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Stolb23 [73]
3 years ago
11

Un niño amarra una soga a una piedra, y la hace girar como en la gráfica. la piedra realiza un M.C.U, girando con una rapidez de

14 rad/s. calcular el ángulo que barre en 4 segundos.
Physics
1 answer:
inessss [21]3 years ago
4 0

Answer:

\Delta \theta = 56\,rad

\Delta \theta \approx 3208.564^{\circ}

Explanation:

El ángulo barrido en el intervalo de tiempo dado es (The covered angle in the given time interval is):

\Delta \theta = \omega \cdot \Delta t

\Delta \theta = \left(14\,\frac{rad}{s} \rjght)\cdot (4\,s)

\Delta \theta = 56\,rad

\Delta \theta \approx 3208.564^{\circ}

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A proton is confined to a space 1 fm wide (about the size of an atomic nucleus). What's the minimum uncertainty in its velocity?
Daniel [21]

Answer:

Minimum uncertainty in velocity of a proton,\Delta v\ge 3.15\times 10^7\ m/s      

Explanation:

It is given that,

A proton is confined to a space 1 fm wide, \Delta x=10^{-15}\ m

We need to find the minimum uncertainty in its velocity. We know that the Heisenberg Uncertainty principle gives the uncertainty between position and the momentum such that,

\Delta p.\Delta x\ge \dfrac{h}{4\pi}

Since, p = mv

\Delta (mv).\Delta x\ge \dfrac{h}{4\pi}

m \Delta v.\Delta x\ge \dfrac{h}{4\pi}

\Delta v\ge \dfrac{h}{4\pi m\Delta x}

\Delta v\ge \dfrac{6.63\times 10^{-34}}{4\pi \times 1.67\times 10^{-27}\times 10^{-15}}

\Delta v\ge 3.15\times 10^7\ m/s

So, the minimum uncertainty in its velocity is greater than 3.15\times 10^7\ m/s. Hence, this is the required solution.

3 0
3 years ago
Suppose you were far from a planet that had a very strong gravitational field, and an emission line spectrum source on the surfa
dangina [55]
The answer would be it will be longer than the 656.3 nm. The reduced mass of positronium is less than hydrogen so the photon energy will be a reduced amount of for positronium than for hydrogen. So this will mean that the wavelength will be lengthier.
5 0
3 years ago
A 71-kg swimmer dives horizontally off a 500-kg raft. If the diver's speed immediately after leaving the raft is 6m/s, what is t
Sloan [31]

Answer:

The answer is below

Explanation:

Momentum is used to measure the quantity of motion in an object. Momentum is the product of mass and velocity.

Momentum = mass * velocity

The principle of conservation of momentum states that momentum cannot be created or destroyed but  can be transferred. Therefore the momentum before and after an action is equal.

Initial momentum = Final momentum

Let m be the mass of the diver, M be the mass of the raft, u be the initial velocity of the diver, U be the initial velocity of the raft, v be the final velocity of the diver and V be the final velocity of the raft.

m = 71 kg, M = 500 kg, v = 6 m/s

Initial both the raft and diver are at rest, hence u and U is zero, hence:

mu + MU = mv + MV

71(0) + 500(0) = 71(6) + 500(V)

0 = 426 + 500(V)

500(V) = -426

V = -426/500

V = -0.852 m/s

5 0
3 years ago
An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea
melisa1 [442]

Answer:

Velocity of the electron at the centre of the ring, v=1.37\times10^7\ \rm m/s

Explanation:

<u>Given:</u>

  • Linear charge density of the ring=0.1\ \rm \mu C/m
  • Radius of the ring R=0.2 m
  • Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C

Potential due the ring at a distance x from the centre of the rings is given by

V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V

Let\Delta U be the change in potential Energy given by

\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s

So the electron will be moving with v=1.37\times10^7\ \rm m/s

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3 years ago
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BabaBlast [244]
Sowers has spent her career fighting for gender equality and it passionate about it .
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