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2 years ago

7

A body of mass 8 kg moves in a (counterclockwise) circular path of radius 10 meters, making one revolution every 10 seconds. You

may assume the circle is in the xy-plane, and so you may ignore the third component. A. Compute the centripetal force acting on the body.

1 answer:

Sav [38]2 years ago

8 0

Answer:

Centripetal force is equal to 31.55 N

Explanation:

We have given mass of the body m = 8 kg

Radius of the circular path r = 10 m

It is given that it makes 1 revolution in 10 seconds

Distance traveled in 10 seconds is equal to $d=2\pi r=2\times 3.14\times 10=62.8m$

Velocity is equal to $velocity=\frac{distance}{time}=\frac{62.8}{10}=6.28m/sec$

We have to find the centripetal force

Centripetal force is equal to $F=\frac{mv^2}{r}=\frac{8\times 6.28^2}{10}=31.55N$

So centripetal force will be equal to 31.55 N

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Answer:

E = k Q / [d(d+L)]

Explanation:

As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field

E = k ∫ dq/ r² r^

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Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant

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λ = dq/dx ⇒ dq = L dx

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dE = k dq / x²2

dE = k λ dx / x²

Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider

E = k $\int\limits^{d+L}_d {\lambda/x^{2}} \, dx$

We take out the constant magnitudes and perform the integral

E = k λ (-1/x) ${(-1/x)}^{d+L} _{d}$

Evaluating

E = k λ [ 1/d - 1/ (d+L)]

Using λ = Q/L

E = k Q/L [ 1/d - 1/ (d+L)]

let's use a bit of arithmetic to simplify the expression

[ 1/d - 1/ (d+L)] = L /[d(d+L)]

The final result is

E = k Q / [d(d+L)]

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1 year ago

The robot arm is elevating and extending simultaneously. At a given instant, θ = 30°, ˙ θ = 10 deg / s = constant θ˙=10 deg/s=co

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Explanation:

The position vector r:

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The velocity vector v:

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The acceleration vector a:

$\overrightarrow{a(t)}}=cos\theta(\ddot{l}-l\dot{\theta}^2)-sin\theta(2\dot{l}\dot{\theta}+l\ddot{\theta})\hat{i}+cos\theta(2\dot{l}\dot{\theta}+l\ddot{\theta})+sin\theta(\ddot{l}-l\dot{\theta}^2)\hat{j}$

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1 year ago

An object moving on the x axis with a constant acceleration increases its x coordinate by 82.9 m in a time of 2.51 s and has a v

Aneli [31]

We are given: Final velocity ( $v_f$ )=20 m/s .

Time t= 2.51 s and

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We know, equation of motion

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Using another equation of motion

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We get,

$20-(20-2.51a)=2*a(82.90)$

Now, we need to solve it for a.

20-20+2.51a=165.8a.

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1 year ago

A transition metal complex in solution has an absorption peak at 450 nm, in the blue region of the visible spectrum. What color

Ivan

Answer:

In the case of a solution transition metal complex that has an absorption peak at 450 nm in the blue region of the visible spectrum, the (complementary) color of this solution is orange (option B).

Explanation:

The portion of UV-visible radiation that is absorbed implies that a portion of electromagnetic radiation is not absorbed by the sample and is therefore transmitted through it and can be captured by the human eye. That is, in the visible region of a complex, the visible color of a solution can be seen and that corresponds to the wavelengths of light it transmits, not absorbs. The absorbing color is complementary to the color it transmits.

So, in the attached image you can see the approximate wavelengths with the colors, where they locate the wavelength with the absorbed color, you will be able to observe the complementary color that is seen or reflected.

<u><em> In the case of a solution transition metal complex that has an absorption peak at 450 nm in the blue region of the visible spectrum, the (complementary) color of this solution is orange (option B).</em></u>

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2 years ago

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el

Alborosie

Answer:

I = 4.75 A

Explanation:

To find the current in the wire you use the following relation:

$J=\frac{E}{\rho}$ (1)

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ρ: resistivity of the material = 2.75×10−8 ohm-meters

J: current density

The current density is also given by:

$J=\frac{I}{A}$ (2)

I: current

A: cross area of the wire = π(d/2)^2

d: diameter of the wire = 0.205 cm = 0.00205 m

You replace the equation (2) into the equation (1), and you solve for the current I:

$\frac{I}{A}=\frac{E(t)}{\rho}\\\\I(t)=\frac{AE(t)}{\rho}$

Next, you replace for all variables:

$I(t)=\frac{\pi (d/2)^2E(t)}{\rho}\\\\I(t)=\frac{\pi(0.00205m/2)^2(0.0004t^2-0.0001t+0.0004)}{2.75*10^{-8}\Omega.m}\\\\I(t)=4.75A$

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1 year ago