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9 days ago

A positive charge moves in the direction of an electric field. Which of the following statements are true?

Physics

2 answers:

ValentinkaMS [1.1K]9 days ago

4 0

Answer:

The potential energy linked with the charge declines

The electric field exerts positive work on the charge.

kicyunya [1K]9 days ago

3 0

Answer:

The potential energy tied to the charge diminishes.

The electric field performs negative work on the charge.

Explanation:

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If a person weighs 882 N on the surface of the earth, at what altitude above the earth’s surface must they be for their weight t

Sav [1095]

Para calcular el peso utilizamos la fórmula:
$F=m*g=882N$

Cuando tenemos dos cuerpos, empleamos la fórmula de la gravedad general:
$F=G* \frac{ M_{p}*M_{E} }{ r^{2} }$
Donde:
G = constante de gravedad = $6.67* 10^{-11} m^{3} kg^{-1} s^{-2}$
Mp = masa de la persona = 882 / 9.81= 89.9kg
ME = masa de la Tierra = $5.97* 10^{24} kg$
r = distancia entre la Tierra y la persona

A partir de estas dos fórmulas, deducimos que los lados izquierdos son equivalentes, por lo tanto, los lados derechos también deben serlo.
$m*g=G* \frac{ M_{p}*M_{E} }{ r^{2} }$

Resolviendo esta ecuación para r:
$r= \sqrt{G* \frac{ M_{p}*M_{E}}{M_{p}*g}$
r=6371116m = 6371.116km

Esta es la distancia desde el centro de la Tierra. El radio de la Tierra es 6370km y la altura sobre la superficie es 6371.116 - 6370 = 1.116km o 1116m.

3 0

1 day ago

A charged wire of negligible thickness has length 2L units and has a linear charge density λ. Consider the electric field E-vect

Softa [913]

Answer:

$E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}$

Explanation:

Consider the following:

Length= 2L

Linear charge density=λ

Distance= d

K=1/(4πε)

The electric field measured at point P

$E=2K\int_{0}^{L}\dfrac{\lambda }{r^2}dx\ sin\theta$

$sin\theta =\dfrac{d}{\sqrt{d^2+x^2}}$

$r^2=d^2+x^2$

Thus,

$E=2K\lambda d\int_{0}^{L}\dfrac{dx }{(x^2+d^2)^{\frac{3}{2}}}$

Now, by applying integration to the equation above

$E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}$

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5 days ago

Your annoying little brother is dropping rocks out of his bedroom window on the 2nd floor. You are on the ground floor and watch

serg [1189]

Answer:

Insufficient details provided; please clarify further.

5 0

13 days ago

A person who weighs 625 N is riding a 98-N mountain bike. Suppose that the entire weight of the rider and bike is supported equa

Keith_Richards [1021]

Answer:

A = 4.76 x 10⁻⁴ m²

Explanation:

Given data:

Person's weight = 625 N

Bike's weight = 98 N

Pressure per tire = 7.60 x 10⁵ Pa

Find: Contact area per tire

Total system weight = 625 + 98 = 723 N

Let F represent the force supported by each tire

2F = 723 N

Therefore, F = 361.5 N

Using the formula F = P × A

$A = \dfrac{F}{P}$

$A = \dfrac{361.5}{7.60 \times 10^5}$

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7 0

15 days ago

As computer structures get smaller and smaller, quantum rules start to create difficulties. Suppose electrons move through a cha

Softa [913]

Answer:

a) ∆x∆v = 5.78*10^-5

∆v = 1157.08 m/s

b) 4.32*10^{-11}

Explanation:

This problem can be addressed using Heisenberg's uncertainty principle, which is expressed as:

$\Delta x\Delta p \geq \frac{\hbar}{2}$

Where h represents Planck’s constant (6.62*10^-34 J s).

Assuming that the electron's mass remains the same, we proceed as follows:

$\Delta x \Delta (m_ev)=m_e\Delta x\Delta v \geq \frac{\hbar}{2}$

Utilizing the electron's mass (9.61*10^-31 kg) and the uncertainty in position (50 nm), we can compute ∆x∆v and ∆v:

$\Delta x \Delta v\geq\frac{\hbar}{2m_e}=\frac{(1.055*10^{-34}Js)}{2(9.1*10^{-31}kg)}=5.78*10^{-5}\ m^2/s$

$\Delta v\geq\frac{5.78*10^{-5}}{50*10^{-9}m}=1157.08\frac{m}{s}$

If we treat the electron like a classic particle, the time required to cross the channel is determined using the upper limit of the uncertainty in velocity:

$t=\frac{x}{v}=\frac{50*10^{-9}m}{1157.08m/s}=4.32*10^{-11}s$

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6 days ago