A particle (q = +5.0 µC) is released from rest when it is 2.0 m from a charged particle which is held at

Physics

1 answer:

PilotLPTM [1.2K]2 years ago

3 0

The answer to your question is A. Have a great day

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A 2kg box is pushed along a flat frictionless surface with an applied force of 53.91newton j+19.62 j were j is horizontal and j

tamaranim1 [39]

Since the box doesn't move vertically at all, no work is done by the vertical component of the force.

If 53.91 Newtons is the horizontal component of the force (very unclear in the question), then the work done is

Work = (force) x (distance)

Work = (53.91 N) x (10 m)

Work = 539.1 Joules

Power = (work done) / (time to do the work)

Power = (539.1 joules) / (90 sec)

Power = (539.1/90) (joule/sec)

Power = 5.99 watts

=====================================

Note: If the mass of the box is only 2 kg, and you push it along the surface with a constant force of 53⁺ Newtons, and the surface really is frictionless, then that box is gonna cover a WHOOOOOLE LOT more than 10 meters in 90 seconds. I get 109,168 meters !

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¿Cuál es la masa aproximada del aire en una habitación de 5.6 m * 3.8 m * 2.8 m?

Nady [450]

It would be 3.15 in meters

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What is the relationship in which the ratio of the manipulated variable and the responding variable is constant? A. inverse prop

Dafna1 [17]

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3 years ago

(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th

julia-pushkina [17]

Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of r from a point charge, Q, is given by

$V = -\dfrac{kQ}{r}$

where $k = 9\times 10^9 \text{ F/m}$

Difference in potential between the points is

$kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}$