The air temperature over a lake decreases linearly with height after sunset, since air cools faster than water. The mean molar m

Answer:

Explanation:

Provided:

mass of the steel ball

initial velocity of the ball

Final velocity of the ball (moving upwards)

The impulse given is determined by the change in the momentum of the object.

Thus, the magnitude of the Impulse is 4 N-s.

Answer:

529.15 m/s

Explanation:

h = Highest point = 70000 m

g = Gravitational acceleration = 2 m/s²

m = Sulfur's mass

Since both potential and kinetic energies are conserved

The velocity at which the liquid sulfur exited the volcano is 529.15 m/s

Answer:

A) 328 m

B) 80.22 m/s

C) 8.18 sec

Explanation:

A)

The rocket's initial acceleration is 2.25 m/s²

Time until engine failure is 15.4 s

Initial velocity u during takeoff = 0 m/s

Distance covered while the engine is functional =?

We apply Newton's laws for this calculation

S = ut + a

Here, S represents the distance traveled under the rocket's thrust

S = (0 x 15.4) + (2.25 x )

S = 0 + 266.81 m = 266.81 m

Before the engine fails, the final velocity can be found using:

v = u + at

v = 0 + (2.25 x 15.4) = 34.65 m/s

Once the engine fails, the rocket decelerates under gravitational pull at g = -9.81 m/s²  (acting downwards)

The upward initial velocity when freefall begins is v = 34.65 m/s

The final velocity is reached at peak height, where the rocket halts, therefore:

u = 0 m/s

The distance covered during this freefall will be s =?

Utilizing the equation

= + 2gs

= + 2(-9.81 x s)

0 = 1200.6 - 19.62s

-1200.6 = -19.62s

s = -1200.6/-19.62 = 61.19 m

Thus,

Maximum Height = 266.81 m  + 61.19 m =  328 m

B)

At maximum height, the rocket’s initial upward velocity drops to 0 m/s (the rocket completely stops)

The descent occurs freely under g = 9.81 m/s² (acting downwards)

The distance covered during the fall will be 328 m

Final velocity v just prior to impact =?

Applying = + 2gs

= + 2(9.81 x 328)

= 0 + 6435.36

v = = 80.22 m/s

The time taken before reaching the pad is found as follows

v = u + gt

80.22 = 0 + 9.81t

t = 80.22/9.81 = 8.18 sec

It's not only the horizontal segments of the loop that extend into the magnetic field. We can ignore the horizontal sections that dip into the magnetic fields since only the sections that are perpendicular to the magnetic field have a significant contribution to it.

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