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3 years ago

Is energy transformation only occurring at only point 3?

Physics

1 answer:

Paul [167]3 years ago

7 0

No, energy transformation is occurring in every point of the motion.

In fact, the ball starts from point 1 with maximum kinetic energy and zero potential energy (taking the hand of the boy as reference level). The kinetic energy converts into gravitational potential energy as it goes higher: in point 2, part of the kinetic energy has converted into potential energy (because the velocity has decreased, while the height has increased), and then when the ball reaches point 3 all the kinetic energy has converted into potential energy (because now the velocity is zero, while the height is maximum). As the ball descends (point 4), the velocity starts to increase again, therefore the kinetic energy increases and the potential energy decreases (because the height is deacreasing now).
Summarizing, energy transformation is occuring in every point of the motion.

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A liquid of density 1270 kg/m3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the f

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${P_2}-P_1=49.99\ KPa$

Explanation:

$v_1=9.43\ m/s$

$d_1=11.7\ cm/s$

$d_2=17.5\ cm/s$

From continuity equation

$A_1v_1=A_2v_2$

$v_1d_1^2=v_2d_2^2$

$v_2=\dfrac{9.43\times 11.7^2}{17.5^2}$

$v_2=4.21\ m/s$

$\dfrac{P_1}{\rho g}+\dfrac{v_1^2}{2g}+y_1=\dfrac{P_2}{\rho g}+\dfrac{v_2^2}{2g}+y_2$

${P_1}+\rho\dfrac{v_1^2}{2}+\rho y_1g={P_2}+\rho\dfrac{v_2^2}{2}+\rho y_2g$

$\rho\dfrac{v_1^2}{2}+\rho y_1g -\rho\dfrac{v_2^2}{2}-\rho y_2g={P_2}-P_1$

$1270\times \dfrac{9.43^2}{2}+1270\times 0\times 10 -1270\times\dfrac{4.21^2}{2}-1270\times 0.175\times 10={P_2}-P_1$

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3 years ago

Why hands are not heated by conduction when placed on a vertical placed copper sheet

Mademuasel [1]

Answer:

A conduction heating material with a cylindrical shape and a 5 cm radius is being heated in high temperature (85 C) air from an initial temperature of 10 C

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3 years ago

A ring of charge of radius r0 lies in the x-y plane with its center at the origin.The charge counterclockwise from θ = 0º to θ =

Sindrei [870]

Answer:

$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{2QR}{(R^2 + r_0^2)^{3/2}}(-\^y)$

The direction of the electric field is in the negative y-direction.

Explanation:

We will separate the both halves of the ring and calculate the E-field separately. We will use the following formula and the method of integration.

$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\^r$

We will choose an arbitrary infinitesimal displacement (ds) on the ring. We will apply the above formula to that small segment, then integrate over the half ring.

The small segment has a length ds. This segment is also equal to Rdθ, because any curve on the ring is equal to radius times angle, and the angle of the infinitesimal displacement is the smallest angle we could choose, dθ.

$dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{R^2 + r_0^2}$

dQ can be written in terms of given variables according to the fact that the charge density of the total ring is equal to any small segment of the ring.

$\frac{Q}{\pi R} = \frac{dQ}{Rd\theta}\\dQ = \frac{Qd\theta}{\pi}$

dE becomes,

$dE = \frac{1}{4\pi\epsilon_0}\frac{Qd\theta}{\pi}\frac{1}{R^2 + r_0^2}$

Before integrating this quantity, we have consider the fact that electric field is a vector, therefore has to be separated into its vertical and horizontal components.

$dE_v = dE\sin{\alpha} = dE \frac{r_0}{\sqrt{R^2 + r_0^2}}\\dE_h = dE\cos{\alpha} = dE \frac{R}{\sqrt{R^2 + r_0^2}}$

These equalities can be deduced from the geometry of the ring.

Now, we can conclude our method by integrating both halves.

$E_1_{v} = \int\limits^{\pi}_0 {dE_1_{v}} \, d\theta = \int\limits^{\pi}_0 {\frac{1}{4\pi\epsilon_0}\frac{Q}{\pi}\frac{r_0}{(R^2 + r_0^2)^{3/2}} \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi}\frac{r_0}{(R^2 + r_0^2)^{3/2}}\int\limits^{\pi}_0 \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi}\frac{r_0}{(R^2 + r_0^2)^{3/2}} \pi$ $E_1_{v} = \frac{1}{4\pi\epsilon_0}\frac{Qr_0}{(R^2 + r_0^2)^{3/2}}\^z$

$E_1_{h} = \int\limits^{\pi}_0 {dE_1_{h}} \, d\theta = \int\limits^{\pi}_0 {\frac{1}{4\pi\epsilon_0}\frac{Q}{\pi}\frac{R}{(R^2 + r_0^2)^{3/2}} \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi}\frac{R}{(R^2 + r_0^2)^{3/2}}\int\limits^{\pi}_0 \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi}\frac{R}{(R^2 + r_0^2)^{3/2}} \pi$ $E_1_{h} = \frac{1}{4\pi\epsilon_0}\frac{QR}{(R^2 + r_0^2)^{3/2}}(-\^y)$

We will repeat these calculations for the bottom half of the ring:

$E_2_{v} = \int\limits^{\pi}_0 {dE_2_{v}} \, d\theta = \int\limits^{\pi}_0 {\frac{1}{4\pi\epsilon_0}\frac{-Q}{\pi}\frac{r_0}{(R^2 + r_0^2)^{3/2}} \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{-Q}{\pi}\frac{r_0}{(R^2 + r_0^2)^{3/2}}\int\limits^{\pi}_0 \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{-Q}{\pi}\frac{r_0}{(R^2 + r_0^2)^{3/2}} \pi$ $E_2_{v} = \frac{1}{4\pi\epsilon_0}\frac{Qr_0}{(R^2 + r_0^2)^{3/2}}(-\^z)$

$E_2_{h} = \int\limits^{\pi}_0 {dE_2_{h}} \, d\theta = \int\limits^{\pi}_0 {\frac{1}{4\pi\epsilon_0}\frac{-Q}{\pi}\frac{R}{(R^2 + r_0^2)^{3/2}} \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{-Q}{\pi}\frac{R}{(R^2 + r_0^2)^{3/2}}\int\limits^{\pi}_0 \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{-Q}{\pi}\frac{R}{(R^2 + r_0^2)^{3/2}} \pi$ $E_2_{h} = \frac{1}{4\pi\epsilon_0}\frac{QR}{(R^2 + r_0^2)^{3/2}}(-\^y)$

Therefore, the net electric field created by the ring is the sum of these four components. Note that the vertical components cancel out each other.

$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{2QR}{(R^2 + r_0^2)^{3/2}}(-\^y)$

5 0

3 years ago