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1 year ago

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Physics

1 answer:

Talja [164]1 year ago

3 0

The correct expression for the maximum speed of the object during its motion is $\frac{FT}{m}$ .

<h3>Maximum speed of the object</h3>

The maximum speed of the object is determined using the following formulas;

v(max) = Aω

where;

A is the amplitude of the motion
ω is angular speed

The maximum speed of the object can also be obtained from the maximum net force on the object,

F = ma

where;

F is the maximum net force
a is the acceleration
m is mass of the object

F = m(v/t)

mv = Ft

v = Ft/m

Thus, the correct expression for the maximum speed of the object during its motion is $\frac{FT}{m}$ .

Learn more about maximum speed here: brainly.com/question/4931057

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Starting from rest, a disk rotates about its central axis with constant angular acceleration. in 6.00 s, it rotates 44.5 rad. du

Klio2033 [76]

a. The disk starts at rest, so its angular displacement at time $t$ is

$\theta=\dfrac\alpha2t^2$

It rotates 44.5 rad in this time, so we have

$44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}$

b. Since acceleration is constant, the average angular velocity is

$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2$

where $\omega_f$ is the angular velocity achieved after 6.00 s. The velocity of the disk at time $t$ is

$\omega=\alpha t$

so we have

$\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}$

making the average velocity

$\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}$

Another way to find the average velocity is to compute it directly via

$\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}$

c. We already found this using the first method in part (b),

$\omega=14.8\dfrac{\rm rad}{\rm s}$

d. We already know

$\theta=\dfrac\alpha2t^2$

so this is just a matter of plugging in $t=12.0\,\mathrm s$ . We get

$\theta=179\,\mathrm{rad}$

Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

$\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2$

Then for $t=6.00\,\rm s$ we would get the same $\theta=179\,\rm rad$ .

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2 years ago

) A striker can give the ball an initial speed of 30m/s. Within what two elevation angles must he kick

sveticcg [70]

Answer:

about 19.6° and 73.2°

Explanation:

The equation for ballistic motion in Cartesian coordinates for some launch angle α can be written ...

y = -4.9(x/s·sec(α))² +x·tan(α)

where s is the launch speed in meters per second.

We want y=2.44 for x=50, so this resolves to a quadratic equation in tan(α):

-13.6111·tan(α)² +50·tan(α) -16.0511 = 0

This has solutions ...

tan(α) = 0.355408 or 3.31806

The corresponding angles are ...

α = 19.5656° or 73.2282°

The elevation angle must lie between 19.6° and 73.2° for the ball to score a goal.

_____

I find it convenient to use a graphing calculator to find solutions for problems of this sort. In the attachment, we have used x as the angle in degrees, and written the function so that x-intercepts are the solutions.

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2 years ago

Principle of electroplating

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1 year ago

Read 2 more answers

A. Draw the electric field lines around a negative charge.

Alborosie

<h2>a. Answer:</h2>

We use Electric field lines for visualizing electric fields, so this helps us to see the problem more real. So an electric field line is an imaginary line or curve drawn through a region of space such that the tangent at any point comes from the direction of the electric-field vector at that point. The electric field lines around a negative charge is shown in the First figure below.

<h2>b. Answer:</h2>

Electric forces can be found by using the Coulomb Law's that states <em>that The magnitude of the electric force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. </em>This can be expressed as follows:

$F=k\frac{\left | q_{1}q_{2} \right |}{r^2} \\ \\ Where: \\ \\ k=9\times 10^9Nm^2/c^2 \\ \\ q_{1}=0.00150 C \ and \ q_{2}=0.00240 C \\ \\ r=0.900 m$

Then:

$F=9\times 10^9\frac{\left | 0.00150 \times 0.00240 \right |}{(0.900)^2} \\ \\ \therefore \boxed{F=40000N}$

This force is repulsive because the two charges are positive and recall that two positive charges or two negative charges repel each other while a positive charge and a negative charge attract each other.

<h2>C. Answer:</h2>

From the statement, we have two charged objects. Let's say that this charges are:

$q_{1} \ and \ q_{2}$

If the amount of charge on one of the objects is tripled, let's say this is the charge $q_{2}$ , then the new charge is:

$q_{N}=3q_{2}$

In the formula of Coulomb:

$F=k\frac{\left | q_{1}q_{N} \right |}{r^2} \\ \\ \therefore F=k\frac{\left | q_{1}(3q_{2}) \right |}{r^2} \\ \\ \therefore \boxed{F=3k\frac{\left | q_{1}q_{2} \right |}{r^2}}$

<em>The conclusion is that if the amount of charge on one of the objects is tripled, the electric force between two charged objects is also tripled</em>

<h2>d. Answer:</h2>

Let's use the Coulomb's Law again to solve this problem. We want to know how the electric force between two charged objects changes if the charges are moved closer together:

$F=k\frac{\left | q_{1}q_{N} \right |}{r^2}$

<em>By saying that the charges are moved closer together, we want to express that r becomes smaller. Since r is in the denominator, this implies that the electric force between these two charged objects becomes greater.</em>

<h2>e. Answer:</h2>

From the figure, we can see a metal sphere on a stand. There we have both positive and negative charges. We can say that the positive charge of this sphere is +10q and the negative and the negative charge is -10q. Since the electric charge is conserved, then the algebraic sum of all the electric charges in any closed system is constant. In conclusion, <em>the sphere has no net charge.</em>

<h2>f. Answer:</h2>

Here we want to know how the negative charges in the same sphere are redistributed when a positively charged rod is brought near it. Therefore, positive charge on rod repels positive charges on the sphere, creating zones of negative and positive charge as indicated in the second Figure.

7 0

2 years ago

Read 2 more answers