inside the spacecraft is .

1) false

2) cross-sectional area of the conductor