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4 years ago

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An object is released from rest high above the surface of the earth. When it has fallen halfway to the surface, its kinetic ener

gy is KE1. After it has fallen twice as far, just before it hits the surface of the earth, its kinetic energy is KE2. 1) How does KE2 compare to twice KE1

1 answer:

aksik [14]4 years ago

4 0

Answer:

There will be an increase in the kinetic energy

Explanation:

A falling object converts the gravitational potential energy to the kinetic energy. The potential energy is then converted to kinetic energy followed by the conversation:

$E_{p} = E_{k}$

where Ep and Ek are potential and kinetic energies respectively.

This potential energy is then converted to kinetic energy. Halfway, the kinetic energy is equal to KE1.

However, the kinetic energy is given by the equation:

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

As the velocity increases, the kinetic energy increases. Hence KE2 will be greater than KE1

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Answer:

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Explanation:

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3 years ago

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Determine the angle of an incline that would yield a constant velocity, given the coefficient of kinetic friction is 0.10.

azamat

Answer:

$\theta=5.71^{o}$

Explanation:

In order to solve this problem, we mus start by drawing a free body diagram of the given situation (See attached picture).

From the free body diagram we can now do a sum of forces in the x and y direction. Let's start with the y-direction:

$\sum F_{y}=0$

$-W_{y}+N=0$

$N=W_{y}$

so:

$N=mgcos(\theta)$

now we can go ahead and do a sum of forces in the x-direction:

$\sum F_{x}=0$

the sum of forces in x is 0 because it's moving at a constant speed.

$-f+W_{x}=0$

$-\mu_{k}N+mg sen(\theta)=0$

$-\mu_{k}mg cos(\theta)+mg sen(\theta)=0$

so now we solve for theta. We can start by factoring mg so we get:

$mg(-\mu_{k} cos(\theta)+sen(\theta))=0$

we can divide both sides into mg so we get:

$-\mu_{k} cos(\theta)+sen(\theta)=0$

this tells us that the problem is independent of the mass of the object.

$\mu_{k} cos(\theta)=sen(\theta)$

we now divide both sides of the equation into $cos(\theta)$ so we get:

$\mu_{k}=\frac{sen(\theta)}{cos(\theta)}$

$\mu_{k}=tan(\theta)$

so we now take the inverse function of tan to get:

$\theta=tan^{-1}(\mu_{k})$

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$\theta=5.71^{o}$

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3 years ago

A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.7 1010 m (inside the orbit

Lubov Fominskaja [6]

Answer:

58515.9 m/s

Explanation:

We are given that

$d_1=4.7\times 10^{10} m$

$v_i=9.5\times 10^4 m/s$

$d_2=6\times 10^{12} m$

We have to find the speed (vf).

Work done by surrounding particles=W=0 Therefore, initial energy is equal to final energy.

$K_i+U_i=K_f+U_f$

$\frac{1}{2}mv^2_i-\frac{GmM}{d_1}=\frac{1}{2}mv^2_f-\frac{GmM}{d_2}$

$\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2}=\frac{1}{2}v^2_f$

$v^2_f=2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})$

$v_f=\sqrt{2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})}$

Using the formula

$v_f=\sqrt{v^2_i+2GM(\frac{1}{d_2}-\frac{1}{d_1})}$

$v_f=\sqrt{(9.5\times 10^4)^2+2\times 6.7\times 10^{-11}\times 1.98\times 10^{30}(\frac{1}{6\times 10^{12}}-\frac{1}{4.7\times 10^{10})}$

Where mass of sun= $M=1.98\times 10^{30} kg$

$G=6.7\times 10^{-11}$

$v_f=58515.9 m/s$

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3 years ago

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Sedbober [7]

Answer:

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Explanation:

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$\vec{dl}x\vec{r} = R*dl*\vec{A}$

Where $\vec{A}$ imply a vector acting perpendicular to both vectors.

Applying Biot-Savart law, we have,

$\vec{V} =\frac{\Gamma}{4\pi}\int\frac{\vec{dl}x\vec{r}}{r^3}$

Substituting the preoviusly equation obtained,

$\vec{V} = \frac{\Gamma}{4\pi}\int\frac{R*dl*\vec{A}}{R^3}$

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$\vec{V} = \frac{\Gamma(2\pi R \vec{A})}{4\pi R^2}$

So we can express the velocity induced is,

$\vec{V} = \frac{\Gamma}{2R}\vec{A}$

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3 years ago

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maks197457 [2]

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3 years ago

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