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HACTEHA [7]
1 year ago
7

A tiny sphere carrying a charge of 6. 5 µc sits in an electric field, at a point where the electric potential is 240 v. what is

the sphere’s potential energy?
Physics
1 answer:
vodka [1.7K]1 year ago
8 0

The sphere’s Electric potential energy is 1.6*10^{-6}J

Given,

q=6. 5 µc, V=240 v,

We know that sphere’s Electric potential energy(E) = qV=6.5*10^{-6} *240=1.6*10^{-6}J

<h3>Electric potential energy</h3>

The configuration of a certain set of point charges within a given system is connected with the potential energy (measured in joules) known as electric potential energy, which is a product of conservative Coulomb forces. Two crucial factors—its inherent electric charge and its position in relation to other electrically charged objects—can determine whether an object has electric potential energy.

In systems with time-varying electric fields, the potential energy is referred to as "electric potential energy," but in systems with time-invariant electric fields, the potential energy is referred to as "electrostatic potential energy."

A tiny sphere carrying a charge of 6. 5 µc sits in an electric field, at a point where the electric potential is 240 v. what is the sphere’s potential energy?

Learn more about Electric potential energy here:

brainly.com/question/24284560

#SPJ4

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Answer:

According to the travellers, Alpha Centauri is <em>c) very slightly less than 4 light-years</em>

<em></em>

Explanation:

For a stationary observer, Alpha Centauri is 4 light-years away but for an observer who is travelling close to the speed of light, Alpha Centauri is <em>very slightly less than 4 light-years. </em>The following expression explains why:

v = d / t

where

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Therefore,

d = v × t

d = (0.999 c)(4 light-years)

d = 3.996  light-years

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3 years ago
An object has a kinetic energy of 275 j and the momentum of magnitude of 25.0 kg m/s. find the speed and mass of the object
lyudmila [28]
<span>Recall formula for Kinetic energy is: KE = 1/2mv^2, where KE = 275J and momentum (which is 25.0 kg m/s) = m*v Therefore substitute for KE and mv in the equation above to get speed => 275 = 0.5 * 25 * v v = 275/12.5 v = 22m/s to get mass m, recall momentum = m*v => 25= 22*m m= 25/22 = 1.3663kg</span>
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Tennis balls traveling at greater than 100 mph routinely bounce off tennis rackets. At some sufficiently high speed, however, th
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Answer:

Probability of tunneling is 10^{- 1.17\times 10^{32}}

Solution:

As per the question:

Velocity of the tennis ball, v = 120 mph = 54 m/s

Mass of the tennis ball, m = 100 g = 0.1 kg

Thickness of the tennis ball, t = 2.0 mm = 2.0\times 10^{- 3}\ m

Max velocity of the tennis ball, v_{m} = 200\ mph = 89 m/s

Now,

The maximum kinetic energy of the tennis ball is given by:

KE = \frac{1}{2}mv_{m}^{2} = \frac{1}{2}\times 0.1\times 89^{2} = 396.05\ J

Kinetic energy of the tennis ball, KE' = \frac{1}{2}mv^{2} = 0.5\times 0.1\times 54^{2} = 154.8\ m/s

Now, the distance the ball can penetrate to is given by:

\eta = \frac{\bar{h}}{\sqrt{2m(KE - KE')}}

\bar{h} = \frac{h}{2\pi} = \frac{6.626\times 10^{- 34}}{2\pi} = 1.0545\times 10^{- 34}\ Js

Thus

\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}

\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}

\eta = 1.52\times 10^{-35}\ m

Now,

We can calculate the tunneling probability as:

P(t) = e^{\frac{- 2t}{\eta}}

P(t) = e^{\frac{- 2\times 2.0\times 10^{- 3}}{1.52\times 10^{-35}}} = e^{-2.63\times 10^{32}}

P(t) = e^{-2.63\times 10^{32}}

Taking log on both the sides:

logP(t) = -2.63\times 10^{32} loge

P(t) = 10^{- 1.17\times 10^{32}}

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