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3 years ago

What does "Time (s)" represent in the graph?

Physics

2 answers:

polet [3.4K]3 years ago

5 0

Answer:

Explanation:

Part 1: Graphing Techniques

In Physics we use a variety of tools – including words, equations, and graphs – to make models of the motion of

objects and the interactions between objects in a system. Graphs are one of the best ways to directly visualize

the quantitative relationship between two variables – in other words, whether the variables are directly

proportional, inversely proportional, not related at all, or something else entirely.

When we construct a graph, we plot the independent variable – the variable that the experimenter controls – on

the x-axis, and the dependent variable – the variable that responds when the independent variable is changed –

on the y-axis. There are also control variables – variables that are kept constant throughout the experiment so

that they do not influence the data. So, for example, if you were trying to determine how the period of a

pendulum changes when the length of the pendulum is varied, the dependent variable would be the pendulum’s

period, and the independent variable would be the pendulum’s length. Controlled variables would include the

pendulum’s mass and the angle at which the pendulum was launched.

An appropriate graph for this experiment is shown below.

Notice that the title lists the dependent variable, which is plotted on the y-axis, first, and the independent

variable, which is plotted on the x-axis, second. The axes are correctly labeled with the appropriate units. The

graph begins at (0, 0) with no “jumps”, and increments are equally spaced.

In this experiment, we can clearly see that as the length of the pendulum increases, the period also increases, but

are the variables directly proportional? In other words, can we write an equation for the relationship in the form

y = mx + b? Excel will draw a trend line for a graph that can help us to determine this.

0 1 2 3 4 5 6 7

( s)

Length (m)

Period vs. Length of a Pendulum

While the graph appears to be somewhat linear, we can see a few problems – first, the majority of the points do

not fall on the line; second, the line does not cross the y-axis at zero, and we would expect it to – after all, a

pendulum with a length very close to zero meters should have a period very close to zero seconds. To

determine the correct relationship between the variables, we will have to linearize the graph.

jenyasd209 [6]3 years ago

3 0

Answer:

Explanation:

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Explanation:

At that point potential energy is zero and kinetic energy is maximum.. P. E=mgh=0

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The 1.5kg ball is launched straight upward with an initial velocity of 7m/s. What is the maximum height it will reach?

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Answer:

h = 2.5 m

Explanation:

Given that,

Mass of a ball, m = 1.5 kg

Initial velocity of the ball, u = 7 m/s

We need to find the maximum height reached by the ball. Let it is be h. Using the conservation of energy to find it such that,

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3 years ago

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3 years ago

If the period of a spring is 5 seconds what is the frequency

Scilla [17]

Answer:

C. 0.2 Hertz

Explanation:

The frequency of a spring is equal to the reciprocal of the period:

$f=\frac{1}{T}$

where

f is the frequency

T is the period

For the spring in this problem,

T = 5 s

therefore, the frequency is

$f=\frac{1}{5 s}=0.2 Hz$

7 0

3 years ago

Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by

Ulleksa [173]

A) $v=\sqrt{\frac{2qV}{m}}$

B) $r=\frac{mv}{qB}$

C) $T=\frac{2\pi m}{qB}$

D) $\omega=\frac{qB}{m}$

E) $r=\frac{\sqrt{2mK}}{qB}$

Explanation:

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

$\Delta U = qV$

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

$\Delta K=\frac{1}{2}mv^2$

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

$qV=\frac{1}{2}mv^2$

And solving for v, we find the speed v at which the particle enters the cyclotron:

$v=\sqrt{\frac{2qV}{m}}$

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

$F=qvB$

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

$F=m\frac{v^2}{r}$

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

$qvB=m\frac{v^2}{r}$

And solving for r, we find the radius of the orbit:

$r=\frac{mv}{qB}$ (1)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference ( $2\pi r$ ) and the velocity of the particle (v):

$T=\frac{2\pi r}{v}$ (2)

From eq.(1), we can rewrite the velocity of the particle as

$v=\frac{qBr}{m}$

Substituting into(2), we can rewrite the period of revolution of the particle as:

$T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}$

And we see that this period is indepedent on the velocity.

The angular frequency of a particle in circular motion is related to the period by the formula

$\omega=\frac{2\pi}{T}$ (3)

where T is the period.

The period has been found in part C:

$T=\frac{2\pi m}{qB}$

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

$\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}$

And we see that also the angular frequency does not depend on the velocity.

For this part, we use again the relationship found in part B:

$v=\frac{qBr}{m}$

which can be rewritten as

$r=\frac{mv}{qB}$ (4)

The kinetic energy of the particle is written as

$K=\frac{1}{2}mv^2$

So, from this we can find another expression for the velocity:

$v=\sqrt{\frac{2K}{m}}$

And substitutin into (4), we find:

$r=\frac{\sqrt{2mK}}{qB}$

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

6 0

3 years ago