# Physics Problems

Published: 2017-11-06 12:07:14 Type of paper: Problem solving Categories: Problem solving Physics Pages: 5 Wordcount: 1110 words
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According to the lecture notes, D = 42163112 meters corresponds to T = 86164 seconds. And according to your calculations from the previous homework, D = 42163212 corresponds to T = 86164.3065. The relative errors in the distance D and the T for one revolution are Δ D relative = 0.00023717% and Δ T relative = 0.000355717%. However, their ratio is suspiciously close to

= 1.49981359129102662

Now prove the following result: if y = f(x) = C.xα is a given function with some constant given number C (whose value is not important), then a relative error of the input variable x of I% leads to a relative error in the output variable y of approximately αI%.

Let’s consider the equation T = D3/2

Multiplying the numerator by D3/2 T =

Cross multiplying T by T. = 2D3/2

Diving both sides by 2 T. /2 = 2D3/2/2

Now the equation becomes = D3/2

Let α be and D be x.

Substituting α and x into this equation = D3/2 it becomes α = x3/2

Applying logarithms to α to base x; Log xα = hence α =

2. A person stands on the ground and has a thin thread hanging from their hand, with a little stone tied to it. This person keeps the thread still until the pendulum motion of the thread with the stone has stopped. Now the thread is a non-moving line. If we extend this line downwards indefinitely, will it go through the earth center? Or will it miss the earth center? And if yes, by how many kilometers will the earth center be missed?

When the line is extended it definitely to the center of the earth without missing it. The reason behind this is due the force of gravity which attracts or pulls all bodies towards the center of the earth. Gravity is a pulling force that works across space, and because of it all objects in the universe attract all other objects. This is what keeps everything there is from flying apart neither things flying off into the space. The attraction is known as the gravitational pull.

3. a) Suppose we stand on the Torre pendent di Pisa at a height H and let there some object drop down. Give a formula for the fall time of this object until it hits the ground. Prove that your formula is correct, and prove that the fall time does not depend on the mass of the object.

Most importantly is to start the derivations assuming while the assumption is acceleration due to the force of gravity is constant.

a = where a is acceleration, is the derivative of velocity and is derivative of time

Since a = g and = g multiplying both sides by we get .g = Integrate over an interval from V= Vi to V= V = V - Vi

Now integrating .g over the interval of t = 0 to t = t .g = gt – 0

Hence the gravity equation for velocity is V = gt + Vi

To derive the time V - Vi = gt

Therefore t =

From the derived equation it is very clear that to determine the time of a free falling body does not depend on its mass but depends on initial and final velocity, and gravitational force.

b) Suppose you are a tourist at an old castle which has a well that contains water deep down. You wish to know how far away the water surface is, so you let a little pebble fall down and measure the time until you hear the echo. Give a formula (with proof) for the distance depending on the elapsed time between dropping the pebble and arrival of the echo.

In order to understand this concept it calls for the kinematics equations of motion;

a = aavg = =

a =

a.dt = Vf - Vi

(1) Vf = Vi + a.dt

Applying average velocity to determine the initial and final positions of the stone

Vavg = and Vavg =

= (Vf + Vi)

Solving for the displacement dx, the equation becomes

(2) dx = (Vf + Vi) dt

Now combining equations (1) and (2) that will describe the motion without the knowledge of the final velocity to obtain equation (3) we can determine the depth of the well using the derived below equation.

dx = (Vi + (Vi + a.dt)) dt

dx = (2Vi + a.dt) dt

(3) dx =Vidt + a (dt)2 where dx is the displacement/ depth derivative and dt is the time derivative, Vf is the final velocity and Vi is the initial velocity.

4. Suppose you have a bottle of shampoo or a bottle of bath foam which is transparent, so you can see what happens inside the bottle. Assume this bottle is half-full. You shake the closed bottle (which produces air bubbles), and then you turn it upside-down. Now the air bubbles in the liquid will rise to the top. The question is how does the rising speed related to the bubble radius.

You assume that the motion quickly becomes un-accelerated, and the relevant forces are buoyant force and the friction, which is in case of a sticky fluid given as

Ffriction = 6η.. V

With radius r being the bubble radius, V the bubble velocity, and η being a viscosity coefficient that depends on the fluid, not the bubble. This η is big for shampoo and honey, and it is small for water.

As long as the radius of the bubble does not become extremely big the drag force Ffriction is controlled just by the viscosity of the shampoo, η, as given in the above formula, where V is the rise speed of the bubble. An equilibrium is reached in which the drag force equals the buoyancy force and, thus the rise speed, V, of the bubble is:

V =

Where mis the density of the shampoo andg is the density of the gas.

From this equation it is apparent that the rise speed of the bubble through the shampoo is proportional to the radius of the bubble i.e., larger bubbles rise disproportionately faster than smaller ones.